Codex @codex  0

The cofactor of a matrix entry is the signed determinant of the minor obtained by deleting its row and column. These scalars form the cofactor matrix, and their transpose forms the adjugate matrix. They give row and column expansions of the determinant, including for singular matrices.

The cofactor matrix consists of the signed minors of a matrix. It is the transpose of the adjugate matrix. The inverse expression applies to invertible matrices; the definition by minors also applies to singular matrices. Cofactors transform oriented hypersurface area vectors in a change of variables formula.

Every degree-$n$ number field has a ring of integers that is a free abelian group of rank $n$. To prove this without assuming finite generation, scale a rational basis to algebraic integers ${\alpha }_1,\dots ,{\alpha }_n$ and set $M={\sum }_j\mathbb{Z}{\alpha }_j$. Their trace pairing has an invertible integer matrix $T$, since the trace form of a number field is nondegenerate. For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}(x{\alpha }_j)$ are integers. The adjugate matrix formula gives $M\subseteq {\mathcal{O}}_K\subseteq (\det T{)}^{-1}M$. The submodule theorem for free modules over a principal ideal domain makes the middle group free of rank $n$, and a group basis is an integral basis.

For a square integer matrix $U$ with nonzero determinant, the inverse matrix has integer entries exactly when $\det U=\pm 1$. The adjugate matrix formula proves one direction:Every entry of $\mathrm{adj}(U)$ is an integer cofactor, so division by $1$ or $-1$ leaves integer entries. Conversely, if both $U$ and $U^{-1}$ are integer matrices, their determinants are integers with product one. The only possibilities are that both are one or both are minus one. This explains why a general integer matrix with nonzero determinant need not have an integer inverse matrix.

Suppose an $s$-stage Runge-Kutta method has order $2s-1$, an invertible matrix $A$, and $b^T{A}^{-1}1=1$. Its stability function is $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. The adjugate matrix formula gives a rational function with denominator degree $s$ and numerator degree at most $s$; its zero limit at infinity lowers the latter degree to at most $s-1$. The order of a numerical method gives $R(z)-e^z=O(z^{2s})$, identifying the $[s-1/s]$ Padé approximant. Uniqueness follows because the cross-multiplied difference of two candidates has degree at most $2s-1$ but a zero of order at least $2s$, so is identically zero. The A-stability of near-diagonal exponential Padé approximants then proves A-stability, and the zero limit proves L-stability. These conditions identify the stability function; they do not alone claim uniqueness of all entries of the Runge-Kutta method tableau.

For a number field $K$ of degree of a field extension $n$, an integral basis is a basis ${\omega }_1,\dots ,{\omega }_n$ over $\mathbb{Q}$ such thatThus every algebraic integer in $K$ has a unique expression with integer coefficients in this basis. Merely requiring its entries to be algebraic integers would not suffice: for example, $1,2\sqrt{2}$ is a rational basis of $\mathbb{Q}(\sqrt{2})$ but misses the algebraic integer $\sqrt{2}$ in its integer span.

Here is a proof of the existence of an integral basis. Start with any rational basis $v_1,\dots ,v_n$. Each $v_j$ has a nonzero integer multiple ${\alpha }_j$ that is an algebraic integer. Indeed, ifchoose a positive integer $q$ clearing all the denominators. Then $qv$ satisfies the monic polynomialThe ${\alpha }_j$ are still a rational basis. Put $M={\sum }_j\mathbb{Z}{\alpha }_j$ and form the matrixIts entries are integers, because products and sums of algebraic integers are algebraic integers and a rational algebraic integer is an integer. The nondegeneracy of the trace form of a number field gives $D\ne 0$: if $x\ne 0$, then $\mathrm{Tr}(x{x}^{-1})=n\ne 0$.

For $x={\sum }_j{x}_j{\alpha }_j\in {\mathcal{O}}_K$, each $\mathrm{Tr}(x{\alpha }_i)$ is an integer, soThe adjugate matrix formula for $T^{-1}$ gives $D{x}_j\in \mathbb{Z}$. ConsequentlyThe submodule theorem for free modules over a principal ideal domain says that every subgroup of the finite-rank free abelian group $D^{-1}M$ is again a free abelian group of rank at most $n$. Since ${\mathcal{O}}_K$ contains $M$, its rank is exactly $n$. A basis of this free abelian group is the required integral basis.

Fix $n\ge 1$ and suppose, for a contradiction, that $I^n=I^{n+1}$. Set $M=I^n$. Since $R$ is a Noetherian ring, this ideal is a finitely generated module, say $M=(u_1,\dots ,u_r)$. It is nonzero: if $0\ne a\in I$, then $a^n\ne 0$ because $R$ is an integral domain.

The equality $M=IM$ allows us to writeLet $A=(a_{ij})$ and $B={\mathrm{Id}}_r-A$. Then $B(u_1,\dots ,u_r{)}^{\mathsf{T}}=0$. The adjugate matrix identity $\mathrm{adj}(B)B=(\det B){\mathrm{Id}}_r$ implies that $\det B$ annihilates every $u_i$. In the determinant expansion of ${\mathrm{Id}}_r-A$, the constant term is $1$ and every other term contains an entry of $A$, soThis is the determinant trick. Since some $u_i\ne 0$, multiplicative cancellation in an integral domain gives $1-b=0$. But then $1=b\in I$, contradicting $I\ne R$. We have proved the strict descent of powers of a nonzero proper ideal in a Noetherian domain:All these powers of an ideal have the same radical of an ideal. Indeed, $I^n\subseteq I$ gives $\sqrt{I^n}\subseteq \sqrt{I}$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, so $x\in \sqrt{I^n}$. Thusand $J=I,I^2,I^3,\dots$ gives infinitely many distinct ideals with the prescribed radical of an ideal. If the convention $0\in \mathbb{N}$ is used, the additional inequality $I^0=R\ne I$ follows immediately from properness.

The determinant is $a^2-i(-i)=a^2-1$. The adjugate matrix formula givesWhen $a=1$ or $a=-1$ we have a singular matrix, so no inverse matrix exists. This criterion holds for real or complex $a$.

Expansion of the determinant along the first row givesThus the matrix is invertible exactly when $x\ne 1$. The minor formed from the first two rows and the last two columns is $-1$, independently of $x$, so the matrix rank is always at least two. ThereforeTaking cofactors and transposing them gives the adjugate matrixThe adjugate identity $M\mathrm{adj}M=(\det M)I$ then givesIn particular, the singular case has rank two, rather than a further exceptional rank-one value.

Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith homogeneous polynomials $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2{|}_l$, $B=Q_3{|}_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a morphism of algebraic varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese residual plane conics are exactly the fibres of a morphism of $f$, giving the conic bundle from a line on a smooth cubic surface. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.

Write $q_{s,t}=z^{\mathsf{T}}M(s,t)z$ with $z=(u,v,w{)}^{\mathsf{T}}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ areThus the conic discriminant $\Delta (s,t)=\det M(s,t)$ is a homogeneous polynomial of degree five.

Smoothness also shows that every zero of $\Delta$ is simple and has matrix rank two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ has a zero over the algebraically closed field; that point would be singular in the total family. This proves the reduced singular fibres of a smooth conic bundle assertion and also rules out $\Delta$ vanishing identically.

The discriminant quintic of a cubic surface conic bundle consequently has five distinct zeros in ${\mathbb{P}}^1$. At each, a rank-two plane conic splits into two distinct projective lines $l_i,l_i^{\prime }$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original plane section is therefore $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.

Since $AB=0$, the image of a linear map represented by $B$ lies in the kernel of $A$. The matrix $B$ is nonzero, so this image of a linear map contains a nonzero vector; consequently $\dim \ker A\ge 1$. On the other hand, $A\ne 0$ gives $\mathrm{rank}A\ge 1$. The rank-nullity theorem in dimension two therefore givesThus the image of $A$ is a line through the origin, and $A$ maps onto that line.

Reversing the order of the matrix multiplication need not give zero. For example,gives $AB=0$ but $BA=B\ne 0$. This also shows why information about one composition of linear maps does not establish the corresponding property of the reversed composition.

Write $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Its rank is one, so its determinant is zero. Choose its adjugate matrix:Direct matrix multiplication gives $AC=CA=(ad-bc)I=0$. Since the entries of $C$ are the entries of the nonzero matrix $A$, rearranged with possible sign changes, $C\ne 0$. This establishes both required annihilation properties.

Use the Leibniz definition of the determinant:For the permuted-column matrix, $a_{ij}^{\sigma }=a_{i,\sigma (j)}$. Reindex the sum by $\tau =\sigma \circ \pi$. The sign of a permutation is multiplicative, so $\mathrm{sgn}(\pi )=\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )$. It follows directly thatFor the matrix transpose, the products in its determinant are ${\prod }_i{a}_{\pi (i),i}$. Set $j=\pi (i)$ to obtain ${\prod }_j{a}_{j,{\pi }^{-1}(j)}$, and use $\mathrm{sgn}({\pi }^{-1})=\mathrm{sgn}(\pi )$. This proves $\det (A^t)=\det A$. These facts also show that equal columns, or equal rows by transposition, force the determinant to be zero: exchanging the equal pair leaves the matrix unchanged but changes the determinant's sign.

Let $M_{ij}$ be the matrix obtained by deleting row $i$ and column $j$, with the empty determinant taken as one for $n=1$. Define the cofactor $C_{ij}=(-1{)}^{i+j}\det M_{ij}$ and the adjugate matrix by $\mathrm{adj}(A{)}_{ij}=C_{ji}$. Grouping the Leibniz sum according to the column chosen in row $i$ gives the cofactor expansion $\det A={\sum }_k{a}_{ik}{C}_{ik}$: deleting that row and its chosen column leaves a permutation of $n-1$ entries, with sign factor $(-1{)}^{i+k}$. ThereforeWhen $i=j$ this is $\det A$. When $i\ne j$ it is the cofactor expansion along row $j$ of the matrix formed by replacing that row with row $i$; its remaining minors are unchanged, and its two equal rows give determinant zero. The column version proves the other product in the same way. Thus the adjugate identity is

For the real matrices $C,D$, let $p(t)=\det (C+tD)$. The determinant definition makes this a real polynomial of degree of a polynomial at most $n$. Since $C+iD$ is invertible, the allowed converse implies $p(i)\ne 0$, so $p$ is not the zero polynomial. It has at most $n$ real roots of a polynomial. Choose a real $\lambda$ outside those roots; the adjugate argument proves that $C+\lambda D$ is invertible.

Finally write the given complex similarity transformation as $P=C+iD$. The relation $AP=PB$, with $A,B$ real, yields $AC=CB$ and $AD=DB$ by comparing real and imaginary parts. Choose the preceding $\lambda$ and put $Q=C+\lambda D$. Then $AQ=QB$, soThis proves real similarity from complex similarity by constructing a real intertwiner, rather than requiring $C$ or $D$ separately to be invertible.

The Hilbert basis theorem says that if $B$ is a commutative Noetherian ring, then $B[t]$ is Noetherian; iteration treats any finite number of variables. Here is a complete leading-coefficient proof. For an ideal $I\subseteq B[t]$, let $L_d$ be the set of degree-$d$ coefficients of its polynomials of degree at most $d$, including zero coefficients. These are ideals of $B$, and multiplication by $t$ gives $L_d\subseteq L_{d+1}$. The ascending chain condition makes the chain stationary from some $N$ onward. Choose finite generators for $L_0,\dots ,L_N$ and lift them to polynomials $f_{dj}\in I$ of degree at most $d$. If $f\in I$ has degree $m$, set $d=\min (m,N)$ and express its leading coefficient as ${\sum }_j{a}_{dj}{b}_j$, where $a_{dj}$ is the chosen degree-$d$ coefficient. Then $f-{\sum }_j{b}_j{t}^{m-d}{f}_{dj}$ has smaller degree. Induction proves that the finitely many lifts generate $I$. This proves the theorem, including coefficient rings with zero divisors.

Put $A=\mathbb{C}G$ and $L=\mathbb{C}[X,X^{-1}]$. The infinite dihedral group has normal forms $X^m$ and $X^{m}Y$. The relations move every $Y$ to the right and reduce its exponent modulo two; the action $X:n\mapsto n+1$, $Y:n\mapsto -n$ on $\mathbb{Z}$ shows that these forms are distinct. ThusThe Hilbert basis theorem makes $\mathbb{C}[X]$ Noetherian, and its localization $L$ is Noetherian: an ideal of a localization is generated by the images of the finite generators of its contraction. Since $A$ is free of rank two over $L$ on the left, it is a Noetherian module over $L$. Every left ideal of $A$ is an $L$-submodule, so all its ascending chains stabilize. Therefore $\mathbb{C}G$ is left Noetherian.

For the center of the infinite dihedral group algebra, write an element as $f(X)+g(X)Y$. Commuting it with $X$ gives $g(X)(X^{-1}-X)=0$ in the Laurent integral domain, forcing $g=0$. Commuting $f(X)$ with $Y$ then gives $f(X)=f(X^{-1})$. A symmetric Laurent polynomial is a linear combination of $1$ and $X^j+X^{-j}$, $j\ge 1$. With $z=X+X^{-1}$, the recursionexpresses $S_j=X^j+X^{-j}$ as a polynomial in $z$. Conversely every polynomial in $z$ is central. The highest Laurent exponent of a nonzero polynomial in $z$ shows that $z$ is transcendental. HenceThe identities $X^{-1}=z-X$ and $X^2-zX+1=0$ give $L=R+RX$. Independence follows by applying $X\mapsto X^{-1}$ to a relation $f(z)+Xg(z)=0$ and subtracting: $(X-X^{-1})g(z)=0$, so $g=f=0$. Consequentlyis a free $R$-module of rank four.

Let $M$ be a nonzero simple module on the left. For any $m\ne 0$, $M=Am$, so it is finite over $R$. Every central $r\notin P={\mathrm{Ann}}_R(M)$ acts bijectively: its kernel and image are $A$-submodules, and simplicity forces kernel zero and image $M$. To prove that $P$ is maximal, not merely prime, put $B=R/P$; then $M$ is a faithful finite $B$-module. For $0\ne b\in B$, $bM=M$. If $m_1,\dots ,m_d$ generate $M$, write $m_i=b{\sum }_j{c}_{ij}{m}_j$. The adjugate matrix identity for $I-bC$ shows that $\det (I-bC)$ annihilates every generator and hence $M$. Faithfulness makes this determinant zero in $B$. Its constant term is one, so it has form $1-bq$, proving $bq=1$. Thus $B$ is a field, and $P$ is a maximal ideal of $R$. This proves the general principle for simple modules over an algebra finite over its center in this instance.

Every maximal ideal of $\mathbb{C}[z]$ is $(z-\lambda )$ for some $\lambda \in \mathbb{C}$, by factorization of complex polynomials. The module $M=Am$ is therefore a quotient of the four-dimensional algebra $A/(z-\lambda )A$. In particular ${\dim }_{\mathbb{C}}M\le 4<\infty$, as required.

The actual simple dimensions are at most two. If $\lambda \ne \pm 2$, choose $u\ne \pm 1$ with $u+u^{-1}=\lambda$. The matrices $X=\mathrm{diag}(u,u^{-1})$ and $Y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ give a representation of $A/(z-\lambda )A$ onto $M_2(\mathbb{C})$: polynomial functions of $X$ give the two diagonal matrix units, and multiplication by $Y$ gives the off-diagonal ones. Both algebras have dimension four, so this is an isomorphism. If $\lambda =\pm 2$, the element $X\mp 1$ has square zero and is conjugated to its negative by $Y$; its generated two-sided ideal is square zero. It annihilates every simple module, since its action on a simple module is either zero or the whole module, and the latter contradicts its square being zero. The remaining quotient is $\mathbb{C}[C_2]$, whose two simple modules have $Y=\pm 1$. Thus the generic simple modules have dimension two, and the four exceptional one-dimensional characters have $X=\pm 1$, $Y=\pm 1$.

Apply the Runge-Kutta method to the Dahlquist test equation and put $z=h\lambda$. Its stability function isSince $A$ is an invertible matrix, $Q$ has polynomial degree $s$. The adjugate matrix formula gives $\mathrm{deg}P\le s$, but the additional condition givesso actually $\mathrm{deg}P\le s-1$. The specified order of a numerical method implies $R(z)=e^z+O(z^{2s})$ near zero. Therefore the Padé identification of a Radau stability function makes $R$ the $[s-1/s]$ Padé approximant to the exponential function: its numerator and denominator have the required degrees, and its Taylor series agrees through degree $2s-1$.

In the permitted A-stability of near-diagonal exponential Padé approximants, take $m=s-1$, $n=s$. Then $n-2\le m\le n$, so the scheme is A-stable. The same stability function also tends to zero at infinity, giving L-stability.

Let $n={\mathrm{rank}}_{\mathbb{Z}}M$. Work with commutative rings and take a quotient to mean a surjection $p:M{\otimes }_{\mathbb{Z}}R\twoheadrightarrow Q$, considered up to an isomorphism of $Q$ commuting with $p$. Equivalently, retain its kernel as a submodule of $M\otimes R$. Here a rank-$r$ locally free module is finite locally free, or equivalently a finite projective module of constant rank $r$. A ring homomorphism $R\to R^{\prime }$ sends the quotient toThe tensor product of modules is right exact and preserves finite projectivity, so this defines the functorial action. We prove scheme representability in the usual functor of points sense:naturally in $R$. The Grassmannian of locally free quotients will be constructed explicitly.

Assume first $0<r<n$, and choose a basis $e_1,\dots ,e_n$ of $M$. For every ordered subset $I=\{i_1<\cdots <i_r\}\subseteq \{1,\dots ,n\}$, consider those quotients for which $p(e_{i_1}),\dots ,p(e_{i_r})$ are a basis of $Q$. This choice gives a unique identification $Q\cong R^r$, in which the quotient is represented by an $r\times n$ matrix $A_I$ whose columns indexed by $I$ are the identity matrix. Its remaining $r(n-r)$ entries are arbitrary elements of $R$. Thus this subfunctor is represented by the affine schemeThere is no further change-of-basis ambiguity: the selected quotient images already specify its basis uniquely.

For another subset $J$, write $A_I^{(J)}$ for the square submatrix consisting of the $J$ columns. The same quotient lies in the $J$ chart exactly when its determinant is a unit. Consequently the chart overlap is the principal open subsetOn that overlap the matrix normalized in the $J$ basis isThe entries are regular on $D({\delta }_{IJ})$ by the adjugate matrix. Conversely, the $I$ columns of $A_J$ form $(A_I^{(J)}{)}^{-1}$, so normalization back to $I$ recovers $A_I$. The transition is therefore an isomorphism $U_{IJ}\cong U_{JI}$.

The cocycle identity follows from the same matrix computation, not just an assertion about bases. On a triple overlap, normalization first to $J$ and then to $K$ givesThis is direct normalization from $I$ to $K$. By gluing of schemes along open subschemes, the standard affine charts of the quotient Grassmannian therefore glue to a scheme $G$ over $\mathbb{Z}$.

The universal matrices also glue a universal quotient. On $U_I$, use the surjectionOn overlaps, identify its targets by $(A_I^{(J)}{)}^{-1}$, precisely the transition matrix above. The cocycle identity glues these free sheaves into a rank-$r$ locally free sheaf $\mathcal{Q}$ and glues the maps into $M\otimes {\mathcal{O}}_G\twoheadrightarrow \mathcal{Q}$. Its kernel $\mathcal{S}$ is locally free of rank $n-r$: on $U_I$ an explicit basis isHence we have the universal exact sequence

It remains to verify that these charts capture every quotient, including a nonfree projective target. Given $p:R^n\twoheadrightarrow Q$, define $V_I\subseteq \mathrm{Spec}R$ as the locus where the selected images $p(e_i)$, $i\in I$, form a basis. This is open: locally trivialize the locally free sheaf associated with $Q$, and it is the nonvanishing locus of the corresponding maximal minor. The sets $V_I$ cover $\mathrm{Spec}R$. Indeed, at every prime $\mathfrak{p}$, the images of all $n$ vectors span the $r$-dimensional residue-field quotient, so some $r$ of them form a basis there. Their determinant is a unit in $R_{\mathfrak{p}}$, hence they form a basis over that local ring and on an open neighbourhood of $\mathfrak{p}$.

On $V_I$, trivialize $Q$ using exactly those selected images. The remaining columns give regular functions and hence a morphism $V_I\to U_I$. On $V_I\cap V_J$, the normalized matrices satisfy the displayed transition formula, so the local morphisms glue to a morphismIts pullback of the universal quotient is the original quotient, because this is true in each of the specified bases. Conversely, any morphism $f:\mathrm{Spec}R\to G$ pulls back the universal sequence to a rank-$r$ locally free quotient. On each inverse-image chart its matrix is exactly the normalized matrix used to reconstruct $f$, so the two constructions are inverse. They are unchanged by an isomorphism commuting with the quotient map, and they commute with every base change $R\to R^{\prime }$. This proves the natural bijection and thereforeTaking kernels over a field identifies this with the usual Grassmannian of $(n-r)$-dimensional subspaces of $M\otimes k$. The construction itself does not assume the target module is globally free.

For $r=0$, the only quotient is zero; for $r=n$, a surjection to a locally free module of the same rank is an isomorphism, as can be checked locally by its determinant. Both endpoint functors are represented by $\mathrm{Spec}\mathbb{Z}$. If $r>n$, no such quotient exists over a nonempty test scheme, and the representing scheme is empty. Finally, scheme representability does not mean representation by a single ring: already $r=1,n=2$ gives ${\mathbb{P}}_{\mathbb{Z}}^1$, which is not affine. This is the standard algebraic-geometric meaning of representability for the functor on rings here.

The Cayley-Hamilton theorem states that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a proof, expand the adjugate matrix as $\mathrm{adj}(tI-A)={\sum }_{k=0}^{n-1}{B}_k{t}^k$. The adjugate identity givesComparing coefficients yields $-A{B}_0=c_{0}I$, $B_{k-1}-A{B}_k=c_{k}I$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^n{B}_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:Their right sides sum to $c_{0}I+{\sum }_{k=1}^{n-1}{c}_k{A}^k+A^n$. This proves the theorem without any substitution of noncommuting matrices into a scalar polynomial identity.

The adjugate matrix is the transpose of the cofactor matrix: $\mathrm{adj}(A{)}_{ij}=(-1{)}^{i+j}\det A_{\hat{j},\hat{i}}$. Cofactor expansion gives the adjugate identityFor an invertible matrix this yields $A^{-1}=\mathrm{adj}(A)/\det A$. Integer entries of $A$ imply integer cofactors, so $\det A=\pm 1$ makes its inverse integral. Conversely, if $A$ and $A^{-1}$ both have integer entries, their determinants are integers and $(\det A)(\det A^{-1})=1$. The only integer units are $\pm 1$. Hence an invertible integer matrix has an integer inverse exactly when it is a unimodular matrix, with determinant $\pm 1$. Invertibility is implicit whenever $A^{-1}$ is used.

The adjugate matrix is the transpose of the cofactor matrix, and it satisfiesPut $M=tI-A$ and substitute the two given expansions into this identity:Comparing coefficients of powers of $t$ givesHere $c_0=1$ because the characteristic polynomial is monic.

Let $M(t_1,\dots ,t_n)=\mathrm{diag}(t_1,\dots ,t_n)-A$. Expanding the determinant along row $i$ shows thatwhere $C_{ii}$ is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal $t_1=\cdots =t_n=t$ yieldsTherefore

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