# Determinants and Their Properties: det as Signed Volume > Defines the determinant both by the Leibniz formula with permutation signs and by cofactor expansion, characterizes it through multilinearity and alternation, proves det(AB)=det(A)det(B) and the invertibility criterion, and reads det as signed volume. > https://rikai.mugen-giken.com/en/mathematics/linear-algebra/determinants ## 0. Key points - Among all maps that assign a number to a list of $n$ column vectors, exactly one is **linear in each column** (multilinearity), **vanishes whenever two columns coincide** (alternation), and **takes the value $1$ at the identity matrix** (normalization): that map is the determinant $\det A$. Both the Leibniz formula and cofactor expansion are merely expressions derived from this characterization. - We have $\det(AB) = \det A \cdot \det B$. This is the algebraic form of the geometric fact that composing transformations multiplies the volume scaling factors. - The conditions "$\det A \ne 0$", "$A$ is invertible", "the columns of $A$ are linearly independent", and "$A\boldsymbol{x} = \boldsymbol{0}$ has only the trivial solution" are all equivalent. The determinant is a device that decides, with a single number, whether space is being collapsed. - $|\det A|$ is the volume of the parallelepiped onto which $A$ maps the unit cube, and the sign records whether the transformation preserves or reverses orientation. - The Leibniz formula (a sum of $n!$ terms) is never used for actual numerical computation. Triangularizing by row operations costs only $O(n^3)$ multiplications and divisions. ## 1. Motivation: telling solvability from a single number Let us solve the system of two linear equations $$ \begin{cases} ax + by = p \\ cx + dy = q \end{cases} $$ by brute force. Multiplying the first equation by $d$ and the second by $b$ and subtracting gives $(ad-bc)x = pd - bq$; multiplying the first by $c$ and the second by $a$ and subtracting gives $(ad-bc)y = aq - pc$. Hence, as soon as $ad - bc \ne 0$, the solution is determined: $$ x = \frac{pd - bq}{ad - bc}, \qquad y = \frac{aq - pc}{ad - bc}. $$ Conversely, when $ad - bc = 0$, the solution fails to be unique (there may be no solution, or infinitely many). **A single number $ad-bc$, built from the coefficients alone, decides solvability without any reference to the right-hand side $p, q$.** Pushing the same computation through for three equations in three unknowns, the deciding number becomes the six-term sum $$ a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33}. $$ Two naive questions arise here. First, for a general system of $n$ linear equations in $n$ unknowns, how should this "deciding number" be written? Second, why does such a number exist at all — how can a complicated web of relations among $n^2$ coefficients be condensed into a single scalar? The first question has had an answer since the end of the seventeenth century. Seki Takakazu treated it in 1683 in *Kaifukudai no Hō* as an expression arising in the course of elimination, and Leibniz handled a similar expression in a 1693 letter to L'Hôpital. Cramer, in a book of 1750, wrote the solution of a system of $n$ linear equations explicitly as a quotient. The general theory was organized, and the word *determinant* acquired its present meaning, through Cauchy's work in the early nineteenth century. In this article we give two answers to the first question: an **explicit formula using signs of permutations (the Leibniz formula)** and an **inductive formula that lowers the size by one (cofactor expansion)**. The two look completely different, yet they describe the same function. The answer to the second question is precisely the identity of that "same function": **multilinearity and alternation**. The determinant is the unique function pinned down by these two properties together with a normalization, and the two formulas are simply different computational recipes for it. In the final sections we shall see that the determinant is, in essence, a **signed volume**. A linear system fails to be uniquely solvable exactly when the linear map defined by the coefficient matrix collapses space into a lower dimension, and "collapsed" means "of volume $0$". The reason $ad - bc$ could serve as a criterion is, in the end, that it was the area of a parallelogram. We take matrix products, inverses, and Gaussian elimination for linear systems as known from [Matrices and Linear Systems](/en/mathematics/linear-algebra/matrices-and-linear-systems) (for the definition of the product see , and for elimination see ). ## 2. Preliminaries: permutations and their signs Throughout this article $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$ (everything below works verbatim over an arbitrary commutative field). We write $M_n(\mathbb{K})$ for the set of $n \times n$ matrices, and for $A = (a_{ij}) \in M_n(\mathbb{K})$ we denote its $j$-th column by $\boldsymbol{a}_j \in \mathbb{K}^n$ and display $A$ column by column as $$ A = (\boldsymbol{a}_1\ \boldsymbol{a}_2\ \cdots\ \boldsymbol{a}_n). $$ Here $\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n$ is the standard basis of $\mathbb{K}^n$, $I_n$ the identity matrix, and $O$ the zero matrix. To define the determinant we need a device for enumerating the rearrangements of $n$ row indices. A bijection from $\{1, 2, \ldots, n\}$ onto itself is called a **permutation** of degree $n$, and the set of all permutations is denoted $S_n$. Under composition $(\sigma\tau)(i) = \sigma(\tau(i))$ the set $S_n$ is a group, and it has $n!$ elements. For distinct $p, q$, the permutation sending $p \mapsto q$ and $q \mapsto p$ and fixing everything else is called a **transposition** and is written $(p\ q)$. For $\sigma \in S_n$, the number of pairs of indices whose order is reversed, $$ \mathrm{inv}(\sigma) = \#\{(i, j) \mid 1 \le i < j \le n,\ \sigma(i) > \sigma(j)\}, $$ is called the **number of inversions** of $\sigma$, and $\mathrm{sgn}(\sigma) = (-1)^{\mathrm{inv}(\sigma)}$ is called the **sign** of $\sigma$. When $\mathrm{sgn}(\sigma) = 1$ we call $\sigma$ even, and when $\mathrm{sgn}(\sigma) = -1$ we call it odd. For instance, take the permutation with $\sigma(1) = 3,\ \sigma(2) = 1,\ \sigma(3) = 2$. The pair $(1,2)$ is inverted since $3 > 1$, the pair $(1,3)$ is inverted since $3 > 2$, and the pair $(2,3)$ is not, since $1 < 2$. Hence $\mathrm{inv}(\sigma) = 2$ and $\mathrm{sgn}(\sigma) = +1$. Everything convenient about signs is contained in the next lemma. Its proof uses the **difference product** $$ \Delta(x_1, \ldots, x_n) = \prod_{1 \le i < j \le n} (x_j - x_i), $$ the simplest polynomial with the property that interchanging variables changes nothing but the sign. Let $n \ge 2$ and $\sigma, \tau \in S_n$. Then the following hold. 1. As an identity of polynomials, $\Delta(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) = \mathrm{sgn}(\sigma)\, \Delta(x_1, \ldots, x_n)$. 2. $\mathrm{sgn}(\sigma\tau) = \mathrm{sgn}(\sigma)\,\mathrm{sgn}(\tau)$. Moreover $\mathrm{sgn}(\mathrm{id}) = 1$ and $\mathrm{sgn}(\sigma^{-1}) = \mathrm{sgn}(\sigma)$. 3. For every transposition $(p\ q)$ we have $\mathrm{sgn}((p\ q)) = -1$. 4. Every $\sigma \in S_n$ can be written as a product of finitely many transpositions, and if $\sigma$ is written as a product of $k$ transpositions then $\mathrm{sgn}(\sigma) = (-1)^k$. In particular the parity of the number of transpositions does not depend on the chosen factorization. **(1)** We have $\Delta(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) = \prod_{i < j}(x_{\sigma(j)} - x_{\sigma(i)})$. Since $\sigma$ is a bijection, the assignment $\{i, j\} \mapsto \{\sigma(i), \sigma(j)\}$ is a bijection of the set of two-element subsets onto itself. Hence, up to sign, the factors on the right correspond one-to-one with the factors of $\Delta(x_1,\ldots,x_n)$. The factor $x_{\sigma(j)} - x_{\sigma(i)}$ (with $i < j$) is literally a factor of $\Delta$ when $\sigma(i) < \sigma(j)$, whereas when $\sigma(i) > \sigma(j)$ it equals $-(x_{\sigma(i)} - x_{\sigma(j)})$, producing one sign. The number of pairs for which the latter happens is exactly $\mathrm{inv}(\sigma)$, so altogether we get $(-1)^{\mathrm{inv}(\sigma)}\Delta = \mathrm{sgn}(\sigma)\Delta$. **(2)** Since (1) is an identity of polynomials, it remains valid under any substitution of the variables. Setting $y_i = x_{\sigma(i)}$ we get $y_{\tau(i)} = x_{\sigma(\tau(i))} = x_{(\sigma\tau)(i)}$, whence $$ \begin{aligned} \mathrm{sgn}(\sigma\tau)\,\Delta(x) &= \Delta(x_{(\sigma\tau)(1)}, \ldots, x_{(\sigma\tau)(n)}) = \Delta(y_{\tau(1)}, \ldots, y_{\tau(n)}) \\ &= \mathrm{sgn}(\tau)\,\Delta(y_1, \ldots, y_n) = \mathrm{sgn}(\tau)\,\Delta(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) = \mathrm{sgn}(\tau)\,\mathrm{sgn}(\sigma)\,\Delta(x). \end{aligned} $$ As $\Delta$ is not the zero polynomial, comparing coefficients on both sides gives $\mathrm{sgn}(\sigma\tau) = \mathrm{sgn}(\sigma)\mathrm{sgn}(\tau)$. The identity has no inversions, so $\mathrm{sgn}(\mathrm{id}) = 1$; combined with $\sigma\sigma^{-1} = \mathrm{id}$ this yields $\mathrm{sgn}(\sigma)\mathrm{sgn}(\sigma^{-1}) = 1$, and since signs take only the values $\pm 1$ we get $\mathrm{sgn}(\sigma^{-1}) = \mathrm{sgn}(\sigma)$. **(3)** Put $\tau = (p\ q)$ with $p < q$ and count inversions directly. There are only three kinds of inverted pairs. The pair $(p, q)$ itself is inverted, since $\tau(p) = q > p = \tau(q)$ (one pair). For each $k$ with $p < k < q$, the pair $(p, k)$ is inverted, since $\tau(p) = q > k = \tau(k)$ ($q - p - 1$ pairs). For the same $k$, the pair $(k, q)$ is inverted, since $\tau(k) = k > p = \tau(q)$ ($q - p - 1$ pairs). Every other pair either has both indices fixed, or has just one index equal to $p$ or $q$ with the other lying outside the interval, and in these cases the order is unchanged. Hence $\mathrm{inv}(\tau) = 2(q - p - 1) + 1$ is odd and $\mathrm{sgn}(\tau) = -1$. **(4)** We prove that $\sigma$ is a product of transpositions by induction on the number of indices moved by $\sigma$. If $\sigma = \mathrm{id}$ we regard it as the empty product (a product of zero transpositions). If $\sigma \ne \mathrm{id}$, choose $m$ with $\sigma(m) \ne m$ and set $\sigma' = (m\ \sigma(m))\,\sigma$; then $\sigma'(m) = m$, and moreover every index fixed by $\sigma$ is still fixed by $\sigma'$. Thus $\sigma'$ moves strictly fewer indices than $\sigma$, so by the induction hypothesis $\sigma'$ is a product of transpositions. Since $(m\ \sigma(m))$ is its own inverse, $\sigma = (m\ \sigma(m))\,\sigma'$ is one as well. The second assertion follows at once from (2) and (3). ## 3. Definition I: the Leibniz formula Looking closely at the six-term sum in the $3 \times 3$ case, each term is a product of entries chosen one from each column, from mutually distinct rows. Writing $\sigma(1), \sigma(2), \ldots, \sigma(n)$ for the row indices chosen for columns $1, 2, \ldots, n$, the assignment $\sigma$ is nothing but a permutation, and the sign attached is the sign of that permutation. We adopt this directly as the definition for general $n$. For $A = (a_{ij}) \in M_n(\mathbb{K})$, the quantity $$ \det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma)\, a_{\sigma(1)1}\, a_{\sigma(2)2} \cdots a_{\sigma(n)n} = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{j=1}^{n} a_{\sigma(j)\,j} $$ is called the **determinant** of $A$. Besides $\det A$ and $|A|$, one also writes the entries enclosed between vertical bars. The sum has $n!$ terms, half of them carrying a plus sign and half a minus sign. Let us write the definition out for small $n$. - $n = 1$: since $S_1 = \{\mathrm{id}\}$, we get $\det (a_{11}) = a_{11}$. - $n = 2$: here $S_2 = \{\mathrm{id}, (1\ 2)\}$ with signs $+1$ and $-1$ respectively, so $$ \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = a_{11}a_{22} - a_{21}a_{12}. $$ This reproduces the quantity $ad - bc$ that decided solvability in §1. - $n = 3$: among the six elements of $S_3$, the identity and the two 3-cycles are even (sign $+1$), while the three transpositions are odd (sign $-1$). Therefore $$ \det A = a_{11}a_{22}a_{33} + a_{13}a_{21}a_{32} + a_{12}a_{23}a_{31} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33}, $$ which again agrees with the expression in §1. For $3 \times 3$ matrices only, this can be memorized as Sarrus' rule: add the three products along the descending diagonals and subtract the three along the ascending ones. **This mnemonic fails from size $4$ onwards** (a $4 \times 4$ determinant has $4! = 24$ terms, and the eight products obtained by criss-crossing are nowhere near enough). Let us compute the determinant of $$ A = \begin{pmatrix} 2 & -1 & 3 \\ 0 & 4 & 1 \\ 5 & 2 & -2 \end{pmatrix} $$ by substituting directly into the six-term formula above. $$ \begin{aligned} a_{11}a_{22}a_{33} &= 2 \cdot 4 \cdot (-2) = -16, \\ a_{13}a_{21}a_{32} &= 3 \cdot 0 \cdot 2 = 0, \\ a_{12}a_{23}a_{31} &= (-1) \cdot 1 \cdot 5 = -5, \\ a_{13}a_{22}a_{31} &= 3 \cdot 4 \cdot 5 = 60, \\ a_{11}a_{23}a_{32} &= 2 \cdot 1 \cdot 2 = 4, \\ a_{12}a_{21}a_{33} &= (-1) \cdot 0 \cdot (-2) = 0. \end{aligned} $$ Adding the first three and subtracting the last three, $$ \det A = (-16 + 0 - 5) - (60 + 4 + 0) = -21 - 64 = -85. $$ We shall confirm this value later by the expansion of . As a computational tool the Leibniz formula is inefficient (as we shall see in §9), but it has the virtue of being **theoretically very transparent**. Here are two properties obtained by merely inspecting the symmetry of the defining formula. Let $A \in M_n(\mathbb{K})$. Then the following hold. 1. $\det A^{\mathsf{T}} = \det A$. That is, the determinant is invariant under transposition, so every property that holds for rows has a counterpart for columns and vice versa. 2. If $A$ is upper triangular ($a_{ij} = 0$ whenever $i > j$) or lower triangular ($a_{ij} = 0$ whenever $i < j$), then $\det A = a_{11}a_{22}\cdots a_{nn}$. In particular $\det I_n = 1$, and the determinant of a diagonal matrix is the product of its diagonal entries. **(1)** Put $B = A^{\mathsf{T}}$, so that $b_{ij} = a_{ji}$. By definition, $$ \det A^{\mathsf{T}} = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{j=1}^{n} b_{\sigma(j)\,j} = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{j=1}^{n} a_{j\,\sigma(j)}. $$ Each term is a product of finitely many factors, so the factors may be reordered. Reindexing by $i = \sigma(j)$, that is $j = \sigma^{-1}(i)$, gives $\prod_{j} a_{j\,\sigma(j)} = \prod_{i} a_{\sigma^{-1}(i)\,i}$. Moreover $\mathrm{sgn}(\sigma) = \mathrm{sgn}(\sigma^{-1})$ by (2), and as $\sigma$ ranges over all of $S_n$ so does $\tau = \sigma^{-1}$. Therefore $$ \det A^{\mathsf{T}} = \sum_{\tau \in S_n} \mathrm{sgn}(\tau) \prod_{i=1}^{n} a_{\tau(i)\,i} = \det A. $$ **(2)** Suppose $A$ is upper triangular. For a term $\prod_j a_{\sigma(j)j}$ to be nonzero we need $a_{\sigma(j)j} \ne 0$ for every $j$, that is $\sigma(j) \le j$. For $j = 1$, from $\sigma(1) \le 1$ we get $\sigma(1) = 1$. For $j = 2$, from $\sigma(2) \le 2$ and $\sigma(2) \ne \sigma(1) = 1$ we get $\sigma(2) = 2$. Proceeding in the same way for successive $j$ (since $\sigma(1), \ldots, \sigma(j-1)$ have used up $1, \ldots, j-1$, the condition $\sigma(j) \le j$ forces $\sigma(j) = j$), we conclude $\sigma = \mathrm{id}$. Only the single term $\mathrm{sgn}(\mathrm{id}) \prod_j a_{jj}$ survives. The lower triangular case reduces to (1), since the transpose is upper triangular. Finally $I_n$ is diagonal with all diagonal entries equal to $1$, so $\det I_n = 1$. ## 4. Two properties that characterize the determinant The Leibniz formula is a perfectly clear definition, but expanding it every time one wants to prove a property is painful work. In fact almost all properties of the determinant come not from the fine structure of the formula but from just the following two properties. Consider a map $D : (\mathbb{K}^n)^n \to \mathbb{K}$, that is, a map assigning a scalar to $n$ vectors $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n \in \mathbb{K}^n$. - $D$ is **multilinear** if, for each $k$, it is linear as a function of $\boldsymbol{v}_k$ when the other arguments are held fixed; that is, for all $\boldsymbol{u}, \boldsymbol{w} \in \mathbb{K}^n$ and $c \in \mathbb{K}$, $$ \begin{aligned} D(\ldots, \boldsymbol{u} + \boldsymbol{w}, \ldots) &= D(\ldots, \boldsymbol{u}, \ldots) + D(\ldots, \boldsymbol{w}, \ldots), \\ D(\ldots, c\,\boldsymbol{u}, \ldots) &= c\, D(\ldots, \boldsymbol{u}, \ldots) \end{aligned} $$ (the arguments in the positions marked $\ldots$ are the same throughout and are held fixed). - $D$ is **alternating** if $D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n) = 0$ whenever $i \ne j$ and $\boldsymbol{v}_i = \boldsymbol{v}_j$. For $A = (\boldsymbol{a}_1\ \cdots\ \boldsymbol{a}_n)$ we regard $\det A = \det(\boldsymbol{a}_1, \ldots, \boldsymbol{a}_n)$ and treat $\det$ as a function of the columns. If $D : (\mathbb{K}^n)^n \to \mathbb{K}$ is multilinear and alternating, then for all $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n \in \mathbb{K}^n$ $$ D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n) = D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n) \cdot \det V, \qquad V = (\boldsymbol{v}_1\ \cdots\ \boldsymbol{v}_n). $$ In particular, $\det$ is the unique multilinear alternating map with $D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n) = 1$. **Step 1 (from alternation to antisymmetry).** Let $D$ be multilinear and alternating. Fix $i < j$ and put the same vector $\boldsymbol{u} + \boldsymbol{w}$ in both the $i$-th and the $j$-th slot; alternation gives $$ 0 = D(\ldots, \boldsymbol{u} + \boldsymbol{w}, \ldots, \boldsymbol{u} + \boldsymbol{w}, \ldots). $$ Expanding the left-hand side by multilinearity, first in the $i$-th slot and then in the $j$-th, produces four terms, of which $D(\ldots, \boldsymbol{u}, \ldots, \boldsymbol{u}, \ldots)$ and $D(\ldots, \boldsymbol{w}, \ldots, \boldsymbol{w}, \ldots)$ both vanish by alternation. What remains is $$ D(\ldots, \boldsymbol{u}, \ldots, \boldsymbol{w}, \ldots) + D(\ldots, \boldsymbol{w}, \ldots, \boldsymbol{u}, \ldots) = 0, $$ that is, **interchanging two arguments changes the sign** (antisymmetry). Next we show that for every $\sigma \in S_n$ $$ D(\boldsymbol{v}_{\sigma(1)}, \ldots, \boldsymbol{v}_{\sigma(n)}) = \mathrm{sgn}(\sigma)\, D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n). $$ By (4) we may write $\sigma = \tau_1 \tau_2 \cdots \tau_k$ with each $\tau_m$ a transposition, and $\mathrm{sgn}(\sigma) = (-1)^k$. We induct on $k$. The case $k = 0$ is trivial. Write $\sigma = \rho\,\tau$ with $\tau$ a transposition and $\rho$ a product of $k-1$ transpositions, and set $\boldsymbol{w}_i = \boldsymbol{v}_{\rho(i)}$. Then $\boldsymbol{v}_{\sigma(j)} = \boldsymbol{v}_{\rho(\tau(j))} = \boldsymbol{w}_{\tau(j)}$, so $D(\boldsymbol{v}_{\sigma(1)}, \ldots, \boldsymbol{v}_{\sigma(n)}) = D(\boldsymbol{w}_{\tau(1)}, \ldots, \boldsymbol{w}_{\tau(n)})$. As $\tau$ interchanges two arguments, the antisymmetry just proved shows this equals $-D(\boldsymbol{w}_1, \ldots, \boldsymbol{w}_n) = -D(\boldsymbol{v}_{\rho(1)}, \ldots, \boldsymbol{v}_{\rho(n)})$, and the induction hypothesis turns this into $-(-1)^{k-1}D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n) = (-1)^k D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n)$. **Step 2 (reduction to the standard basis).** Write $\boldsymbol{v}_j = \sum_{i=1}^{n} v_{ij}\boldsymbol{e}_i$ in coordinates and expand by multilinearity, starting from the first argument: $$ D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n) = \sum_{i_1 = 1}^{n} \sum_{i_2 = 1}^{n} \cdots \sum_{i_n = 1}^{n} v_{i_1 1} v_{i_2 2} \cdots v_{i_n n}\, D(\boldsymbol{e}_{i_1}, \boldsymbol{e}_{i_2}, \ldots, \boldsymbol{e}_{i_n}). $$ Among the index tuples $(i_1, \ldots, i_n)$, those in which some value occurs twice or more contribute $D(\boldsymbol{e}_{i_1}, \ldots, \boldsymbol{e}_{i_n}) = 0$ by alternation and drop out. What remains are the tuples with $i_1, \ldots, i_n$ pairwise distinct, that is, those for which $j \mapsto i_j$ is a permutation $\sigma$. By Step 1 we have $D(\boldsymbol{e}_{\sigma(1)}, \ldots, \boldsymbol{e}_{\sigma(n)}) = \mathrm{sgn}(\sigma)D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n)$, so $$ D(\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n) = \left( \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{j=1}^{n} v_{\sigma(j)\,j} \right) D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n) = \det V \cdot D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n), $$ which is the asserted identity. **Step 3 ($\det$ really does satisfy the conditions).** Uniqueness alone might be vacuous, so we check that $\det$ itself is multilinear and alternating with $\det I_n = 1$. *Multilinearity.* Each term $\mathrm{sgn}(\sigma)\prod_j a_{\sigma(j)j}$ of the defining formula contains exactly one entry $a_{\sigma(k)k}$ from the $k$-th column, and no other factor depends on that column. Hence each term is homogeneous of degree one in the entries of the $k$-th column, and so is their sum $\det$; that is, $\det$ is linear in the $k$-th column. *Alternation.* Suppose the $p$-th and $q$-th columns coincide ($p \ne q$), that is, $a_{ip} = a_{iq}$ for all $i$. Using the transposition $\tau = (p\ q)$, consider the assignment $\sigma \mapsto \sigma\tau$; it is a fixed-point-free involution of $S_n$ (a bijection which is its own inverse), so $S_n$ splits into $n!/2$ pairs $\{\sigma, \sigma\tau\}$. Comparing the two terms of a pair: for $j \ne p, q$ we have $a_{\sigma\tau(j)j} = a_{\sigma(j)j}$; for $j = p$ we have $a_{\sigma\tau(p)p} = a_{\sigma(q)p} = a_{\sigma(q)q}$; and for $j = q$ we have $a_{\sigma\tau(q)q} = a_{\sigma(p)q} = a_{\sigma(p)p}$ (in the second equality of each we used that the two columns agree). The products are therefore exactly equal, and only the sign is reversed, by (2) and (3). The contributions of each pair cancel, so $\det A = 0$. *Normalization.* The identity $\det I_n = 1$ is (2). This theorem is used in two ways. First, it lets us **start from the properties**, saying "the determinant is the function with these properties". Second, once we have constructed a multilinear alternating map, we can **identify** it automatically as a constant multiple of $\det$. The latter shows its power in the product theorem of §6. First, though, let us record a corollary that bears directly on everyday computation. Let $A = (\boldsymbol{a}_1\ \cdots\ \boldsymbol{a}_n) \in M_n(\mathbb{K})$. 1. Interchanging two columns multiplies the determinant by $-1$. 2. Multiplying one column by $c \in \mathbb{K}$ multiplies the determinant by $c$. In particular $\det(cA) = c^n \det A$. 3. Adding a scalar multiple of one column to another leaves the determinant unchanged: for $p \ne q$ and $c \in \mathbb{K}$, $$ \det(\ldots, \boldsymbol{a}_p + c\,\boldsymbol{a}_q, \ldots) = \det(\ldots, \boldsymbol{a}_p, \ldots). $$ 4. If some column is the zero vector, then the determinant is $0$. 5. All of the above remain true with "column" replaced by "row". Statement (1) is precisely Step 1 (antisymmetry) of the proof of . Statement (2) is the homogeneity half of multilinearity; multiplying all $n$ columns by $c$ produces $c$ a total of $n$ times, giving $\det(cA) = c^n \det A$. For (3), additivity in the $p$-th argument gives $$ \det(\ldots, \boldsymbol{a}_p + c\,\boldsymbol{a}_q, \ldots) = \det(\ldots, \boldsymbol{a}_p, \ldots) + c \det(\ldots, \boldsymbol{a}_q, \ldots), $$ and the second term on the right vanishes by alternation, since there the $p$-th and $q$-th columns are both $\boldsymbol{a}_q$. For (4), take $c = 0$ in (2): $\det A = \det(\ldots, 0 \cdot \boldsymbol{0}, \ldots) = 0 \cdot \det(\ldots) = 0$. Finally (5) follows because (1) allows rows and columns to exchange roles. Item (3) is the important one. **The operation used in Gaussian elimination — adding a constant multiple of one row to another — does not change the determinant at all**, which is why determinant computations go so well with elimination. We put this to use in §9. ## 5. Definition II: cofactor expansion and the adjugate The Leibniz formula settles the $n \times n$ case in one stroke, but there is also a recursion reducing size $n$ to size $n-1$. This is cofactor expansion; theoretically it yields a closed formula for the inverse, and in practice it is used to lighten hand computations by picking a row or column with many zeros. For $A \in M_n(\mathbb{K})$ with $n \ge 2$, let $A_{ij}$ denote the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and the $j$-th column, and call its determinant $\det A_{ij}$ the $(i, j)$ **minor** of $A$. The quantity $$ \widetilde{a}_{ij} = (-1)^{i+j} \det A_{ij} $$ is called the $(i, j)$ **cofactor**. The matrix obtained by arranging the cofactors **transposed**, $$ \mathrm{adj}\,A = \begin{pmatrix} \widetilde{a}_{11} & \widetilde{a}_{21} & \cdots & \widetilde{a}_{n1} \\ \widetilde{a}_{12} & \widetilde{a}_{22} & \cdots & \widetilde{a}_{n2} \\ \vdots & & & \vdots \\ \widetilde{a}_{1n} & \widetilde{a}_{2n} & \cdots & \widetilde{a}_{nn} \end{pmatrix}, \qquad (\mathrm{adj}\,A)_{ij} = \widetilde{a}_{ji}, $$ is called the **adjugate** of $A$. Note carefully the interchange of the indices. Let $n \ge 2$ and $A \in M_n(\mathbb{K})$. 1. For each fixed $j \in \{1, \ldots, n\}$ (expansion along the $j$-th column), $$ \det A = \sum_{i=1}^{n} a_{ij}\, \widetilde{a}_{ij} = \sum_{i=1}^{n} (-1)^{i+j} a_{ij} \det A_{ij}. $$ 2. For each fixed $i \in \{1, \ldots, n\}$ (expansion along the $i$-th row), $$ \det A = \sum_{j=1}^{n} a_{ij}\, \widetilde{a}_{ij} = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} \det A_{ij}. $$ **Step 1.** Decompose the $j$-th column along the standard basis. Since $\boldsymbol{a}_j = \sum_{i=1}^{n} a_{ij}\boldsymbol{e}_i$, linearity in the $j$-th column (verified in ) gives $$ \det A = \sum_{i=1}^{n} a_{ij} \det A^{(i,j)}, $$ where $A^{(i,j)}$ is $A$ with its $j$-th column replaced by $\boldsymbol{e}_i$. It therefore suffices to prove $\det A^{(i,j)} = (-1)^{i+j}\det A_{ij}$. **Step 2 (moving the column and the row to the edge).** In $A^{(i,j)}$, move the $j$-th column to the last position by $n - j$ interchanges with neighbouring columns (the relative order of the other columns is preserved). Then move the $i$-th row to the bottom by $n - i$ interchanges with neighbouring rows. By (1) and (5), these operations multiply the determinant by $(-1)^{(n-j)+(n-i)} = (-1)^{2n - i - j} = (-1)^{i+j}$. Calling the resulting matrix $B$, we get $$ \det B = (-1)^{i+j} \det A^{(i,j)}. $$ Let us determine the shape of $B$. The last column was originally $\boldsymbol{e}_i$, whose only $1$ sat in the $i$-th row; since that row has moved to the bottom, the last column of $B$ is $\boldsymbol{e}_n$. Moreover the upper-left $(n-1) \times (n-1)$ block consists of the entries of $A$ with the $i$-th row and $j$-th column deleted, in their original order, that is, $A_{ij}$. Hence $$ B = \begin{pmatrix} A_{ij} & \boldsymbol{0} \\ \boldsymbol{c}^{\mathsf{T}} & 1 \end{pmatrix} $$ where $\boldsymbol{c}^{\mathsf{T}}$ is the $i$-th row of $A$ with its $j$-th entry removed. **Step 3 ($\det B = \det A_{ij}$).** We return to the definition. In $\det B = \sum_{\sigma} \mathrm{sgn}(\sigma) \prod_{k=1}^{n} b_{\sigma(k)k}$, the $n$-th column is $\boldsymbol{e}_n$, so $b_{\sigma(n)n} \ne 0$ only when $\sigma(n) = n$, in which case $b_{nn} = 1$. The permutations $\sigma$ with $\sigma(n) = n$ correspond bijectively to permutations $\sigma'$ of $\{1, \ldots, n-1\}$, and since $n$ is fixed no additional inversions appear, so $\mathrm{sgn}(\sigma) = \mathrm{sgn}(\sigma')$. Furthermore, for $k \le n-1$ with $\sigma(k) \le n-1$, the entry $b_{\sigma(k)k}$ lies in the upper-left block, that is, it is the $(\sigma'(k), k)$ entry of $A_{ij}$. Therefore $$ \det B = \sum_{\sigma' \in S_{n-1}} \mathrm{sgn}(\sigma') \prod_{k=1}^{n-1} (A_{ij})_{\sigma'(k)\,k} = \det A_{ij}. $$ Combining with Step 2 gives $\det A^{(i,j)} = (-1)^{i+j}\det B = (-1)^{i+j}\det A_{ij}$, which proves (1). **Step 4 (expansion along rows).** Apply (1) to $A^{\mathsf{T}}$. The $(j, i)$ submatrix of $A^{\mathsf{T}}$ is the transpose of the $(i, j)$ submatrix of $A$, so by (1) its determinant equals $\det A_{ij}$; also $(A^{\mathsf{T}})_{ji} = a_{ij}$. Hence the expansion of $A^{\mathsf{T}}$ along its $i$-th column is exactly the expansion of $A$ along its $i$-th row. Cofactor expansion pays off the more zeros the chosen column (or row) has. The practical technique is to create zeros by elementary operations first and then expand. For $x_1, x_2, x_3 \in \mathbb{K}$ we compute the determinant of $$ V = \begin{pmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \end{pmatrix}. $$ First **subtract $x_1$ times the second row from the third row**, and then **subtract $x_1$ times the first row from the second row** (the order matters: if we changed the second row first, the row used in the third-row operation would already have changed). By (3) and (5) the determinant is unchanged. The third row becomes $(x_1^2 - x_1 \cdot x_1,\ x_2^2 - x_1 x_2,\ x_3^2 - x_1 x_3) = (0,\ x_2(x_2 - x_1),\ x_3(x_3 - x_1))$, and the second row becomes $(x_1 - x_1,\ x_2 - x_1,\ x_3 - x_1) = (0,\ x_2 - x_1,\ x_3 - x_1)$. That is, $$ \det V = \det \begin{pmatrix} 1 & 1 & 1 \\ 0 & x_2 - x_1 & x_3 - x_1 \\ 0 & x_2(x_2 - x_1) & x_3(x_3 - x_1) \end{pmatrix}. $$ Applying along the first column, the only nonzero entry is the $1$ in position $(1,1)$, whose sign is $(-1)^{1+1} = +1$, so $$ \det V = \det \begin{pmatrix} x_2 - x_1 & x_3 - x_1 \\ x_2(x_2 - x_1) & x_3(x_3 - x_1) \end{pmatrix}. $$ Factoring $x_2 - x_1$ out of the first column and $x_3 - x_1$ out of the second ( (2)), $$ \det V = (x_2 - x_1)(x_3 - x_1) \det \begin{pmatrix} 1 & 1 \\ x_2 & x_3 \end{pmatrix} = (x_2 - x_1)(x_3 - x_1)(x_3 - x_2). $$ This is exactly the difference product $\Delta(x_1, x_2, x_3)$ of §2: the polynomial used in the theory of signs reappears as a determinant in its own right. Repeating the same computation in size $n$ proves $\det V_n = \prod_{i < j}(x_j - x_i)$. In particular, if $x_1, \ldots, x_n$ are pairwise distinct then $\det V_n \ne 0$, which is equivalent to the existence of exactly one polynomial of degree at most $n-1$ through $n$ distinct points (uniqueness of Lagrange interpolation). Another consequence of cofactor expansion is a closed formula for the inverse. For $n \ge 2$ and $A \in M_n(\mathbb{K})$, $$ A \cdot \mathrm{adj}\,A = (\mathrm{adj}\,A) \cdot A = (\det A)\, I_n . $$ In particular, if $\det A \ne 0$ then $A$ is invertible with $A^{-1} = \dfrac{1}{\det A}\,\mathrm{adj}\,A$. We have $(A \cdot \mathrm{adj}\,A)_{ik} = \sum_{j=1}^{n} a_{ij} (\mathrm{adj}\,A)_{jk} = \sum_{j=1}^{n} a_{ij}\widetilde{a}_{kj}$. When $i = k$, this is exactly the expansion along the $i$-th row from (2), hence equals $\det A$. Now suppose $i \ne k$. Let $A'$ be the matrix obtained from $A$ by replacing its $k$-th row with its $i$-th row. Then $A'$ has equal $i$-th and $k$-th rows, so $\det A' = 0$ by alternation (Step 3 of , which holds for rows as well by (1)). On the other hand the $(k, j)$ cofactor of $A'$ is formed by deleting the $k$-th row, so it coincides with the $(k, j)$ cofactor $\widetilde{a}_{kj}$ of $A$ (since $A$ and $A'$ agree outside the $k$-th row), while the $(k, j)$ entry of $A'$ is $a_{ij}$. Expanding $A'$ along its $k$-th row therefore gives $$ 0 = \det A' = \sum_{j=1}^{n} a_{ij}\widetilde{a}_{kj}. $$ Altogether $A \cdot \mathrm{adj}\,A = (\det A)I_n$. The identity $(\mathrm{adj}\,A)\cdot A = (\det A)I_n$ follows in the same way, using expansion along columns ( (1)) and alternation in the columns. The last assertion follows by dividing both sides by $\det A$ when $\det A \ne 0$; since both the left and the right product equal $I_n$, the matrix $\mathrm{adj}\,A / \det A$ really is the inverse. When $\det A \ne 0$, the solution of $A\boldsymbol{x} = \boldsymbol{b}$ is $\boldsymbol{x} = A^{-1}\boldsymbol{b} = (\det A)^{-1}(\mathrm{adj}\,A)\boldsymbol{b}$, so componentwise $$ x_j = \frac{1}{\det A}\sum_{i=1}^{n} \widetilde{a}_{ij} b_i = \frac{\det A_j}{\det A}, $$ where $A_j$ is $A$ with its $j$-th column replaced by $\boldsymbol{b}$ (the last equality is precisely the expansion of $A_j$ along its $j$-th column: the $(i,j)$ cofactor of $A_j$ is formed by deleting the $j$-th column, so it is the same $\widetilde{a}_{ij}$ as for $A$). Written out for $n = 2$, this agrees with $x = (pd - bq)/(ad - bc)$, computed by hand in §1. The formula is theoretically beautiful, but it requires computing $n+1$ determinants of size $n$, so it is not used in numerical work. ## 6. The product theorem The determinant of a product of matrices is the product of the determinants. This fact is the heart of the theory of determinants, and its proof is a model illustration of how is used to "identify" a map. For $A, B \in M_n(\mathbb{K})$, $$ \det(AB) = \det A \cdot \det B . $$ Fix $A$. Writing $B = (\boldsymbol{b}_1\ \cdots\ \boldsymbol{b}_n)$ column by column, the definition of the matrix product shows that the $j$-th column of $AB$ is $A\boldsymbol{b}_j$. So consider the map $$ D(\boldsymbol{b}_1, \ldots, \boldsymbol{b}_n) := \det(A\boldsymbol{b}_1,\ A\boldsymbol{b}_2,\ \ldots,\ A\boldsymbol{b}_n) = \det(AB) $$ and verify that it is multilinear and alternating. *Multilinearity.* Since $\boldsymbol{b} \mapsto A\boldsymbol{b}$ is linear, we have $A(\boldsymbol{u} + \boldsymbol{w}) = A\boldsymbol{u} + A\boldsymbol{w}$ and $A(c\boldsymbol{u}) = c(A\boldsymbol{u})$; inserting these in the $k$-th column slot of $\det$ and using the linearity of $\det$ in the $k$-th column (Step 3 of ) shows that $D$ is linear in the $k$-th argument. *Alternation.* If $\boldsymbol{b}_p = \boldsymbol{b}_q$ with $p \ne q$, then $A\boldsymbol{b}_p = A\boldsymbol{b}_q$, so the $p$-th and $q$-th columns inside $\det$ coincide, and $D = 0$ by the alternation of $\det$. Hence applies and gives $$ \det(AB) = D(\boldsymbol{b}_1, \ldots, \boldsymbol{b}_n) = D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n)\cdot \det B . $$ Finally $D(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n) = \det(A\boldsymbol{e}_1, \ldots, A\boldsymbol{e}_n)$, and $A\boldsymbol{e}_j$ is nothing but the $j$-th column $\boldsymbol{a}_j$ of $A$, so this equals $\det A$. Therefore $\det(AB) = \det A \det B$. The following consequences are immediate. If $A$ is invertible, so that there is $A^{-1}$ with $AA^{-1} = I_n$, taking determinants of both sides gives $\det A \cdot \det(A^{-1}) = \det I_n = 1$. Since $\mathbb{K}$ is a field, this identity forces $\det A \ne 0$ and at the same time yields $$ \det(A^{-1}) = \frac{1}{\det A}. $$ Also, for an invertible $P$ we get $\det(P^{-1}AP) = \det(P^{-1})\det A \det P = \det A$, so **similar matrices have the same determinant**. This means that the determinant is a quantity attached to the linear map itself, independent of the choice of basis, and it is the foundation for the relation with eigenvalues discussed later (see [Eigenvalues and Eigenvectors](/mathematics/linear-algebra/eigenvalues)). Finally, $\det(A^k) = (\det A)^k$ follows by induction. ## 7. Determinants as a criterion for invertibility We can now give a complete answer to the question raised in §1: can solvability be read off from a single number? For $A = (\boldsymbol{a}_1\ \cdots\ \boldsymbol{a}_n) \in M_n(\mathbb{K})$, the following five conditions are equivalent. 1. $\det A \ne 0$. 2. $A$ is invertible; that is, there exists $B \in M_n(\mathbb{K})$ with $AB = BA = I_n$. 3. The columns $\boldsymbol{a}_1, \ldots, \boldsymbol{a}_n$ are linearly independent. 4. The only solution of the homogeneous equation $A\boldsymbol{x} = \boldsymbol{0}$ is $\boldsymbol{x} = \boldsymbol{0}$. 5. For every $\boldsymbol{b} \in \mathbb{K}^n$, the equation $A\boldsymbol{x} = \boldsymbol{b}$ has exactly one solution. We prove the implications in the order $(1) \Rightarrow (2) \Rightarrow (5) \Rightarrow (4) \Leftrightarrow (3) \Rightarrow (1)$. **$(1) \Rightarrow (2)$** By , the matrix $B = (\det A)^{-1}\mathrm{adj}\,A$ satisfies $AB = BA = I_n$. (For $n = 1$ we have $A = (a_{11})$ with $a_{11} \ne 0$, and we may take $B = (a_{11}^{-1})$; the case $n = 1$ can be checked directly in the same way in what follows.) **$(2) \Rightarrow (5)$** Setting $\boldsymbol{x} = B\boldsymbol{b}$ gives $A\boldsymbol{x} = AB\boldsymbol{b} = \boldsymbol{b}$, so a solution exists. For uniqueness, if $A\boldsymbol{x} = A\boldsymbol{x}' = \boldsymbol{b}$ then multiplying on the left by $B$ gives $\boldsymbol{x} = BA\boldsymbol{x} = BA\boldsymbol{x}' = \boldsymbol{x}'$. **$(5) \Rightarrow (4)$** Apply (5) with $\boldsymbol{b} = \boldsymbol{0}$. Since $A\boldsymbol{0} = \boldsymbol{0}$, the vector $\boldsymbol{x} = \boldsymbol{0}$ is one solution of $A\boldsymbol{x} = \boldsymbol{0}$, and by (5) there is only one solution, so it is the only one. **$(4) \Leftrightarrow (3)$** By the definition of the matrix-vector product, $A\boldsymbol{x} = x_1\boldsymbol{a}_1 + \cdots + x_n\boldsymbol{a}_n$. Hence the statement "the only solution of $A\boldsymbol{x} = \boldsymbol{0}$ is $\boldsymbol{x} = \boldsymbol{0}$" is the same as "$x_1\boldsymbol{a}_1 + \cdots + x_n\boldsymbol{a}_n = \boldsymbol{0}$ implies $x_1 = \cdots = x_n = 0$", which is the definition of linear independence (). **$(3) \Rightarrow (1)$** The vectors $\boldsymbol{a}_1, \ldots, \boldsymbol{a}_n$ are $n$ linearly independent vectors in the $n$-dimensional vector space $\mathbb{K}^n$, hence a basis of $\mathbb{K}^n$ (; see [Vector Spaces and Linear Maps](/en/mathematics/linear-algebra/vector-spaces)). Consequently each standard basis vector can be written $\boldsymbol{e}_j = \sum_{i=1}^{n} b_{ij}\boldsymbol{a}_i$. Assembling these coefficients into a matrix $B = (b_{ij})$, these $n$ identities together read $AB = I_n$. Taking determinants of both sides and using gives $\det A \cdot \det B = \det I_n = 1$, so $\det A \ne 0$.
|build the inverse from the adjugate| D2["A is invertible"] D2 -->|x = Bb and uniqueness| D3["Ax = b always has a unique solution"] D3 -->|set b = 0| D4["Ax = 0 only for x = 0"] D4 -->|Ax is a combination of the columns| D5["the columns are linearly independent"] D5 -->|pass to a basis: AB = I, product theorem| D1`} />
The contrapositive of $(3) \Rightarrow (1)$, namely "if the columns are linearly dependent then $\det A = 0$", can be proved directly from multilinearity without invoking the theory of bases. Indeed, if $\boldsymbol{a}_k = \sum_{j \ne k} c_j \boldsymbol{a}_j$, then linearity in the $k$-th argument gives $$ \det A = \det(\ldots, \textstyle\sum_{j \ne k} c_j\boldsymbol{a}_j, \ldots) = \sum_{j \ne k} c_j \det(\ldots, \boldsymbol{a}_j, \ldots), $$ and each term on the right vanishes by alternation, since there the $k$-th and $j$-th columns are both $\boldsymbol{a}_j$. In practice one most often uses this direction, reading "$\det = 0$" as "there is a linear relation among the columns (or rows)". Thanks to this theorem, the determinant acts as a detector at key points of the theory. For example, $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ when there exists $\boldsymbol{x} \ne \boldsymbol{0}$ with $A\boldsymbol{x} = \lambda\boldsymbol{x}$, that is, when $(\lambda I_n - A)\boldsymbol{x} = \boldsymbol{0}$ has a nontrivial solution. By $(1) \Leftrightarrow (4)$ of , this is equivalent to $$ \varphi_A(\lambda) = \det(\lambda I_n - A) = 0. $$ It is thanks to this equivalence that questions about eigenvalues translate into "find the roots of a polynomial of degree $n$". The polynomial $\varphi_A$ is called the ; it is treated in detail in [Eigenvalues and Eigenvectors](/mathematics/linear-algebra/eigenvalues). ## 8. Geometric meaning: signed volume Why can the determinant, treated so far purely algebraically, measure whether space is being collapsed? The answer is area and volume. Let us first prove this completely in two dimensions. Let $\boldsymbol{u} = \begin{pmatrix} a \\ c \end{pmatrix}$ and $\boldsymbol{v} = \begin{pmatrix} b \\ d \end{pmatrix}$ be vectors in $\mathbb{R}^2$ and put $A = (\boldsymbol{u}\ \boldsymbol{v})$. Then the area of the parallelogram spanned by $\boldsymbol{u}, \boldsymbol{v}$, $$ P = \{ s\boldsymbol{u} + t\boldsymbol{v} \mid 0 \le s \le 1,\ 0 \le t \le 1 \}, $$ equals $|\det A| = |ad - bc|$. If $\boldsymbol{u} = \boldsymbol{0}$ or $\boldsymbol{v} = \boldsymbol{0}$, then $P$ is a segment or a point, of area $0$, and the determinant is $0$ by (4), so the claim holds. Assume now $\boldsymbol{u}, \boldsymbol{v} \ne \boldsymbol{0}$ and let $\theta \in [0, \pi]$ be the angle between them. The area of the parallelogram is "base times height", $S = \|\boldsymbol{u}\| \cdot \|\boldsymbol{v}\|\sin\theta$ (note $\sin\theta \ge 0$). Using the relation $\langle \boldsymbol{u}, \boldsymbol{v}\rangle = \|\boldsymbol{u}\|\|\boldsymbol{v}\|\cos\theta$ between the inner product and the angle (for the definition of the angle see ), $$ S^2 = \|\boldsymbol{u}\|^2\|\boldsymbol{v}\|^2 \sin^2\theta = \|\boldsymbol{u}\|^2\|\boldsymbol{v}\|^2 - \|\boldsymbol{u}\|^2\|\boldsymbol{v}\|^2\cos^2\theta = \|\boldsymbol{u}\|^2\|\boldsymbol{v}\|^2 - \langle \boldsymbol{u}, \boldsymbol{v}\rangle^2 . $$ In coordinates, $\|\boldsymbol{u}\|^2 = a^2 + c^2$, $\|\boldsymbol{v}\|^2 = b^2 + d^2$ and $\langle \boldsymbol{u}, \boldsymbol{v}\rangle = ab + cd$, so $$ \begin{aligned} S^2 &= (a^2 + c^2)(b^2 + d^2) - (ab + cd)^2 \\ &= a^2b^2 + a^2d^2 + c^2b^2 + c^2d^2 - (a^2b^2 + 2abcd + c^2d^2) \\ &= a^2d^2 - 2abcd + b^2c^2 = (ad - bc)^2 . \end{aligned} $$ Since $S \ge 0$, we get $S = |ad - bc| = |\det A|$. The same holds for general $n$. Let $P(\boldsymbol{a}_1, \ldots, \boldsymbol{a}_n) = \{\sum_i t_i \boldsymbol{a}_i \mid 0 \le t_i \le 1\}$ be the parallelepiped spanned by $\boldsymbol{a}_1, \ldots, \boldsymbol{a}_n \in \mathbb{R}^n$, and write $\mathrm{vol}$ for its volume. Listing naively the properties that a notion of "volume" ought to satisfy, we get the following. - **Normalization**: the unit cube $P(\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n)$ has volume $1$. - **Homogeneity**: scaling one edge by $c > 0$ scales the volume by $c$. - **Shear invariance**: adding a constant multiple of one edge to another leaves the volume unchanged (the base and the height are unaffected). - **Degeneration**: if the edges are linearly dependent, the parallelepiped is squashed into a lower dimension and the volume is $0$. If we now pass to the quantity $\widetilde{\mathrm{vol}}$ carrying a sign (**signed volume**: positive for a right-handed frame, negative for a left-handed one), then homogeneity holds for $c < 0$ as well, and together with additivity this becomes multilinearity. Shear invariance and degeneration are alternative expressions of alternation. In other words, $\widetilde{\mathrm{vol}}$ is a multilinear alternating map giving the value $1$ to the unit cube, so by it must satisfy $$ \widetilde{\mathrm{vol}}(\boldsymbol{a}_1, \ldots, \boldsymbol{a}_n) = \det(\boldsymbol{a}_1\ \cdots\ \boldsymbol{a}_n). $$ **The determinant is just another name for signed volume.** Rigorously, this is stated as the identity $\mu(A(E)) = |\det A|\,\mu(E)$ for the Lebesgue measure $\mu$ on $\mathbb{R}^n$ and measurable sets $E$, and its proof is given in analysis as part of the change-of-variables formula for multiple integrals.
e₁ e₂ area 1 Ae₁ Ae₂ area |det A| A
The right-hand side of the figure is the image under $$ A = \begin{pmatrix} 2 & 1 \\ \tfrac{1}{2} & \tfrac{3}{2} \end{pmatrix}, \qquad \det A = 2\cdot\tfrac{3}{2} - 1\cdot\tfrac{1}{2} = 3 - \tfrac{1}{2} = \tfrac{5}{2}. $$ The unit square (of area $1$) is mapped to a parallelogram of area $5/2$. Applying with $\boldsymbol{u} = A\boldsymbol{e}_1 = (2, 1/2)^{\mathsf{T}}$ and $\boldsymbol{v} = A\boldsymbol{e}_2 = (1, 3/2)^{\mathsf{T}}$ gives the same value, $|2 \cdot 3/2 - 1 \cdot 1/2| = 5/2$. Lining up some typical transformations makes it plain what the determinant measures. Consider the following five matrices. $$ S_k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}, \quad R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, \quad F = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad E = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}, \quad \Pi = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$ | Matrix | Transformation | Determinant | Geometric meaning | |---|---|---|---| | $S_k$ | shear | $1 \cdot 1 - k \cdot 0 = 1$ | only tilts; preserves area and orientation | | $R_\theta$ | rotation | $\cos^2\theta + \sin^2\theta = 1$ | preserves area and orientation | | $F$ | swap of axes (reflection) | $0 \cdot 0 - 1 \cdot 1 = -1$ | preserves area but reverses orientation | | $E$ | stretching in one direction | $3 \cdot 1 - 0 \cdot 0 = 3$ | multiplies area by 3 | | $\Pi$ | projection onto the first axis | $1 \cdot 0 - 0 \cdot 0 = 0$ | collapses onto a line; area disappears | The last row is the geometric content of . That $\det A = 0$ means the image of the unit cube "loses thickness"; in that case $A$ discards information, so no inverse transformation can exist. The product theorem too can be read geometrically. If $B$ multiplies volume by $\det B$ and $A$ multiplies volume by $\det A$, then performing them one after the other should multiply volume by $\det A \cdot \det B$. is the precise form of this intuition. Likewise $\det(A^{-1}) = 1/\det A$ reads as "to undo an expansion, shrink by the same factor". ## 9. Computation and appearances in data analysis The theoretical definition and the actual computational procedure are different things. The Leibniz formula is a sum of $n!$ terms, each involving $n-1$ multiplications, so it needs roughly $n! \cdot (n-1)$ multiplications. By contrast, reducing to upper triangular form by row operations and then multiplying the diagonal entries (Gaussian elimination, LU decomposition) needs only about $n^3/3$ multiplications and divisions. | $n$ | Number of terms $n!$ in the Leibniz formula | Multiplications/divisions in elimination (about $n^3/3$) | |---|---|---| | $5$ | $120$ | about $42$ | | $10$ | $3628800$ | about $333$ | | $20$ | about $2.4 \times 10^{18}$ | about $2667$ | | $50$ | about $3.0 \times 10^{64}$ | about $41667$ | Already at $n = 20$, a machine evaluating $10^9$ terms per second would need more than 70 years for the Leibniz formula. One must not implement the defining formula literally. $$ A = \begin{pmatrix} 0 & 1 & 2 & 1 \\ 2 & 1 & 3 & 1 \\ 6 & 5 & 16 & 5 \\ 2 & 3 & 9 & 4 \end{pmatrix} $$ **Method 1: row operations.** First interchange the first and second rows. By (1) and (5) this multiplies the determinant by $-1$, so if $A'$ denotes the interchanged matrix then $\det A' = -\det A$. $$ A' = \begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 1 & 2 & 1 \\ 6 & 5 & 16 & 5 \\ 2 & 3 & 9 & 4 \end{pmatrix} $$ Next subtract $3$ times the first row from the third, and $1$ times the first row from the fourth. By (3) and (5) the determinant is unchanged. $$ \begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 2 & 7 & 2 \\ 0 & 2 & 6 & 3 \end{pmatrix} $$ Then subtract $2$ times the second row from the third and fourth rows, and finally subtract $2/3$ times the third row from the fourth. $$ \begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 2 & 1 \end{pmatrix} \longrightarrow \begin{pmatrix} 2 & 1 & 3 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ The matrix is now upper triangular, so by (2) its determinant is $2 \cdot 1 \cdot 3 \cdot 1 = 6$. This equals $\det A'$, whence $\det A = -\det A' = -6$. **Method 2: cofactor expansion.** The first row contains a $0$, so it is a good choice for expansion. Using (2) with $i = 1$, the signs are $(-1)^{1+j}$, so $$ \det A = -1 \cdot \det A_{12} + 2 \cdot \det A_{13} - 1 \cdot \det A_{14} $$ (the term with $a_{11} = 0$ drops out). We compute the three minors by Sarrus' rule. $$ \det A_{12} = \det\begin{pmatrix} 2 & 3 & 1 \\ 6 & 16 & 5 \\ 2 & 9 & 4 \end{pmatrix} = 2(64 - 45) - 3(24 - 10) + 1(54 - 32) = 38 - 42 + 22 = 18, $$ $$ \det A_{13} = \det\begin{pmatrix} 2 & 1 & 1 \\ 6 & 5 & 5 \\ 2 & 3 & 4 \end{pmatrix} = 2(20 - 15) - 1(24 - 10) + 1(18 - 10) = 10 - 14 + 8 = 4, $$ $$ \det A_{14} = \det\begin{pmatrix} 2 & 1 & 3 \\ 6 & 5 & 16 \\ 2 & 3 & 9 \end{pmatrix} = 2(45 - 48) - 1(54 - 32) + 3(18 - 10) = -6 - 22 + 24 = -4. $$ Hence $\det A = -18 + 8 + 4 = -6$, in agreement with Method 1. Even at size $4$, cofactor expansion requires computing three determinants of size $3$, and it becomes rapidly less competitive as the size grows. Numerical libraries also use LU decomposition internally. ```python import numpy as np A = np.array([ [0.0, 1.0, 2.0, 1.0], [2.0, 1.0, 3.0, 1.0], [6.0, 5.0, 16.0, 5.0], [2.0, 3.0, 9.0, 4.0], ]) print(np.linalg.det(A)) # a value very close to -6.0 (with floating-point error) sign, logabsdet = np.linalg.slogdet(A) print(sign, np.exp(logabsdet)) # values very close to -1.0 and 6.0 ``` Here `slogdet` returns the sign and $\log|\det A|$ separately. In high dimensions $\det$ itself easily overflows or underflows (it is a product of diagonal entries, so multiplying a thousand copies of $0.1$ gives $10^{-1000}$), which is why statistics and machine learning work with $\log\det$ rather than $\det$. For a symmetric positive definite matrix $\Sigma$, the Cholesky decomposition $\Sigma = LL^{\mathsf{T}}$ gives $$ \log\det\Sigma = \log\big((\det L)^2\big) = 2\sum_{i=1}^{n}\log L_{ii} $$ (we used and (1) and (2)). Suppose that from bivariate data we obtain the sample covariance matrix $$ S = \begin{pmatrix} 4 & 2 \\ 2 & 3 \end{pmatrix}. $$ Here $\det S = 4 \cdot 3 - 2 \cdot 2 = 8$, and this value is called the **generalized variance**. Let us see what it means. First compute the eigenvalues. We have $\varphi_S(\lambda) = \det(\lambda I_2 - S) = (\lambda - 4)(\lambda - 3) - 4 = \lambda^2 - 7\lambda + 8$, with roots $\lambda = (7 \pm \sqrt{17})/2$, numerically $\lambda_1 \approx 5.562$ and $\lambda_2 \approx 1.438$. Their product is $\lambda_1\lambda_2 = 8 = \det S$ and their sum is $\lambda_1 + \lambda_2 = 7 = \mathrm{tr}\,S$. The agreement of the products is no accident. In the characteristic polynomial $\varphi_A(\lambda) = \det(\lambda I_n - A)$, the term of the Leibniz formula with $\sigma = \mathrm{id}$ is $\prod_i (\lambda - a_{ii})$ while all other terms have degree at most $n-2$, so $\varphi_A$ is a monic polynomial of degree $n$ in $\lambda$. Hence over the complex numbers it factors as $\varphi_A(\lambda) = \prod_{i=1}^{n}(\lambda - \lambda_i)$. Substituting $\lambda = 0$, the left-hand side is $\det(-A) = (-1)^n\det A$ ( (2)) and the right-hand side is $(-1)^n\prod_i \lambda_i$, so $$ \det A = \lambda_1\lambda_2\cdots\lambda_n $$ holds in general. **The determinant is the product of the eigenvalues** (; see also [Diagonalization and Jordan Normal Form](/mathematics/linear-algebra/diagonalization-and-jordan-form)). The geometric meaning is equally clear. The ellipse $E = \{\boldsymbol{x} \in \mathbb{R}^2 \mid \boldsymbol{x}^{\mathsf{T}}S^{-1}\boldsymbol{x} \le 1\}$ (the equiprobability ellipse describing the scatter of the data) is the image of the unit disc under $S^{1/2}$, so by the generalization of its area is $\pi \cdot \det(S^{1/2}) = \pi\sqrt{\det S} = 2\sqrt{2}\,\pi \approx 8.886$. Thus the generalized variance measures the "volume of the spread" of the data cloud. If the two variables were perfectly correlated, the ellipse would collapse to a segment and $\det S = 0$. This viewpoint leads directly to principal component analysis (PCA). PCA diagonalizes the covariance matrix by an orthogonal matrix to extract the directions of largest variance, and $\det S = \prod_i \lambda_i$ is the product of the variances of the principal components, that is, the volume of the overall spread. If some $\lambda_i$ is extremely small, there is almost no information in that direction, and little is lost by reducing the dimension. Likewise the term $-\frac{1}{2}\log\det\Sigma$ appearing in the log-likelihood of a multivariate normal distribution acts as a penalty on this "volume of spread". The general theory of diagonalization by orthogonal matrices is treated in [Inner Product Spaces and Gram-Schmidt Orthogonalization](/mathematics/linear-algebra/inner-product-spaces) and [The Spectral Theorem](/mathematics/linear-algebra/spectral-theorem) (that a real symmetric matrix is diagonalizable by an orthogonal matrix is ). ## 10. Exercises Using determinants, find the area of the triangle in the plane with vertices $P(1, 2)$, $Q(4, 3)$, $R(2, 6)$. Determine also whether the circuit $P \to Q \to R$ runs counterclockwise or clockwise. Taking $P$ as base point, $\overrightarrow{PQ} = (3, 1)^{\mathsf{T}}$ and $\overrightarrow{PR} = (1, 4)^{\mathsf{T}}$. By , the area of the parallelogram spanned by these two vectors is $$ \left| \det\begin{pmatrix} 3 & 1 \\ 1 & 4 \end{pmatrix} \right| = |3 \cdot 4 - 1 \cdot 1| = 11. $$ The triangle is half of this parallelogram, so its area is $11/2$. As for the sign, $\det = 11 > 0$, so $(\overrightarrow{PQ}, \overrightarrow{PR})$ has the same orientation as the standard basis $(\boldsymbol{e}_1, \boldsymbol{e}_2)$, that is, the circuit is counterclockwise. As stated in §8, the absolute value of the determinant carries the area and its sign carries the orientation. Let $n$ be odd and let $A \in M_n(\mathbb{R})$ be skew-symmetric ($A^{\mathsf{T}} = -A$). Show that $\det A = 0$. Show also, by an example with $n = 2$, that the conclusion fails when $n$ is even. By (1) we have $\det A = \det A^{\mathsf{T}}$. Substituting the hypothesis $A^{\mathsf{T}} = -A$ gives $\det A = \det(-A)$. Now $-A$ is obtained from $A$ by multiplying all $n$ columns by $-1$, so $\det(-A) = (-1)^n \det A$ by (2). Since $n$ is odd, $(-1)^n = -1$, and therefore $$ \det A = -\det A . $$ Adding $\det A$ to both sides gives $2\det A = 0$. As $2 \ne 0$ in $\mathbb{R}$, we conclude $\det A = 0$. When $n$ is even we have $(-1)^n = 1$ and the argument above says nothing. Indeed, for $n = 2$ the matrix $$ A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$ is skew-symmetric, satisfying $A^{\mathsf{T}} = -A$, yet $\det A = 0 \cdot 0 - 1 \cdot (-1) = 1 \ne 0$. So the hypothesis that $n$ is odd was essential. Let $p, q \ge 1$ and $n = p + q$. Let $A \in M_p(\mathbb{K})$, $D \in M_q(\mathbb{K})$, let $B$ be a $p \times q$ matrix and $O$ the $q \times p$ zero matrix, and put $$ M = \begin{pmatrix} A & B \\ O & D \end{pmatrix} \in M_n(\mathbb{K}). $$ Show that $\det M = \det A \cdot \det D$. Following the definition (), consider $\det M = \sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)\prod_{j=1}^{n} m_{\sigma(j)\,j}$. The entries of $M$ satisfy $m_{ij} = 0$ when $i > p$ and $j \le p$ (the position of the lower-left block $O$). For a term to be nonzero we need $\sigma(j) \le p$ for every $j \le p$. This says that $\sigma$ maps $\{1, \ldots, p\}$ into $\{1, \ldots, p\}$, and since $\sigma$ is injective and the set is finite, it maps $\{1, \ldots, p\}$ onto itself. Consequently $\sigma$ also maps the remainder $\{p+1, \ldots, n\}$ onto itself. Defining $\alpha \in S_p$ to be the restriction of $\sigma$ to $\{1, \ldots, p\}$ and $\beta \in S_q$ by $\beta(k) = \sigma(p + k) - p$, we get a one-to-one correspondence between $\sigma$ and the pair $(\alpha, \beta)$. Now compare signs. Inverted pairs fall into three kinds. The number of inversions among pairs with both indices in $\{1,\ldots,p\}$ is $\mathrm{inv}(\alpha)$, and the number among pairs with both indices in $\{p+1,\ldots,n\}$ is $\mathrm{inv}(\beta)$. For a pair with $i \le p < j$ we have $\sigma(i) \le p < \sigma(j)$, so it is never inverted. Hence $\mathrm{inv}(\sigma) = \mathrm{inv}(\alpha) + \mathrm{inv}(\beta)$, that is, $\mathrm{sgn}(\sigma) = \mathrm{sgn}(\alpha)\mathrm{sgn}(\beta)$. As for the products, $m_{\sigma(j)j} = a_{\alpha(j)j}$ for $j \le p$, and $m_{\sigma(j)j} = d_{\beta(k)k}$ for $j = p + k$ with $k \le q$. Putting all of this together, $$ \det M = \sum_{\alpha \in S_p}\sum_{\beta \in S_q} \mathrm{sgn}(\alpha)\mathrm{sgn}(\beta) \prod_{j=1}^{p} a_{\alpha(j)j} \prod_{k=1}^{q} d_{\beta(k)k} = \left(\sum_{\alpha}\mathrm{sgn}(\alpha)\prod_j a_{\alpha(j)j}\right)\left(\sum_{\beta}\mathrm{sgn}(\beta)\prod_k d_{\beta(k)k}\right), $$ and the right-hand side is $\det A \cdot \det D$. Note that the upper-right block $B$ does not appear in the result at all. This is the block version of the formula for triangular matrices in (2). Let $n \ge 2$. Show that $\det(\mathrm{adj}\,A) = (\det A)^{n-1}$ for every $A \in M_n(\mathbb{K})$ (including the case $\det A = 0$). By we have $A \cdot \mathrm{adj}\,A = (\det A)I_n$. Taking determinants of both sides and using together with (2) (so that $\det(cI_n) = c^n \det I_n = c^n$), $$ \det A \cdot \det(\mathrm{adj}\,A) = (\det A)^n \qquad \cdots (\ast). $$ **Case 1: $\det A \ne 0$.** Dividing both sides of $(\ast)$ by $\det A$ gives $\det(\mathrm{adj}\,A) = (\det A)^{n-1}$. **Case 2: $\det A = 0$.** Here the right-hand side is $(\det A)^{n-1} = 0$ since $n - 1 \ge 1$, so it suffices to show $\det(\mathrm{adj}\,A) = 0$ (the relation $(\ast)$ merely reads $0 = 0$ and gives no information, so a different argument is needed). Suppose, for contradiction, that $\det(\mathrm{adj}\,A) \ne 0$. By , the matrix $\mathrm{adj}\,A$ is invertible and has an inverse $(\mathrm{adj}\,A)^{-1}$. On the other hand, together with $\det A = 0$ gives $A \cdot \mathrm{adj}\,A = O$, so multiplying on the right by $(\mathrm{adj}\,A)^{-1}$ yields $A = O$. But if $A = O$, then all its $(i,j)$ submatrices are zero matrices of size $n - 1 \ge 1$, whose determinants are $0$ ( (4)), so all cofactors vanish, that is, $\mathrm{adj}\,A = O$. This contradicts the invertibility of $\mathrm{adj}\,A$. Hence $\det(\mathrm{adj}\,A) = 0 = (\det A)^{n-1}$. (The hypothesis $n \ge 2$ is essential in Case 2. For $n = 1$ one sets $\mathrm{adj}\,A = (1)$ by convention, so $\det(\mathrm{adj}\,A) = 1 = (\det A)^0$ and the conclusion still holds, but the argument above does not apply as it stands.) ## References - Saito Masahiko, *Senkei Daisū Nyūmon*, University of Tokyo Press, 1966 (in Japanese) — the chapter on determinants. The standard route from signs of permutations through the Leibniz formula to cofactor expansion. - Satake Ichiro, *Senkei Daisūgaku*, Shokabo (Sūgaku Sensho), reissued edition 2015 (in Japanese) — the chapter on determinants. A lucid treatment of the determinant as an alternating multilinear form. - S. Lang, *Linear Algebra*, 3rd ed., Springer, 1987 — the chapter on determinants. Built around an axiomatic characterization like in this article. - Sugiura Mitsuo, *Kaiseki Nyūmon II*, University of Tokyo Press, 1985 (in Japanese) — the change-of-variables formula for multiple integrals. The analytic proof that $|\det|$ appears as the volume scaling factor. - G. H. Golub, C. F. Van Loan, *Matrix Computations*, 4th ed., Johns Hopkins University Press, 2013 — the chapters on Gaussian elimination and LU decomposition. Numerical computation of determinants and the handling of overflow. - M. Kline, *Mathematical Thought from Ancient to Modern Times*, Oxford University Press, 1972 — the chapters on determinants and matrices. The historical route from Seki Takakazu and Leibniz through Cramer to Cauchy.