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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]

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 an invertible matrix $A\in {\mathbb{R}}^{n\times n}$ and real vectors $u,v\in {\mathbb{R}}^n$, the matrix determinant lemma statesThe expression $v^T{A}^{-1}u$ is a scalar, so a rank-one matrix update changes the determinant by a single multiplicative factor. One need not expand the updated determinant entry by entry.

For the preceding example with $A=I$, the dot product is $v^{T}u=3-2=1$. Hence the updated determinant is $1+1=2$. Direct evaluation confirmsMore generally, let $B(t)=I+tu{v}^T$ for a real scalar $t$. The same calculation gives $\det B(t)=1+t$. The determinant is one at $t=0$, two at $t=1$, and zero at $t=-1$.

A compact proof explains the scalar reduction. Put $w=A^{-1}u$ and factor $A+u{v}^T=A(I+w{v}^T)$. If $w\ne 0$, choose a basis starting with $w$. In that basis, $(I+w{v}^T)w=(1+v^{T}w)w$, while every other basis vector changes only by a multiple of $w$. Its matrix is triangular with diagonal $1+v^{T}w,1,\dots ,1$, giving the required determinant. If $w=0$, the correction vanishes. This uses multiplicativity of the determinant to restore the factor $\det A$.

The denominator in the Sherman–Morrison formula is exactly this factor. The rank-one update singularity criterion therefore identifies when the updated matrix inverse ceases to exist. Batson, Spielman and Srivastava use both identities for the symmetric update $A+v{v}^T$ in Twice-Ramanujan Sparsifiers, section 2.3, Lemmas 2.1–2.2.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.

The real invertible matrices with positive determinant form a subgroup of the general linear group. The identity has determinant one, multiplicativity of the determinant preserves positivity under products, and $\det (A^{-1})=1/\det A>0$ preserves it under inversion. More precisely this is the kernel of a group homomorphism $A\mapsto \mathrm{sgn}(\det A)$ to $\{1,-1\}$, so it is a normal subgroup of index two for $n\ge 1$. Its elements preserve an orientation of a vector space.

Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable mean value theorem applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,for some $0<\theta ,\eta <1$. The same formula with the absent segment omitted covers zero increments. Continuity of both partial derivatives at $p$ makes the remainder after subtracting $ah+bk$ at most $\epsilon (|h|+|k|)$ for small increments, and hence at most $\sqrt{2}\epsilon \sqrt{h^2+k^2}$. Therefore the Fréchet derivative is $(h,k)\mapsto ah+bk$. Continuity throughout the disc is not required; continuity at the point suffices.

For the determinant, expand by multilinearity of the determinant in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\mathrm{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\Vert H{\Vert }^2)$. HenceFor invertible $A$, multiplicativity of the determinant gives $\det (A+H)=\det A\det (I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the derivative of the determinantTo justify an open invertible neighborhood, let $\Vert K\Vert <1$. Completeness and submultiplicativity make the Neumann series $S={\sum }_{j=0}^{\infty }(-K{)}^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K{)}^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus $I+K$ is invertible, and the series gives $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$ near zero.

Apply the first-derivative formula at $I+H$ to a test matrix $K$:The last estimate is uniform for $\Vert K\Vert \le 1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The second derivative of the determinant is the symmetric bilinear mapIts symmetry follows from $\mathrm{tr}(HK)=\mathrm{tr}(KH)$.

For the determinant of a complex block representation, putThe change-of-basis matrix $T$ is invertible, and direct block multiplication givesThus $T^{-1}MT$ is block diagonal. Using multiplicativity of the determinant and the preceding block formula,The identity follows from this constant change of basis for arbitrary complex $A,B$.

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

Adding the diagonal entries gives $\mathrm{tr}A=4+4+2=10$. Expansion along the first row gives $\det A=4(8-4)+2(0-8)=0$. For the remaining trace, direct matrix multiplication givesThus $\mathrm{tr}(A^2)=20+20+12=52$. By multiplicativity of the determinant, $\det (A^2)=(\det A{)}^2=0$. Altogether,

A Cartesian tensor of order $n$ is a multilinear geometric object whose components under an orthogonal change of axes obey the tensor component transformation law,An isotropic tensor has the same components under every proper rotation. For the Kronecker delta,Thus ${\delta }_{ij}$ is isotropic, even under improper orthogonal transformations.

To prove multiplicativity of the determinant from the given formula, first consider $C_{pqr}={ϵ}_{ijk}{A}_{ip}{A}_{jq}{A}_{kr}$. Exchanging any two of $p,q,r$ changes its sign, so it is proportional to the Levi-Civita symbol. Its $123$ component is the determinant, as follows by expanding the given double-epsilon expression. ThereforeSubstitute $(AB{)}_{il}=A_{ip}{B}_{pl}$ into that same determinant formula:HenceNo invertibility assumption was used, so this identity includes singular matrices.

Now differentiate the double-epsilon formula for a differentiable $A(t)$. Its three differentiated terms are equal after renaming the paired dummy indices: the simultaneous swaps in both epsilon symbols cancel their signs. Thuswhere $C_{il}$ is the cofactor of entry $A_{il}$. For an invertible matrix, the adjugate matrix identity gives $C_{il}=(\det A)(A^{-1}{)}_{li}$. Consequently the Jacobi determinant derivative formula isThe order of the factors inside the trace follows from the reversed cofactor indices; no commutativity of $A$ and $\dot{A}$ is assumed. The cofactor version is also Jacobi's formula and remains valid when $A$ is singular, although the inverse version then cannot be used.

A group homomorphism is a multiplication-preserving map $\varphi :G_1\to G_2$, so $\varphi (gh)=\varphi (g)\varphi (h)$. A group isomorphism is a bijective homomorphism; its inverse is automatically multiplication-preserving as well.

For the real general linear group, the identity matrix has nonzero determinant. If $A,B$ have nonzero determinants, multiplicativity of the determinant gives $\det (AB)=\det A\det B\ne 0$. Associativity is inherited from matrix multiplication, and every such matrix has an inverse, with $\det A^{-1}=1/\det A\ne 0$. Hence the given set $\mathrm{GL}(n,\mathbb{R})$ is a group.

For the special orthogonal group, the identity is included andwhen both factors are special orthogonal. Moreover $A^{-1}=A^T$ is again orthogonal and has determinant one. Thus $\mathrm{SO}(n)$ is a subgroup. The orientation-preserving general linear group is also a subgroup: the identity has positive determinant, products preserve positivity, and inversion replaces the determinant by its positive reciprocal.

Now take an element $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ of the special orthogonal group in two dimensions. Orthogonality and determinant one implyHence $d=a$, $b=-c$, and the unit length of the first column gives $a^2+c^2=1$. There is an angle $\theta$, unique modulo $2\pi$, such that $a=\cos \theta$, $c=\sin \theta$. ThereforeDefine the map from the circle group byIf $z=a+ic$ and $w=p+iq$, matrix multiplication gives $\varphi (z)\varphi (w)=\left(\begin{array}{cc}ap-cq& -(aq+cp)\\ aq+cp& ap-cq\end{array}\right)=\varphi (zw)$. Thus it is a homomorphism. The entries determine $a,c$ uniquely, proving injectivity, and the parametrization proves surjectivity. This establishesThe use of the unit complex number, rather than an unrestricted real angle, removes the $2\pi$ ambiguity: angles add modulo $2\pi$, exactly as unit complex numbers multiply.