The Centraliser and the Diagonal Peirce Part

Introduction

The eigenvalue decomposition of an element splits the algebra into two-sided Peirce pieces $e_i A e_j$. The pieces with $i = j$ form a subalgebra, the diagonal Peirce part, and the pieces with $i \neq j$ form the off-diagonal part. This article proves that the diagonal part is exactly the centraliser of the element, that is, the set of elements commuting with it. The identification is the structural reason the centraliser of a diagonalisable element is a direct sum of matrix algebras, and it is the first step in the Peirce analysis of noncentral idempotents.

The article assumes the eigenvalue decomposition of The Eigenvalue Decomposition of an Element, the Chinese remainder decomposition of The Algebra Generated by an Element and the Chinese Remainder Decomposition, and the Peirce decomposition of Unital Algebras, §Idempotents and the Peirce Decomposition. The off-diagonal part is treated separately in Off-Diagonal Peirce Components and Noncentral Idempotents; the commutator operator is The Commutator Operator; the double centraliser theorem for endomorphism rings is Algebras of Endomorphisms. The base is a field $k$ and $A$ is a finite-dimensional unital associative $k$-algebra. The element $a \in A$ is split separable with eigenvalue decomposition $a = \sum_{i=1}^{r} \lambda_i e_i$ unless stated otherwise.

The Centraliser

Definition

The centraliser of $a$ in $A$ is

$$ C(a) = \{ x \in A : ax = xa \}. $$

It is a unital subalgebra of $A$: it contains $1_A$, it is closed under addition and multiplication, and it is a linear subspace. It contains $k[a]$, because every polynomial in $a$ commutes with $a$. More generally, $C(a) = C(k[a])$, the centraliser of the subalgebra $k[a]$: an element commutes with $a$ exactly when it commutes with every polynomial in $a$, and conversely.

The centraliser is also the kernel of the commutator operator

$$ \operatorname{ad}_a : A \longrightarrow A, \qquad \operatorname{ad}_a(x) = ax - xa = (L_a - R_a)(x), $$

where $L_a$ and $R_a$ are the left and right multiplications. Thus $C(a) = \ker \operatorname{ad}_a$. The image of $\operatorname{ad}_a$ is the set of commutators $[a,x]$; the decomposition of $A$ into the kernel and the image is the subject of The Commutator Operator and, for a single idempotent, of Off-Diagonal Peirce Components and Noncentral Idempotents.

The theorem

Theorem. Let $a = \sum_{i=1}^{r} \lambda_i e_i$ be an eigenvalue decomposition with the $\lambda_i$ distinct. Then

$$ \boxed{\;C(a) = \bigoplus_{i=1}^{r} e_i A e_i\;} . $$

Proof. By the two-sided Peirce decomposition of The Eigenvalue Decomposition of an Element, every $x \in A$ has a unique expansion

$$ x = \sum_{j,k=1}^{r} e_j x e_k . $$

Compute

$$ ax = \sum_{j,k} \lambda_j e_j x e_k, \qquad xa = \sum_{j,k} \lambda_k e_j x e_k . $$

The condition $ax = xa$ is therefore

$$ \sum_{j,k} (\lambda_j - \lambda_k)e_j x e_k = 0 . $$

The summands $e_j x e_k$ are independent, because they lie in the direct summands of the two-sided Peirce decomposition, so

$$ (\lambda_j - \lambda_k)e_j x e_k = 0 $$

for every $j,k$. If $j \neq k$, the eigenvalues are distinct and $\lambda_j - \lambda_k \neq 0$, so $e_j x e_k = 0$. Hence $x = \sum_j e_j x e_j$ lies in the diagonal part.

Conversely, if $x = \sum_j e_j x e_j$, then

$$ ax = \sum_j \lambda_j e_j x e_j = xa, $$

so every diagonal element commutes with $a$. This proves the equality.

The theorem uses only the eigenvalue decomposition; the distinctness of the $\lambda_i$ is essential, because it makes the off-diagonal coefficients nonzero. The diagonal part is a subalgebra with identity $1_A = \sum_i e_i$, and each $e_i A e_i$ is a subalgebra with identity $e_i$; the direct sum is a direct sum of algebras.

The dimension

From the theorem,

$$ \dim_k C(a) = \sum_{i=1}^{r} \dim_k e_i A e_i . $$

For $A = M_n(k) = \operatorname{End}_k(V)$, the projector $e_i$ has rank $m_i$, the algebraic multiplicity of $\lambda_i$, and $e_i A e_i \cong M_{m_i}(k)$. Hence

$$ \boxed{\;\dim_k C(a) = \sum_{i=1}^{r} m_i^2, \qquad \sum_{i=1}^{r} m_i = n\;} . $$

Since $\sum_i m_i^2 \ge \sum_i m_i = n$, with equality exactly when every $m_i$ is one, the centraliser has dimension at least $n$. The lower bound is attained exactly when all the eigenvalues are simple; then each $e_i$ has rank one, each corner is one-dimensional, and $C(a)$ has dimension $n$. The maximum is $n^2$, attained when $r=1$ and $m_1=n$, that is, when $a$ is a scalar matrix.

For a general algebra there is no characteristic polynomial and no rank, but the formula $\dim C(a) = \sum_i \dim e_i A e_i$ remains valid.

The double centraliser in a matrix algebra

For $A = M_n(k)$, the centraliser has a second centraliser,

$$ C(C(a)) = \{ x \in A : xy = yx \text{ for all } y \in C(a) \}, $$

and it recovers the algebra generated by $a$.

Theorem. For $A = M_n(k)$ and any split separable $a \in A$,

$$ \boxed{\;C(C(a)) = k[a]\;}. $$

Proof. By the theorem, $C(a) = \bigoplus_i M_{m_i}(k)$ in the decomposition $V = \bigoplus_i V_i$ of the underlying vector space into the eigenspaces. The matrices commuting with every block matrix $\bigoplus_i M_{m_i}(k)$ are exactly the block scalar matrices $\bigoplus_i \alpha_i I_{m_i}$, that is, the linear span of the $e_i$. The span of the $e_i$ is $k[a]$: the $e_i$ are polynomials in $a$ by the Lagrange formula of The Lagrange–Sylvester Eigenprojector Formula, and $a = \sum_i \lambda_i e_i$ is a linear combination of the $e_i$. Hence $C(C(a)) = \operatorname{span}\{e_i\} = k[a]$.

The theorem is the matrix case of the double centraliser theorem; the general form for endomorphism rings is in Algebras of Endomorphisms.

For a general algebra the theorem can fail. Let $A = k \times k \times k$ and $a = (1,0,0)$. Then $A$ is commutative, so $C(a) = A$ and $C(C(a)) = A$, while $k[a] = \{(\alpha,\beta,\beta) : \alpha,\beta \in k\}$ is a proper subalgebra. The failure is possible because $A$ has extra factors that commute with $a$ without being generated by it. In a full matrix algebra there is no such extra factor: the bicommutant theorem for a single matrix says that the only matrices commuting with everything that commutes with $a$ are the polynomials in $a$.

The Diagonal Peirce Part

The decomposition into diagonal and off-diagonal parts

Keep the eigenvalue decomposition $a = \sum_i \lambda_i e_i$ and put

$$ D = \bigoplus_{i=1}^{r} e_i A e_i, \qquad N = \bigoplus_{i \neq j} e_i A e_j . $$

Then $A = D \oplus N$ as $k$-vector spaces. The subspace $D$ is the diagonal Peirce part and $N$ is the off-diagonal Peirce part. The theorem of §The theorem says $D = C(a)$.

The two parts have complementary algebraic behaviour.

$D$ is a unital subalgebra: $1_A \in D$, and $D \cdot D \subseteq D$ because $e_i A e_i \cdot e_j A e_j = 0$ for $i \neq j$ and $e_i A e_i \cdot e_i A e_i \subseteq e_i A e_i$.
$N$ is a $D$-bimodule: $D \cdot N \subseteq N$ and $N \cdot D \subseteq N$, because the Peirce multiplication rule is $e_i A e_j \cdot e_k A e_l \subseteq \delta_{jk} e_i A e_l$.
$N \cdot N$ is contained in $A$, and its diagonal part is contained in $D$: a product $e_i A e_j \cdot e_k A e_l$ is nonzero only when $j = k$, and then it lies in $e_i A e_l$, which is diagonal exactly when $i = l$. For $r = 2$ there are only two indices, so $i = l$ is forced and $N \cdot N \subseteq D$; for $r \ge 3$ the product can be off-diagonal, as in $e_1 A e_2 \cdot e_2 A e_3 = e_1 A e_3$.
$N$ is not a subalgebra in general, and it need not be an ideal.

The projection onto the diagonal part

Define

$$ \pi : A \longrightarrow D, \qquad \pi(x) = \sum_{i=1}^{r} e_i x e_i . $$

This is the projection onto $D$ along $N$. It is $k$-linear, unital ($\pi(1_A) = \sum_i e_i = 1_A$), and a $D$-bimodule map: for $d, d' \in D$, $\pi(dxd') = d\pi(x)d'$. It is the diagonal part map.

The map $\pi$ is not in general an algebra homomorphism. Write $x = \pi(x) + n_x$ and $y = \pi(y) + n_y$ with $n_x, n_y \in N$. Then

$$ xy = \pi(x)\pi(y) + \pi(x)n_y + n_x\pi(y) + n_x n_y . $$

The cross terms lie in $N$ (because $D \cdot N \subseteq N$ and $N \cdot D \subseteq N$), so applying $\pi$ kills them. The product $n_x n_y$ has its diagonal part in $D$ and its off-diagonal part in $N$, so $\pi(n_x n_y)$ is the diagonal part of $n_x n_y$. Therefore

$$ \pi(xy) = \pi(x)\pi(y) + \pi(n_x n_y). $$

Consequently $\pi$ is multiplicative exactly when $\pi(n_x n_y) = 0$ for all $x, y \in A$, that is, exactly when the diagonal part of $N^2$ vanishes.

In the case of a single idempotent, $r = 2$ and $N^2 \subseteq D$, so the condition is simply $N^2 = 0$. The upper triangular algebra $T_2(k)$ with $a = e = E_{11}$ is an example. The eigenvalue decomposition of $e$ is $e = 1\cdot e + 0\cdot(1-e)$; the diagonal part is the algebra of diagonal matrices, and the off-diagonal part is $N = kE_{12}$, which squares to zero because $E_{12}E_{12} = 0$. The diagonal projection is multiplicative, but $e$ is not central in $T_2(k)$: $E_{12}e = 0$ while $eE_{12} = E_{12}$. Thus $\pi$ can be an algebra homomorphism without $a$ being central; what $N^2 = 0$ says is that the off-diagonal part is square-zero, not that it vanishes.

For $r \ge 3$ the diagonal part of $N^2$ can be nonzero even when $N^2$ has off-diagonal components, and the condition is genuinely $\pi(N^2) = 0$. In $M_3(k)$ with $a = \operatorname{diag}(1,2,3)$ and $e_i = E_{ii}$, the off-diagonal part is the space of off-diagonal matrix units, and $N^2 = A$: for instance $E_{12}E_{23} = E_{13}$ is off-diagonal, while $E_{12}E_{21} = E_{11}$ is diagonal. The diagonal projection is not multiplicative, because $\pi(E_{12}E_{21}) = E_{11}$ while $\pi(E_{12})\pi(E_{21}) = 0$.

If $a$ is central in $A$, then $N = 0$, $D = A$, and $\pi$ is the identity, hence multiplicative. The converse fails: $N^2 = 0$ can hold with $N \neq 0$, as the $T_2(k)$ example shows.

The centraliser as an intersection of commutants

The eigenprojectors are polynomials in $a$, so

$$ C(a) = \bigcap_{i=1}^{r} C(e_i). $$

Indeed, $C(a) \subseteq C(e_i)$ because $e_i \in k[a]$, and conversely an element commuting with every $e_i$ commutes with $a = \sum_i \lambda_i e_i$.

For a single idempotent $e_i$, the Peirce decomposition by $e_i$ has diagonal part

$$ C(e_i) = e_i A e_i \oplus (1-e_i)A(1-e_i), $$

because an element commutes with $e_i$ exactly when it preserves the two complementary subspaces $e_i A$ and $(1-e_i)A$ (or, equivalently, when its off-diagonal pieces vanish). The intersection over $i$ of these diagonal parts is exactly $\bigoplus_i e_i A e_i$: a block $e_j A e_k$ with $j \neq k$ is off-diagonal for $e_j$, so it is excluded by $C(e_j)$; a diagonal block $e_j A e_j$ is diagonal for every $e_i$, because $e_i$ is orthogonal to $e_j$ when $i \neq j$. This gives a second proof of the theorem and exhibits $D$ as the simultaneous commutant of the eigenprojectors.

The case of one idempotent

The theorem of this article contains the case of a single idempotent as the case $r=2$ with eigenvalues $1$ and $0$. If $e$ is a nontrivial idempotent, its eigenvalue decomposition is $e = 1\cdot e + 0\cdot(1-e)$, and

$$ C(e) = eAe \oplus (1-e)A(1-e). $$

The off-diagonal part is

$$ N = eA(1-e) \oplus (1-e)Ae = [e,A], $$

the image of the commutator with $e$. The diagonal part is the centraliser, and the off-diagonal part is the set of elements that fail to commute with $e$. This is the starting point of Off-Diagonal Peirce Components and Noncentral Idempotents; the general case of this article is the simultaneous version for a complete family of idempotents.

Examples

A scalar element

If $a = \lambda 1_A$, then $r = 1$, $e_1 = 1_A$, $D = A$, $N = 0$, and $C(a) = A$. Every element commutes with a scalar, and the projection $\pi$ is the identity. The double centraliser is $C(C(a)) = Z(A)$, while $k[a] = k 1_A$. In $M_n(k)$ they agree, because $Z(M_n(k)) = k 1$, but in a commutative algebra with more than one factor they differ: for $A = k \times k$ and $a = \lambda(1,1)$, $C(C(a)) = A$ while $k[a] = k(1,1)$ is proper.

A matrix with distinct eigenvalues

Let $a \in M_n(k)$ have $n$ distinct eigenvalues. Then $a$ is diagonalisable and cyclic, each $e_i$ has rank one, and

$$ C(a) = k[a], \qquad \dim_k C(a) = n . $$

The companion matrix of $x^3-x$ is an example: its centraliser is $\mathbb{Q}[C]$, of dimension three, and the diagonal part is the span of the three idempotents $e_0, e_1, e_{-1}$. The off-diagonal part has dimension $9 - 3 = 6$, the sum of the six one-dimensional corners $e_i \mathbb{Q}[C] e_j$ with $i \neq j$.

A matrix with multiplicities

Let $a = \operatorname{diag}(1,1,2)$ in $M_3(k)$. The eigenvalues are $1$ and $2$, with $e_1 = \operatorname{diag}(1,1,0)$ of rank two and $e_2 = \operatorname{diag}(0,0,1)$ of rank one. The centraliser is

$$ C(a) = M_2(k) \oplus k, $$

of dimension $4 + 1 = 5$. The diagonal part is this algebra; the off-diagonal part is

$$ N = e_1 A e_2 \oplus e_2 A e_1, $$

of dimension $2 \cdot 2 = 4$ (the two mixed blocks, each $2 \times 1$ or $1 \times 2$). The total is $5 + 4 = 9 = \dim M_3(k)$. Here $N^2 = D \neq 0$, so the diagonal projection is not multiplicative: the product of two mixed blocks can be a nonzero diagonal block.

The upper triangular algebra

Let $A = T_2(k)$ be the algebra of upper triangular $2 \times 2$ matrices, and let $a = e = E_{11}$. The eigenvalues are $1$ and $0$, with $e_1 = e$ and $e_2 = 1-e = E_{22}$. The diagonal part is

$$ D = eAe \oplus (1-e)A(1-e) = kE_{11} \oplus kE_{22}, $$

the algebra of diagonal matrices, of dimension two. The off-diagonal part is $N = eA(1-e) = kE_{12}$, of dimension one, and $N^2 = 0$. The centraliser $C(e) = D$ consists of the diagonal matrices, and the diagonal projection is multiplicative. The element $e$ is not central, because $eE_{12} = E_{12}$ while $E_{12}e = 0$. The example shows that the diagonal projection can be multiplicative without the element being central; the condition is $N^2 = 0$, not $N = 0$.

A central element

If $a \in Z(A)$, then every eigenprojector $e_i = h_i(a)$ is central, so $N = 0$, $D = A$, and $C(a) = A$. The Chinese remainder decomposition of $k[a]$ is then a decomposition of $A$ itself:

$$ A \cong \prod_{i=1}^{r} A e_i, $$

with $A e_i$ the algebra of the $i$-th component. The split-complex algebra $\mathbb{D} = \mathbb{R}[j]/(j^2-1)$ with $a = j$ is an example: the idempotents $(1 \pm j)/2$ are central, and $\mathbb{D} \cong \mathbb{R} \times \mathbb{R}$. In a noncommutative algebra, a central split separable element is precisely one whose eigenvalue decomposition is a product decomposition of the whole algebra.

Summary

The centraliser of $a$ is the set $C(a) = \{x : ax = xa\}$, a unital subalgebra containing $k[a]$, equal to the kernel of the commutator operator $\operatorname{ad}_a = L_a - R_a$. If $a = \sum_i \lambda_i e_i$ is an eigenvalue decomposition with distinct eigenvalues, then

$$ C(a) = \bigoplus_i e_i A e_i, $$

the diagonal Peirce part. The proof expands $x$ in the two-sided Peirce decomposition and uses the distinctness of the eigenvalues to kill the off-diagonal coefficients. In $M_n(k)$, $C(a) = \bigoplus_i M_{m_i}(k)$ with $m_i$ the algebraic multiplicity, so $\dim C(a) = \sum_i m_i^2 \ge n$, with equality exactly when all eigenvalues are simple. For a split separable element of a full matrix algebra, the double centraliser recovers the algebra generated by $a$: $C(C(a)) = k[a]$; in a general algebra this can fail.

The diagonal part $D = \bigoplus_i e_i A e_i$ is a unital subalgebra, the off-diagonal part $N = \bigoplus_{i \neq j} e_i A e_j$ is a $D$-bimodule, and $A = D \oplus N$. The product $N^2$ has its diagonal part in $D$ and may have off-diagonal parts when $r \ge 3$; for $r = 2$ one has $N^2 \subseteq D$. The projection $\pi$ onto $D$ is unital and $D$-bimodule linear; it is multiplicative exactly when the diagonal part of $N^2$ vanishes, a condition that is weaker than $a$ being central. The centraliser is the intersection of the commutants of the eigenprojectors, $C(a) = \bigcap_i C(e_i)$, and for one idempotent the theorem reduces to $C(e) = eAe \oplus (1-e)A(1-e)$.

Summary of Notation

Symbol	Meaning
$C(a)$	The centraliser $\{x \in A : ax = xa\}$
$C(k[a])$	The centraliser of the subalgebra generated by $a$; equal to $C(a)$
$\operatorname{ad}_a$	The commutator operator $x \mapsto ax - xa$; $C(a) = \ker \operatorname{ad}_a$
$e_i$	The eigenprojector of $\lambda_i$
$D = \bigoplus_i e_i A e_i$	The diagonal Peirce part; equal to $C(a)$
$N = \bigoplus_{i \neq j} e_i A e_j$	The off-diagonal Peirce part
$\pi : A \to D$	The diagonal projection $x \mapsto \sum_i e_i x e_i$
$m_i = \operatorname{rank} e_i$	In $M_n(k)$, the algebraic multiplicity of $\lambda_i$
$C(C(a))$	The double centraliser; equal to $k[a]$ for a split separable $a$ when $A = M_n(k)$
$T_2(k)$	The algebra of upper triangular $2 \times 2$ matrices; the example with $N^2 = 0$ but $N \neq 0$