Matrix Algebras

Introduction

The algebra $M_n(R)$ of $n\times n$ matrices over a commutative ring $R$ is the model by which the general theory of associative algebras is tested. It is simple when $R$ is a field, its centre is the scalar matrices, its units are the matrices of invertible determinant, and it carries a basis — the matrix units — whose multiplication table is the clearest possible display of a non-commutative structure. This article develops those facts, together with the identification of $M_n(R)$ with the endomorphism algebra of the free module $R^n$ and the tensor product formula $M_m(k)\otimes_k M_n(k) \cong M_{mn}(k)$.

Throughout, $R$ is a commutative ring with identity $1 \neq 0$, $k$ is a field, and $n \geq 1$. The matrix algebra $M_n(R)$ is associative and unital with unit $I_n$, the identity matrix. For $n \geq 2$ it is non-commutative; for $n = 1$ it is the ring $R$ itself. The general theory of ideals and quotients used below is that of Ideals and Quotients of Algebras, and the centre is as in Centre, Units, Zero Divisors and Division Algebras.

Matrix Units

Definition. For $1 \leq i, j \leq n$ let $E_{ij}$ be the matrix whose only nonzero entry is a $1$ in position $(i,j)$. The matrix units $E_{ij}$ form an $R$-basis of $M_n(R)$, so that

$$ \dim_R M_n(R) = n^2, \qquad A = \sum_{i,j=1}^{n} A_{ij}E_{ij} . $$

Proposition (multiplication of matrix units). For all $i,j,k,l$,

$$ E_{ij}E_{kl} = \delta_{jk}E_{il}, $$

where $\delta_{jk}$ is $1$ if $j = k$ and $0$ otherwise.

Proof. The matrix $E_{ij}E_{kl}$ has entry in position $(p,q)$ equal to $\sum_r (E_{ij})_{pr}(E_{kl})_{rq} = \sum_r \delta_{ip}\delta_{jr}\delta_{kr}\delta_{lq} = \delta_{jr}\delta_{kr}\delta_{ip}\delta_{lq}$. The sum over $r$ collapses to $\delta_{jk}$, so the entry is $\delta_{jk}\delta_{ip}\delta_{lq}$, which is the $(p,q)$-entry of $\delta_{jk}E_{il}$.

Two consequences are used constantly. First, $E_{ij}E_{jl} = E_{il}$, so any matrix unit is the product of two others; second, $E_{ij}E_{kl} = 0$ whenever $j \neq k$, so there are many zero-divisor pairs as soon as $n \geq 2$, and $M_n(R)$ is not a domain. The matrix units satisfy $E_{ij} = E_{ik}E_{kj}$ for every $k$, which is the identity that makes the ideal theory of $M_n(R)$ trivial.

The Algebra as an Endomorphism Algebra

Let $R^n$ be the free $R$-module of column vectors of length $n$. Every matrix $A \in M_n(R)$ acts on $R^n$ by left multiplication, and the map

$$ M_n(R) \longrightarrow \operatorname{End}_R(R^n), \qquad A \longmapsto (x \mapsto Ax), $$

is an isomorphism of $R$-algebras. It is injective because $Ae_j$ is the $j$-th column of $A$, so a matrix acting as zero is zero, and it is surjective because an $R$-linear endomorphism is determined by the images of the standard basis, which are the columns of a matrix. Consequently

$$ M_n(R) \cong \operatorname{End}_R(R^n) . $$

For $n \geq 2$ the algebra is non-commutative, since $E_{12}E_{21} = E_{11}$ while $E_{21}E_{12} = E_{22}$, and these differ.

Trace and Determinant

Definition. The trace of $A \in M_n(R)$ is $\operatorname{Tr} A = \sum_{i=1}^n A_{ii} \in R$, and the determinant is

$$ \det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n A_{i,\sigma(i)} \in R . $$

Both are polynomial functions of the entries, so they are defined over any commutative ring.

Proposition (properties). For $A, B \in M_n(R)$:

$\operatorname{Tr}(A+B) = \operatorname{Tr}A + \operatorname{Tr}B$ and $\operatorname{Tr}(\lambda A) = \lambda\operatorname{Tr}A$;
$\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$, and more generally the trace is invariant under cyclic permutation of a product;
$\det(AB) = \det A \cdot \det B$ and $\det I_n = 1$;
the characteristic polynomial $\chi_A(t) = \det(tI_n - A)$ has constant term $(-1)^n\det A$ and coefficient of $t^{n-1}$ equal to $-\operatorname{Tr}A$.

Proof. The trace identities follow from the definitions, since $\operatorname{Tr}(AB) = \sum_{i,j}A_{ij}B_{ji} = \sum_{i,j}B_{ji}A_{ij} = \operatorname{Tr}(BA)$. The multiplicativity of the determinant is the classical theorem, proved by expanding both sides as sums over $S_n$ and reindexing; over a general commutative ring it follows from the universal polynomial identity over $\mathbb{Z}$, which is then evaluated in $R$. The two coefficients of $\chi_A$ are read off from the Leibniz formula.

Theorem (unit criterion). A matrix $A \in M_n(R)$ is a unit if and only if $\det A \in R^\times$.

Proof. If $AB = I_n$ then $\det A \cdot \det B = \det I_n = 1$, so $\det A$ is a unit. Conversely, if $\det A$ is a unit, let $\operatorname{adj}(A)$ be the adjugate matrix, whose entries are the cofactors; the Laplace expansion gives $A\operatorname{adj}(A) = \operatorname{adj}(A)A = (\det A)I_n$. Hence $A^{-1} = (\det A)^{-1}\operatorname{adj}(A)$ is a two-sided inverse.

Over a field the criterion reads: $A$ is a unit if and only if $\det A \neq 0$, which is the usual statement that the units are $\mathrm{GL}_n(k)$. Over a general commutative ring the determinant can be a nonzero non-unit: the matrix $\operatorname{diag}(2,1) \in M_2(\mathbb{Z})$ has determinant $2$, so it is not invertible, and it is not a zero divisor either, since $2$ is not a zero divisor in $\mathbb{Z}$.

Ideals and Simplicity

Theorem (simplicity). Let $k$ be a field. Then $M_n(k)$ has exactly two two-sided ideals, $0$ and $M_n(k)$; that is, $M_n(k)$ is simple.

Proof. Let $I \neq 0$ be a two-sided ideal and let $0 \neq A \in I$ with $A_{ij} \neq 0$. For any $p, q$,

$$ E_{pq} = \frac{1}{A_{ij}}\, E_{pi}\, A\, E_{jq} \in I, $$

since $E_{pi}A E_{jq}$ has $(p,q)$-entry $A_{ij}$ and all other entries zero. As the matrix units span, $I = M_n(k)$.

Theorem (left ideals). Let $k$ be a field. The left ideals of $M_n(k)$ are exactly the sets

$$ I_V = \{ A \in M_n(k) : \operatorname{col}(A) \subseteq V \}, $$

where $V \subseteq k^n$ is a subspace and $\operatorname{col}(A)$ is the span of the columns of $A$.

Proof. Each $I_V$ is a left ideal, because left multiplication replaces the columns by linear combinations of them. Conversely, if $I$ is a left ideal, let $V$ be the span of the columns of all matrices in $I$; then $I \subseteq I_V$. For the reverse inclusion, choose $A \in I$ whose columns span $V$. Every vector of $V$ is $Ay$ for some $y$, so every matrix $B$ with columns in $V$ has the form $B = AC$ for a suitable $C$; since $I$ is a left ideal, $AC \in I$ and hence $B \in I$.

The left ideals are therefore in bijection with the subspaces of $k^n$, the two trivial ideals corresponding to $V = 0$ and $V = k^n$ and the others to the proper subspaces; over $\mathbb{R}$ and $n = 2$ the minimal ones form the projective line. The minimal left ideals are the $I_V$ with $\dim V = 1$, each isomorphic to $k^n$ as a left module. The right ideals are described by the rows in the same way, and the two-sided ideals are the two trivial ones, in agreement with simplicity.

Corollary (the centre). For a commutative ring $R$, $Z(M_n(R)) = R I_n$, and $M_n(R)$ is a central $R$-algebra.

Proof. If $A$ commutes with every matrix unit, then $AE_{ij} = E_{ij}A$ for all $i,j$. Comparing the $(k,l)$-entries gives $A_{ki}\delta_{jl} = \delta_{ik}A_{jl}$. Taking $j = l$ yields $A_{ki} = \delta_{ik}A_{jj}$ for all $i,k,j$: with $i = k$ this gives $A_{ii} = A_{jj}$, so all diagonal entries are equal, and with $i \neq k$ it gives $A_{ki} = 0$, so all off-diagonal entries vanish. Hence $A = \lambda I_n$ with $\lambda \in R$.

Tensor Products of Matrix Algebras

Theorem (tensor product formula). For a field $k$ and positive integers $m, n$,

$$ M_m(k) \otimes_k M_n(k) \cong M_{mn}(k). $$

Proof. Define

$$ \Phi\left(E_{ij} \otimes E_{kl}\right) = E_{(i,k),(j,l)}, $$

where the right-hand index $(i,k)$ is read as the integer $(i-1)n + k$ and correspondingly for $(j,l)$; extend $k$-bilinearly. The matrix units $E_{ij}\otimes E_{kl}$ form a basis of the tensor product, and the target matrix units $E_{(i,k),(j,l)}$ form a basis of $M_{mn}(k)$, so $\Phi$ is a linear isomorphism. It is multiplicative because

$$ \Phi\bigl((E_{ij}\otimes E_{kl})(E_{i'j'}\otimes E_{k'l'})\bigr) = \delta_{ji'}\delta_{lk'}\, \Phi\bigl(E_{ij'}\otimes E_{kl'}\bigr) = \delta_{ji'}\delta_{lk'}\, E_{(i,k),(j',l')}, $$

which is exactly the product $E_{(i,k),(j,l)}E_{(i',k'),(j',l')}$; and it sends $I_m\otimes I_n$ to $I_{mn}$.

The formula is the algebraic statement that the tensor product of two matrix algebras is again a matrix algebra, of size the product of the sizes; the two factors embed as the subalgebras of matrices acting on the two tensor legs, and the two images commute. Iterating, $M_{n_1}(k)\otimes\dots\otimes M_{n_r}(k) \cong M_{n_1\cdots n_r}(k)$. The construction is the matrix case of the tensor product of algebras in Tensor Products of Algebras.

The Regular Module

$M_n(k)$ is a left module over itself in the evident way, the regular module. The standard basis of $k^n$ as a column vector gives

$$ M_n(k) = I_{V_1} \oplus \dots \oplus I_{V_n}, $$

where $V_j = ke_j$ is the $j$-th coordinate line and $I_{V_j}$ is the minimal left ideal of matrices supported in the $j$-th column. Each $I_{V_j}$ is isomorphic to the simple module $k^n$ (column vectors), so the regular module is a direct sum of $n$ copies of the unique simple left module; this is the concrete form of the Wedderburn structure theory: $M_n(k)$ is the matrix algebra over the division ring $k$, and its regular module is $n$ copies of the unique simple module $k^n$. The same decomposition with $k$ replaced by a division ring $D$ describes $M_n(D)$, the general simple finite-dimensional algebra, whose simple module is $D^n$.

The Case $n = 2$

For $n = 2$ the algebra is four-dimensional, with basis $E_{11}, E_{12}, E_{21}, E_{22}$ and multiplication

$$ E_{11}^2 = E_{11}, \quad E_{22}^2 = E_{22}, \quad E_{12}^2 = E_{21}^2 = 0, \quad E_{11}E_{22} = E_{22}E_{11} = 0, $$

$$ E_{12}E_{21} = E_{11}, \quad E_{21}E_{12} = E_{22}, \quad E_{11}E_{12} = E_{12}, \quad E_{12}E_{22} = E_{12}, \quad E_{22}E_{21} = E_{21}, \quad E_{21}E_{11} = E_{21}. $$

The matrix units $E_{12}$ and $E_{21}$ are nilpotent and are zero divisors; the identity $E_{11}E_{22} = 0$ exhibits another zero-divisor pair. The centre is $kI_2$, of dimension $1$, so the algebra is central and simple of dimension $4$. Its units are the matrices with nonzero determinant, and

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} . $$

As a left module over itself $M_2(k)$ decomposes as the sum of the two minimal left ideals $I_{ke_1} = kE_{11} \oplus kE_{21}$ and $I_{ke_2} = kE_{12} \oplus kE_{22}$, each a copy of the simple module $k^2$.

The idempotents $E_{11}$ and $E_{22}$ are orthogonal and complete, and the Peirce decomposition of $M_2(k)$ with respect to $E_{11}$ is the matrix-unit decomposition

$$ M_2(k) = kE_{11} \oplus kE_{12} \oplus kE_{21} \oplus kE_{22}, $$

the four one-dimensional Peirce spaces being the matrix units themselves. The off-diagonal spaces are $E_{11}M_2E_{22} = kE_{12}$ and $E_{22}M_2E_{11} = kE_{21}$. The idempotent $E_{11}$ is not central, since $E_{11}E_{12} = E_{12}$ while $E_{12}E_{11} = 0$, so the two off-diagonal spaces do not vanish and the decomposition is not a product of algebras; consistently with the simplicity of $M_2(k)$, none of the four spaces is a two-sided ideal. The general theorem, with the product decomposition that a central idempotent gives instead, is Unital Algebras, §Idempotents and the Peirce Decomposition. The column decomposition above groups $E_{11}$ with $E_{21}$ and the Peirce decomposition groups $E_{11}$ with $E_{12}$: the two groupings are the left-ideal and the corner decompositions of the same four matrix units.

The Trace Form and Skolem–Noether

Proposition (the trace form). The bilinear form on $M_n(k)$ defined by

$$ \langle A, B\rangle = \operatorname{Tr}(AB) $$

is symmetric, associative in the sense that $\langle AB, C\rangle = \langle A, BC\rangle$, and non-degenerate.

Proof. Symmetry is $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$. Associativity is $\operatorname{Tr}((AB)C) = \operatorname{Tr}(A(BC))$. For non-degeneracy, suppose $\langle A, B\rangle = 0$ for all $B$; taking $B = E_{ji}$ gives $\operatorname{Tr}(AE_{ji}) = A_{ij} = 0$ for all $i,j$, hence $A = 0$.

The trace form is the concrete case of the non-degenerate associative form that makes $M_n(k)$ a symmetric Frobenius algebra, and it can be used in place of the matrix units in many arguments, for instance to identify the dual of $M_n(k)$ with itself.

Theorem (Skolem–Noether, standard). Every $k$-algebra automorphism $\varphi$ of $M_n(k)$ is inner: there is $P \in \mathrm{GL}_n(k)$ with

$$ \varphi(A) = PAP^{-1}, \qquad A \in M_n(k), $$

and the assignment $P \mapsto (A \mapsto PAP^{-1})$ induces an isomorphism

$$ \operatorname{Aut}_k(M_n(k)) \cong \mathrm{PGL}_n(k) = \mathrm{GL}_n(k)/k^\times. $$

Proof (sketch). The algebra $M_n(k)$ has a unique simple left module $k^n$ up to isomorphism, and $\varphi$ transports the module structure to an isomorphic one; an isomorphism of the transported module with $k^n$ is a matrix $P$, and comparing the actions gives $\varphi(A) = PAP^{-1}$. The kernel of $P \mapsto (A\mapsto PAP^{-1})$ is $k^\times I_n$.

Corollary (derivations). Every $k$-derivation of $M_n(k)$ is inner, $\delta(A) = [H, A]$ for some $H \in M_n(k)$.

This is the finite-dimensional case of the innerness theorem for von Neumann algebr, and it is the worked case promised in Automorphisms and Derivations of Algebras: for a matrix algebra the automorphism group is $\mathrm{PGL}_n(k)$, so the outer automorphism group $\operatorname{Out}_k(M_n(k)) = \operatorname{Aut}_k(M_n(k))/\operatorname{Inn}_k(M_n(k))$ is trivial, and the space of outer derivations is zero.

Summary

The matrix algebra $M_n(R)$ over a commutative ring $R$ has $R$-basis the matrix units $E_{ij}$, $1 \leq i,j \leq n$, with multiplication $E_{ij}E_{kl} = \delta_{jk}E_{il}$ and dimension $n^2$. It is isomorphic to the endomorphism algebra $\operatorname{End}_R(R^n)$, it is associative with unit $I_n$, and it is non-commutative for $n \geq 2$. Trace is $R$-linear and satisfies $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$, and determinant is multiplicative; a matrix is a unit if and only if its determinant lies in $R^\times$, with $A^{-1} = (\det A)^{-1}\operatorname{adj}(A)$. Over a field, $M_n(k)$ is simple: its only two-sided ideals are $0$ and itself, proved by writing each matrix unit as $E_{pi}AE_{jq}$ scaled; its left ideals correspond to subspaces of $k^n$, and its centre is $kI_n$. The tensor product formula $M_m(k)\otimes_k M_n(k)\cong M_{mn}(k)$ holds, and the regular module is a direct sum of $n$ copies of the simple module $k^n$. The trace form $\langle A,B\rangle = \operatorname{Tr}(AB)$ is symmetric, associative and non-degenerate, and by the Skolem–Noether theorem every automorphism of $M_n(k)$ is inner, $\varphi(A) = PAP^{-1}$, so that $\operatorname{Aut}_k(M_n(k)) \cong \mathrm{PGL}_n(k)$ and every derivation is inner.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with identity $1 \neq 0$
$k$	Field
$M_n(R)$	Algebra of $n\times n$ matrices over $R$
$I_n$	Identity matrix
$E_{ij}$	Matrix unit, $E_{ij}E_{kl} = \delta_{jk}E_{il}$
$A_{ij}$	Entries of $A$
$\operatorname{Tr}A$	Trace
$\det A$	Determinant
$\operatorname{adj}(A)$	Adjugate, $A\operatorname{adj}(A) = (\det A)I_n$
$\mathrm{GL}_n(k)$	Units of $M_n(k)$, $= \{A : \det A \neq 0\}$
$\operatorname{End}_R(R^n)$	Endomorphism algebra, $\cong M_n(R)$
$I_V$	Left ideal of matrices with columns in $V$
$M_n(D)$	Matrix algebra over a division ring $D$
$\langle A,B\rangle = \operatorname{Tr}(AB)$	Trace form
$\operatorname{Aut}_k(M_n(k)) \cong \mathrm{PGL}_n(k)$	Automorphisms, all of them inner; $\operatorname{Out}_k(M_n(k)) = 1$
$\otimes_k$	Tensor product of algebras