Determinant ($\det$)

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

In the Leibniz formula for a determinant of a block diagonal matrix, a term is zero unless its permutation maps every block's row indices onto that same block's column indices. The surviving permutations are precisely independent permutations within the blocks; their signs and products of entries factor over the blocks. Summing therefore gives the product of the individual determinants. Applied to $tI-T$, this also proves that the characteristic polynomial of a block diagonal linear operator is the product of the characteristic polynomials of its blocks.

Two side vectors span a parallelogram whose signed area is their determinant. Its two congruent triangles give the factor one-half, and the absolute value removes orientation. Reversing the vertex order changes the sign but not the area; collinear vertices give zero.

A determinantal function on $M_n(F)$ is an alternating multilinear map of the $n$ columns, normalized to take value one at the identity matrix. It is uniquely the determinant. Indeed, expand each column in the standard basis. Terms with a repeated basis vector vanish, and all remaining terms differ from the value on the standard basis only by the sign of a permutation. Thus an alternating multilinear map of $n$ columns is determined by its value on the identity matrix.

For planar vectors $u=(u_x,u_y)$ and $v=(v_x,v_y)$, the scalar $u\times v=u_x{v}_y-u_y{v}_x$ is the determinant of their coordinate columns. It equals the perpendicular component of the three-dimensional cross product after embedding the plane in three dimensions. Its sign records orientation, and its absolute value is twice the area of the triangle spanned by $u,v$.

For planar points $A,B,P$, the sign of $(B-A)\times (P-A)$, using the two-dimensional cross product, determines which side of the directed line from $A$ to $B$ contains $P$. Zero means collinearity. A point is strictly inside a nondegenerate triangle exactly when these signs for its three cyclically ordered sides are all strictly positive or all strictly negative.

For $A=B=I_2$, $\det (A+B)=\det (2I_2)=4$, while $\det A+\det B=2$. The determinant is multilinear in its separate columns, rather than a linear function of the matrix as a whole. Expanding the columns of a sum introduces mixed terms that the proposed additive identity omits.

Expand the determinant along any fixed row $i$ using the displayed formula, where $A^{(i,j)}$ deletes row $i$ and column $j$. The signed matrix minors are its cofactors. Applying the same formula to the transpose gives the column expansion. The row formula follows from the Leibniz formula for determinants by grouping permutations according to the selected column in row $i$. For a lower triangular matrix, expanding along the first row reduces the determinant to $A_{11}$ times a smaller triangular determinant; induction gives the product of all diagonal entries, even when some vanish.

For a weighted acyclic directed graph, let $M_{ij}$ be the sum of path weights from $A_j$ to $B_i$. Expanding $\det M$ gives a signed sum of path families. At the first intersection in a fixed topological order, swap the two path tails. The operation preserves weight and reverses the endpoint permutation's sign, so intersecting families cancel. Only vertex-disjoint families remain. If planarity forces these families to join identically indexed endpoints, their sum has positive sign. Applied to northeast paths with horizontal weight $x_r$ at height $r$, this gives the skew Jacobi–Trudi identity.

Let $L$ be the triangular matrix with $L_{ij}=1$ for $j\le i$ and zero otherwise. Then $(L{L}^T{)}_{ij}=\min (i,j)$ and $\det L=1$, so the determinant of $L{L}^T$ is one. More generally, for $0=t_0<t_1<\cdots <t_n$, the matrix with entries $\min (t_i,t_j)$ factors as $L\mathrm{diag}(t_k-t_{k-1})L^T$ and has determinant ${\prod }_k(t_k-t_{k-1})$. This is the finite-dimensional structure behind the Brownian covariance kernel.

For matrices $A\in {\mathbb{R}}^{r\times m}$ and $B\in {\mathbb{R}}^{m\times r}$ with $r\le m$,where the indices of $S$ are taken in increasing order. To prove it, write the $j$th column of $AB$ as ${\sum }_q{B}_{qj}{A}_{:,q}$ and expand its determinant by multilinearity. Terms with a repeated $q$ vanish by alternation. For each remaining index set $S=\{s_1<\cdots <s_r\}$, collect all permutations $q_j=s_{\sigma (j)}$. The factor from $A$ is $\mathrm{sgn}(\sigma )\det A_{:,S}$; the sum of the products $\mathrm{sgn}(\sigma ){\prod }_j{B}_{s_{\sigma (j)},j}$ is $\det B_{S,:}$. This proves the formula. Applying it to selected rows and columns proves that a product of compatible totally nonnegative matrices is totally nonnegative.

For an invertible square matrix $A$ and $Ax=b$, replace column $i$ by $b$ to form $A_i$. Multilinearity of the determinant and $b={\sum }_j{x}_j{A}_{\cdot j}$ give $\det A_i=x_i\det A$: all other summands have repeated columns and vanish. Division gives the formula. With integer coefficients this supplies explicit rational solutions and, together with the Hadamard determinant inequality, bounds their numerators and denominators.

For a square matrix with row vectors $r_i$, the absolute determinant is at most the product of the Euclidean norms of its rows. If the rows are dependent the determinant is zero. Otherwise Gram-Schmidt process leaves the determinant unchanged and replaces each row by its component orthogonal to its predecessors. The resulting orthogonal rows have lengths no greater than the original rows, and their volume is their product. This proves the inequality; equality holds for mutually orthogonal rows.

For an antisymmetric $2n\times 2n$ matrix, the Pfaffian is $\mathrm{Pf}(A)=(2^{n}n!{)}^{-1}{\sum }_{\sigma \in S_{2n}}\mathrm{sgn}(\sigma ){\prod }_{j=1}^n{A}_{\sigma (2j-1),\sigma (2j)}$. Equivalently, it is the coefficient of the top exterior power of the associated alternating form, with the usual factorial normalization. Its square is the determinant. In size four it is $a_{12}{a}_{34}-a_{13}{a}_{24}+a_{14}{a}_{23}$, so its vanishing characterizes the simple nonzero bivectors of the Klein quadric.

The block matrix $M=\left(\begin{array}{cc}A& -B\\ B& A\end{array}\right)$ is similar over $\mathbb{C}$ to $\mathrm{diag}(A+iB,A-iB)$ via $T=\left(\begin{array}{cc}I& I\\ -iI& iI\end{array}\right)$. Therefore its determinant is $\det (A+iB)\det (A-iB)$ for arbitrary square matrices $A,B$. The argument uses a constant change-of-basis matrix and multiplicativity of the determinant.

For a differentiable invertible matrix $A(t)$, $\frac{d}{dt}\log \det A=\mathrm{tr}(A^{-1}\dot{A})$ wherever a consistent logarithm is defined. In particular a positive spatial metric tensor satisfies $h^{ij}{\dot{h}}_{ij}=\dot{h}/h$. This separates its local volume change from shape change.

For an invertible matrix $A$ and any matrix $H$,Indeed $\det (A+tH)=\det (A)\det (I+t{A}^{-1}H)$, and multilinearity shows that the linear coefficient of $\det (I+tB)$ is $\mathrm{tr}B$. At a singular matrix the derivative is instead written $\mathrm{tr}(\mathrm{adj}(A)H)$; this polynomial formula holds for all $A$.

For an arbitrary square matrix $A$, the derivative of the determinant in direction $H$ is the displayed expression. Differentiating the determinant's multilinear expansion gives the sum of entries of $H$ weighted by their cofactors, equal to the displayed trace. If $A$ is invertible, $\mathrm{adj}(A)=(\det A)A^{-1}$ gives $(\det A)\mathrm{tr}(A^{-1}H)$, and division by $\det A$ gives the Jacobi determinant derivative formula for the logarithmic derivative. The adjugate expression remains valid at singular matrices.

The derivative of the determinant at an invertible matrix is $D{\det }_A(K)=\det A\mathrm{tr}(A^{-1}K)$. At $A=I+H$, use $\det (I+H)=1+\mathrm{tr}H+O(\Vert H{\Vert }^2)$ and $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$. Differentiating the first derivative gives the displayed symmetric bilinear map. More generally $D^2{\det }_A(H,K)=\det A[\mathrm{tr}(A^{-1}H)\mathrm{tr}(A^{-1}K)-\mathrm{tr}(A^{-1}H{A}^{-1}K)]$.

On positive-definite matrices, the log-determinant has differential $d\log \det A=\mathrm{Tr}(A^{-1}dA)$ and is a strictly concave function.

The determinant is linear in each row and in each column when all the others are held fixed.

The determinant is multiplicative: for square matrices of the same size, $\det (AB)=\det (A)\det (B)$.

For an $n$ by $n$ matrix,

For diagonal entries $a_i$, superdiagonal entries $b_i$, and subdiagonal entries $c_i$, the leading principal determinants satisfy

For $N$ equal masses joined to fixed walls by equal springs, the stiffness matrix has diagonal $2k$ and adjacent entries $-k$. Its characteristic determinant obeys $D_N=d{D}_{N-1}-k^2{D}_{N-2}$, $d=2k-m{\omega }^2$, $D_0=1$, $D_1=d$. If ${\beta }_1+{\beta }_2=d$ and ${\beta }_1{\beta }_2=k^2$, then $D_N=({\beta }_1^{N+1}-{\beta }_2^{N+1})/({\beta }_1-{\beta }_2)$ for distinct roots. At a repeated root $\beta =\pm k$, the correct limit is $(N+1){\beta }^N$, which is nonzero. Thus the apparent ratio solution ${\beta }_1/{\beta }_2=1$ must be excluded. Writing $d=2k\cos q$ gives $D_N=k^N\sin [(N+1)q]/\sin q$, the Chebyshev polynomial of the second kind. There are exactly $N$ distinct positive normal frequencies $2\sqrt{k/m}\sin [j\pi /(2(N+1))]$, $1\le j\le N$.

For $N$ identical masses between fixed walls with equal springs, the dimensionless stiffness matrix has diagonal entries $2$ and adjacent entries $-1$. Its normalized eigenvectors and eigenvalues areExtend each vector by $a_0=a_{N+1}=0$. The sine addition identity gives $2a_i-a_{i-1}-a_{i+1}={\lambda }_j{a}_i$, including the fixed-end rows. The finite sine orthogonality identity supplies the normalization and makes distinct vectors orthogonal. The normal mode angular frequencies are ${\omega }_j=\sqrt{k{\lambda }_j/m}$ for mass $m$ and spring constant $k$. This complements the fixed-end spring-chain determinant and its repeated-root limit with the spatial mode shapes.