The Eigenvalue Decomposition of an Element

Introduction

This article treats the writing of an element of an algebra as a sum of scalar multiples of orthogonal idempotents, $a = \sum_i \lambda_i e_i$. It is the algebraic form of diagonalisation: the $\lambda_i$ are the eigenvalues, the $e_i$ are the spectral projections, and the identity $\sum_i e_i = 1$ says that the projections resolve the identity. The writing exists exactly when the minimal polynomial is split and separable, and then it is unique.

The article assumes the algebra generated by an element and the Chinese remainder decomposition of The Algebra Generated by an Element and the Chinese Remainder Decomposition, the notions of separability and splitting of Separable Elements and Split Minimal Polynomials, and the Peirce decomposition of Unital Algebras, §Idempotents and the Peirce Decomposition. The explicit formula for the idempotents is The Lagrange–Sylvester Eigenprojector Formula; the centraliser and the off-diagonal part are The Centraliser and the Diagonal Peirce Part and Off-Diagonal Peirce Components and Noncentral Idempotents; the version with a nilpotent part is Nilpotent Elements and the Jordan–Chevalley Decomposition. The matrix algorithm and the characteristic polynomial are Eigenvalues and Diagonalisation; this article gives the structural statement behind them.

The base is a field $k$ and $A$ is a finite-dimensional unital associative $k$-algebra throughout.

The Eigenvalues of an Element

The definition

Let $a \in A$. A scalar $\lambda \in k$ is an eigenvalue of $a$ if the element $a - \lambda 1_A$ is not invertible in $A$. The set of eigenvalues is the spectrum of $a$, written $\sigma(a)$.

In a finite-dimensional algebra, non-invertibility and being a zero divisor agree: an element is invertible if and only if its left multiplication is invertible, and a linear map on a finite-dimensional space is invertible if and only if it has trivial kernel. The definition is therefore equivalent to the existence of a nonzero $x \in A$ with $(a-\lambda)x = 0$, or of a nonzero $y$ with $y(a-\lambda) = 0$. These are the left and right eigenvalue equations; in finite dimension the two notions coincide with the spectrum.

The eigenvalue criterion

Theorem. Let $m_a$ be the minimal polynomial of $a$. Then

$$ \boxed{\;\lambda \in \sigma(a) \quad \Longleftrightarrow \quad m_a(\lambda) = 0\;} . $$

Proof. Suppose first that $m_a(\lambda) = 0$. Write $m_a(x) = (x-\lambda)g(x)$ with $\deg g = \deg m_a - 1$. Evaluating at $a$ gives $(a-\lambda)g(a) = 0$. The element $g(a)$ is nonzero, because $\deg g < \deg m_a$ and $m_a$ is the minimal polynomial. If $a-\lambda$ were invertible, multiplication by its inverse would give $g(a) = 0$, a contradiction. Hence $a-\lambda$ is not invertible and $\lambda$ is an eigenvalue.

Conversely, suppose that $a-\lambda$ is not invertible and that $m_a(\lambda) \neq 0$. Then $x-\lambda$ and $m_a(x)$ are coprime in $k[x]$, so there exist $u, v \in k[x]$ with

$$ u(x)(x-\lambda) + v(x)m_a(x) = 1 . $$

Evaluating at $a$ gives $u(a)(a-\lambda) = 1$, so $a-\lambda$ is invertible, a contradiction. Hence $m_a(\lambda) = 0$.

The spectrum is therefore the set of roots of $m_a$ in $k$. It is finite, of cardinality at most $\deg m_a$, and it may be empty when $k$ is not algebraically closed: the rotation matrix of The Algebra Generated by an Element and the Chinese Remainder Decomposition, §Examples, has no eigenvalue in $\mathbb{R}$ and two eigenvalues in $\mathbb{C}$.

Multiplicities

Let $\lambda$ be a root of $m_a$ and let $e_\lambda$ be its multiplicity as a root, that is, the largest $e$ with $(x-\lambda)^e \mid m_a$. This number is the nilpotent index of the block: the restriction of $a-\lambda$ to the generalized eigenspace is nilpotent of index $e_\lambda$, and $e_\lambda$ is the size of the largest Jordan block for $\lambda$ when $A$ is a matrix algebra. It is not the algebraic multiplicity of $\lambda$ as an eigenvalue of a matrix, which is the dimension of the generalized eigenspace and is the rank of the spectral projection.

For $A = M_n(k)$ the two multiplicities are related but different. If $a$ has the eigenvalue decomposition of this article, all the multiplicities $e_\lambda$ are one, while the algebraic multiplicities may be greater than one: the scalar matrix $\lambda I$ has minimal polynomial $x-\lambda$ and algebraic multiplicity $n$. The multiplicity in the minimal polynomial measures the thickness of the nilpotent part; the algebraic multiplicity measures the size of the eigenspace. Even in the semisimple case the two need not agree: $\operatorname{diag}(1,1,2)$ has minimal polynomial $(x-1)(x-2)$, so both root multiplicities are one, while the algebraic multiplicities are two and one.

The Eigenvalue Decomposition

The definition

An eigenvalue decomposition of $a$ is a writing

$$ a = \sum_{i=1}^{r} \lambda_i e_i $$

with $\lambda_1, \dots, \lambda_r \in k$ pairwise distinct, $e_1, \dots, e_r \in A$ nonzero idempotents satisfying

$$ e_i e_j = \delta_{ij} e_i, \qquad \sum_{i=1}^{r} e_i = 1_A . $$

The $e_i$ are the eigenprojectors, the spectral projections or the Peirce idempotents of $a$; the $\lambda_i$ are the eigenvalues. The condition $\sum_i e_i = 1_A$ is the resolution of the identity. The number $r$ is the number of distinct eigenvalues, and it is at most $\deg m_a$.

Immediate consequences

The definition has several immediate consequences.

• The element acts by $\lambda_i$ on $e_i$. Multiplying the decomposition by $e_i$ on the left or on the right gives

$$ a e_i = e_i a = \lambda_i e_i . $$

Thus $e_i$ is an eigenvector of the left multiplication $L_a$ and of the right multiplication $R_a$ with eigenvalue $\lambda_i$. The left and the right eigenvalues coincide here because $\lambda_i$ is central in $A$. - The idempotents are polynomials in $a$. The Lagrange formula of The Lagrange–Sylvester Eigenprojector Formula gives

$$ \boxed{\;e_i = \prod_{j \neq i} \frac{a - \lambda_j}{\lambda_i - \lambda_j}\;}, $$

so $e_i \in k[a]$. In particular the $e_i$ commute with $a$ and with one another, and the decomposition is internal to the commutative algebra $k[a]$. - The decomposition is the Chinese remainder decomposition. The polynomial $m_a$ is $\prod_i (x-\lambda_i)$, the factors are pairwise coprime, and the $e_i$ are the canonical idempotents of The Algebra Generated by an Element and the Chinese Remainder Decomposition. The external form is $k[a] \cong k^r$. - The eigenvalues are exactly the $\lambda_i$. The spectrum of $a$ is $\{\lambda_1, \dots, \lambda_r\}$, because $m_a = \prod_i (x-\lambda_i)$ and the roots are the $\lambda_i$. No other scalar can be an eigenvalue.

Examples of decompositions

• The scalar. If $a = \lambda 1_A$, then $m_a = x-\lambda$, $r=1$, $e_1 = 1_A$, and the decomposition is $a = \lambda 1_A$. The scalar is the degenerate case in which there is only one eigenvalue and only one projection.
• The diagonal matrix. If $a = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ with the $\lambda_i$ not necessarily distinct, the eigenvalues are the distinct diagonal entries, and the eigenprojector of $\lambda$ is the diagonal matrix with ones exactly at the positions where the diagonal entry is $\lambda$. If the entries are distinct, the projectors are the matrix units $E_{ii}$.
• The two-by-two matrix. For $a = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$, the eigenvalues are $1$ and $2$, and

$$ e_1 = \frac{a - 2I}{1 - 2} = 2I - a = \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}, \qquad e_2 = \frac{a - I}{2 - 1} = a - I = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}. $$

One checks $e_1^2 = e_1$, $e_2^2 = e_2$, $e_1 e_2 = 0$ and $e_1 + e_2 = I$, and $a = 1\cdot e_1 + 2\cdot e_2$. - The projection. An idempotent $e$ has minimal polynomial $x(x-1)$ when $e \neq 0,1$, so its eigenvalue decomposition is $e = 1\cdot e + 0\cdot (1-e)$. The two projections are $e$ and $1-e$; this is the smallest nontrivial eigenvalue decomposition, and it is the one-element case of the Peirce decomposition.

Existence and Uniqueness

The theorem

Theorem. Let $A$ be a finite-dimensional unital associative $k$-algebra and let $a \in A$. The following are equivalent:

1. $a$ has an eigenvalue decomposition;
2. $m_a$ is split with distinct roots;
3. $m_a(x) = \prod_{i=1}^{r}(x-\lambda_i)$ with $\lambda_i \in k$ pairwise distinct;
4. the left multiplication $L_a$ is diagonalisable;
5. the right multiplication $R_a$ is diagonalisable;
6. $a$ is split and separable in the sense of Separable Elements and Split Minimal Polynomials.

When these conditions hold, $r = \deg m_a$, the eigenvalues are the roots of $m_a$, and the eigenprojectors are unique.

The proof

The equivalence of 2 and 3 is the definition of a split polynomial with distinct roots, and the equivalence of 3 and 6 is the definition of split and separable. It remains to relate 1, 2, 4 and 5.

From 2 to 1. If $m_a = \prod_i (x-\lambda_i)$ with the $\lambda_i$ distinct, the factors are pairwise coprime. The Chinese remainder theorem of The Algebra Generated by an Element and the Chinese Remainder Decomposition gives idempotents $e_i \in k[a]$ with $e_i \equiv 1 \pmod{x-\lambda_i}$ and $e_i \equiv 0 \pmod{x-\lambda_j}$ for $j \neq i$. The image of $a$ in the $i$-th factor is $\lambda_i$, so the image of $a$ in the product is $\sum_i \lambda_i e_i$; since the map $k[a] \to \prod_i k[a]e_i$ is injective, $a = \sum_i \lambda_i e_i$.

From 1 to 2. Suppose $a = \sum_i \lambda_i e_i$. On the one hand, each $\lambda_i$ is an eigenvalue: $a e_i = \lambda_i e_i$ with $e_i \neq 0$, so $a - \lambda_i$ is a zero divisor and is not invertible, and the criterion of §The eigenvalue criterion gives $m_a(\lambda_i) = 0$. On the other hand, the polynomial $\prod_i (x-\lambda_i)$ vanishes at $a$, because on the component $e_j$ it has the factor $x-\lambda_j$. Hence $m_a$ divides $\prod_i (x-\lambda_i)$. Since $m_a$ is divisible by each $x-\lambda_i$ and the $\lambda_i$ are distinct, $m_a$ is divisible by the product, so $m_a$ equals the product up to a unit; both are monic, so $m_a = \prod_i (x-\lambda_i)$.

From 2 to 4 and 5. Consider the map $k[x]/(m_a) \to \operatorname{End}_k(A)$, $f \mapsto L_{f(a)}$. It is injective: if $L_{f(a)} = 0$, then $f(a) = L_{f(a)}(1_A) = 0$, so $m_a$ divides $f$ and $f$ is zero in the quotient. Hence the minimal polynomial of the operator $L_a$ is $m_a$. An operator on a finite-dimensional space is diagonalisable exactly when its minimal polynomial is split with distinct roots, so $L_a$ is diagonalisable exactly when $m_a$ is. The same argument applies to $R_a$, using $R_{f(a)}(1_A) = f(a)$.

From 4 or 5 to 2. This is the same equivalence read backwards: the minimal polynomial of $L_a$ is $m_a$, so if $L_a$ is diagonalisable then $m_a$ is split with distinct roots. The same for $R_a$.

Uniqueness

Suppose that $a = \sum_i \lambda_i e_i = \sum_j \mu_j f_j$ are two eigenvalue decompositions. By the criterion, the sets $\{\lambda_i\}$ and $\{\mu_j\}$ are both the set of roots of $m_a$, so they coincide; write the common eigenvalues as $\lambda_1, \dots, \lambda_r$. For each $i$, the Lagrange polynomial

$$ h_i(x) = \prod_{j \neq i} \frac{x - \lambda_j}{\lambda_i - \lambda_j} $$

satisfies $h_i(\lambda_j) = \delta_{ij}$. Evaluating at $a$ on the first decomposition gives $h_i(a) = \sum_j h_i(\lambda_j)e_j = e_i$; evaluating on the second gives $h_i(a) = f_i$. Hence $e_i = f_i$. The decomposition is unique.

The uniqueness also shows that the eigenprojectors are intrinsic to $a$ and do not depend on the choice of a basis: they are the Lagrange polynomials in $a$.

The Eigenspaces

The eigenspaces of the multiplications

Theorem. Let $a = \sum_i \lambda_i e_i$ be an eigenvalue decomposition. Then

$$ \boxed{\;\ker(L_a - \lambda_i) = e_i A, \qquad \ker(R_a - \lambda_i) = A e_i\;} . $$

Proof. Since $a e_i = \lambda_i e_i$, every element $e_i x$ satisfies $a(e_i x) = \lambda_i e_i x$, so $e_i A \subseteq \ker(L_a - \lambda_i)$. Conversely, let $x \in \ker(L_a - \lambda_i)$. Write $x = \sum_j e_j x$, using $\sum_j e_j = 1_A$. Then

$$ 0 = (a - \lambda_i)x = \sum_j (\lambda_j - \lambda_i)e_j x . $$

The summands $e_j x$ lie in the direct sum $\bigoplus_j e_j A$, because the $e_j$ are orthogonal idempotents, so each summand vanishes. For $j \neq i$ the scalar $\lambda_j - \lambda_i$ is nonzero, so $e_j x = 0$; hence $x = e_i x \in e_i A$. The proof for $R_a$ is the same with the factors on the other side.

The simultaneous eigenspaces

The two multiplications commute, $L_a R_a = R_a L_a$, because both are multiplications in the commutative algebra $k[a]$. They are therefore simultaneously diagonalisable, and their common eigenspaces are

$$ \ker(L_a - \lambda_i) \cap \ker(R_a - \lambda_j) = e_i A e_j . $$

Indeed, $x \in e_i A e_j$ satisfies $a x = \lambda_i x$ and $x a = \lambda_j x$; conversely, if $a x = \lambda_i x$ and $x a = \lambda_j x$, the expansion $x = \sum_{p,q} e_p x e_q$ and the comparison of the coefficients of $e_p x e_q$ give $p = i$ and $q = j$, because the $\lambda$'s are distinct.

The simultaneous eigenspaces give the two-sided Peirce decomposition of $A$ relative to the complete orthogonal family $e_1, \dots, e_r$:

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

The diagonal summands $e_i A e_i$ are the corners on which $L_a$ and $R_a$ both act by $\lambda_i$; they form the centraliser of $a$, as proved in The Centraliser and the Diagonal Peirce Part. The off-diagonal summands $e_i A e_j$ with $i \neq j$ mix the two eigenspaces and form the off-diagonal Peirce part of Off-Diagonal Peirce Components and Noncentral Idempotents. The eigenvalues of $L_a$ on $e_i A e_j$ are $\lambda_i$ and those of $R_a$ are $\lambda_j$; the pair $(\lambda_i, \lambda_j)$ labels the summand.

The matrix case

Let $A = M_n(k) = \operatorname{End}_k(V)$ and let $a$ be a diagonalisable matrix with distinct eigenvalues. The eigenprojector $e_i$ is the projection of $V$ onto the eigenspace $V_i = \ker(a - \lambda_i)$ along the sum of the other eigenspaces, and its rank $m_i$ is the algebraic multiplicity of $\lambda_i$. The matrix eigenspace $V_i$ is the image $e_i V$, of dimension $m_i$. The eigenspace of the left multiplication is the left ideal $e_i A$, of dimension $n m_i$: it consists of the operators whose image lies in $V_i$. The eigenspace of the right multiplication is the right ideal $A e_i$, of dimension $n m_i$: it consists of the operators that vanish on the sum of the other eigenspaces. The simultaneous eigenspace $e_i A e_j$ has dimension $m_i m_j$ and consists of the operators mapping $V_j$ to $V_i$ and killing the other eigenspaces.

The matrix eigenspaces and the algebra eigenspaces are therefore different objects of different dimensions, and the notation $e_i A$ must not be read as the matrix eigenspace. The matrix eigenspace is $e_i V$; the algebra eigenspace is $e_i A$. The distinction is the reason the left multiplication has eigenvalues with multiplicity $n m_i$ while the matrix has eigenvalues with algebraic multiplicity $m_i$.

The extremal cases

• Distinct eigenvalues. If all the eigenvalues are simple, then $r = n$ for a matrix, each $e_i$ has rank one, the corners $e_i A e_i$ are one-dimensional, and the centraliser is $k[a]$, of dimension $n$. The matrix is cyclic, and its minimal polynomial has degree $n$.
• A single eigenvalue. If $r = 1$, then $e_1 = 1_A$ and $a = \lambda 1_A$. The decomposition is trivial. A non-scalar element with a single eigenvalue cannot have an eigenvalue decomposition; its minimal polynomial has a repeated root, and the best one can do is the generalized decomposition of §The Generalized Eigenvalue Decomposition.

The Generalized Eigenvalue Decomposition

Split but not separable

Let $m_a$ be split but not necessarily separable, and write

$$ m_a(x) = \prod_{i=1}^{r}(x-\lambda_i)^{e_i}, \qquad \lambda_i \in k \text{ distinct}, \quad e_i \ge 1 . $$

The factors $(x-\lambda_i)^{e_i}$ are pairwise coprime, so the Chinese remainder theorem gives idempotents $\varepsilon_i \in k[a]$, the generalized eigenprojectors, with $\varepsilon_i \equiv 1 \pmod{(x-\lambda_i)^{e_i}}$ and $\varepsilon_i \equiv 0 \pmod{(x-\lambda_j)^{e_j}}$ for $j \neq i$. Put $n_i = (a - \lambda_i)\varepsilon_i$. Then

$$ a = \sum_{i=1}^{r} \left( \lambda_i \varepsilon_i + n_i \right), \qquad n_i^{e_i} = 0, \qquad n_i n_j = 0 \ (i \neq j), $$

and the $\varepsilon_i$ are orthogonal idempotents summing to $1_A$. The scalar $\lambda_i$ is the eigenvalue of the block, $\varepsilon_i$ is the projection onto the generalized eigenspace, and $n_i$ is the nilpotent part of the block. This is the generalized eigenvalue decomposition. It exists whenever $m_a$ is split, without any separability hypothesis.

In the semisimple case $e_i = 1$ for every $i$, the nilpotent parts vanish, $\varepsilon_i = e_i$, and the generalized decomposition reduces to the eigenvalue decomposition. In the other direction, the eigenvalue decomposition is the special case in which the generalized eigenprojectors are the eigenprojectors. The pair $(\varepsilon_i, n_i)$ contains the same information as the block $a\varepsilon_i$, whose minimal polynomial is $(x-\lambda_i)^{e_i}$.

Relation to the Jordan–Chevalley decomposition

The semisimple part of $a$ is

$$ s = \sum_i \lambda_i \varepsilon_i, $$

and the nilpotent part is

$$ n = \sum_i n_i . $$

They commute, $sn = ns$, because both are polynomials in $a$, and $a = s + n$ with $s$ split semisimple and $n$ nilpotent. This is the Jordan–Chevalley decomposition of Nilpotent Elements and the Jordan–Chevalley Decomposition, read in the split case. The exponents $e_i$ are the nilpotency indices of the blocks, and $\max_i e_i$ is the nilpotency index of $n$. The eigenvalue decomposition of this article is the case $n = 0$.

Over a non-split field

If $m_a$ has an irreducible factor of degree greater than one, there is no eigenvalue in $k$ for that factor, and the eigenvalue decomposition over $k$ does not exist. What always exists is the Chinese remainder decomposition into the primary factors

$$ m_a = f_1^{e_1} \cdots f_r^{e_r}, \qquad f_i \text{ irreducible and distinct}, $$

with idempotents $\varepsilon_i$ and blocks $a_i = a\varepsilon_i$ of minimal polynomial $f_i^{e_i}$. The block $a_i$ has eigenvalues only in the field extension $k[x]/(f_i)$; over $k$ it has none. The decomposition separates the irreducible factors but does not separate the roots inside a factor. The eigenvalue decomposition over $k$ is the special case in which every $f_i$ is linear, and the eigenvalue decomposition over an algebraic closure is the case in which the base field is extended until every $f_i$ splits.

The rotation matrix is the model case: over $\mathbb{R}$ its minimal polynomial $x^2+1$ is irreducible, there is no eigenvalue and no nontrivial idempotent; over $\mathbb{C}$ it splits, and the eigenprojectors appear with complex entries. The passage from $k$ to its algebraic closure is the subject of Splitting Fields and Algebraic Closure.

Worked Examples

The two-by-two matrix

For $a = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ the minimal polynomial is $(x-1)(x-2)$, and the decomposition is

$$ a = 1\cdot e_1 + 2\cdot e_2, \qquad e_1 = \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}, \quad e_2 = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}. $$

The left eigenspace of $1$ is $e_1 A$, the set of matrices with second row zero; the left eigenspace of $2$ is $e_2 A$, the set of matrices with equal rows. The right eigenspace of $1$ is $A e_1$, the set of matrices whose second column is the negative of the first; the right eigenspace of $2$ is $A e_2$, the set of matrices with equal columns. The four corners $e_i A e_j$ are one-dimensional, because the two eigenvalues are simple.

The diagonal matrix

For $a = \operatorname{diag}(1,2)$ the eigenprojectors are $e_1 = E_{11}$ and $e_2 = E_{22}$. The four corners are

$$ e_1 A e_1 = k E_{11}, \quad e_1 A e_2 = k E_{12}, \quad e_2 A e_1 = k E_{21}, \quad e_2 A e_2 = k E_{22}, $$

so the Peirce decomposition is the decomposition of a matrix into its four entries. The centraliser is the algebra of diagonal matrices, of dimension two, which is $k[a]$. The off-diagonal part is nonzero, because $a$ is not central in $M_2(k)$.

The companion matrix of $x^3 - x$

For $C = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$ the eigenvalues are $0, 1, -1$, and the eigenprojectors are those of The Algebra Generated by an Element and the Chinese Remainder Decomposition, §Worked Examples:

$$ e_0 = I - C^2, \qquad e_1 = \frac{C^2 + C}{2}, \qquad e_{-1} = \frac{C^2 - C}{2}. $$

The decomposition is $C = 0\cdot e_0 + 1\cdot e_1 + (-1)\cdot e_{-1}$. The eigenvalues are simple, so each corner $e_i A e_j$ is one-dimensional and the centraliser is $\mathbb{Q}[C]$, of dimension three. The left eigenspaces $e_i A$ have dimension three each; the right eigenspaces $A e_i$ have dimension three each; the algebra $M_3(\mathbb{Q})$ has dimension nine, the sum of the nine one-dimensional corners.

A Jordan block

For $a = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ the minimal polynomial is $(x-1)^2$, with a repeated root. There is no eigenvalue decomposition: the only idempotents in $k[a]$ are $0$ and $1$, and neither writing $a = 1\cdot 1$ nor $a = 0\cdot 0$ holds. The generalized decomposition is

$$ a = 1\cdot \varepsilon + n, \qquad \varepsilon = 1, \quad n = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad n^2 = 0, $$

which is the trivial decomposition into the scalar $1$ and the nilpotent $n$. The example shows that split but not separable gives a nilpotent part and no nontrivial projection. The Jordan–Chevalley form of the same statement is $a = 1 + n$.

A biquaternion

Under the matrix model $\mathbb{B} \cong M_2(\mathbb{C})$ of Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, the element $a = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ has the eigenvalue decomposition with the two idempotents $e_1, e_2$ read as biquaternions. The example is not different from the matrix one; it records that the eigenvalue decomposition is a statement about the algebra, not about a particular number system. The biquaternion spectral theory of The Biquaternion Eigenvalue Decomposition treats the additional structure that the conjugations impose on the spectrum.

Summary

An eigenvalue of $a$ is a scalar $\lambda$ for which $a-\lambda$ is not invertible; in finite dimension this is equivalent to $m_a(\lambda)=0$, and the spectrum is the root set of the minimal polynomial. Multiplicities in $m_a$ measure the nilpotent thickness of the blocks, not the dimensions of the matrix eigenspaces.

An eigenvalue decomposition $a = \sum_i \lambda_i e_i$ is a writing with distinct $\lambda_i$, orthogonal idempotents $e_i$ and $\sum_i e_i = 1$. It exists exactly when $m_a$ is split with distinct roots, equivalently when $a$ is split separable, equivalently when $L_a$ (or $R_a$) is diagonalisable. The eigenvalues are the roots of $m_a$, the eigenprojectors are the Lagrange polynomials in $a$, and the decomposition is unique. The left eigenspace of $\lambda_i$ is the left ideal $e_i A$, the right eigenspace is the right ideal $A e_i$, and the simultaneous eigenspace is the Peirce piece $e_i A e_j$. The two-sided Peirce decomposition $A = \bigoplus_{i,j} e_i A e_j$ is the simultaneous diagonalisation of $L_a$ and $R_a$; its diagonal part is the centraliser of $a$, and its off-diagonal part measures the failure of $a$ to be central.

When $m_a$ is split but not separable, the generalized eigenprojectors still exist and $a = \sum_i(\lambda_i \varepsilon_i + n_i)$ with $n_i$ nilpotent; this is the generalized eigenvalue decomposition and, after collecting the scalar parts, the Jordan–Chevalley decomposition. When $m_a$ has a non-linear irreducible factor, the decomposition over $k$ separates the irreducible factors but gives no eigenvalue in $k$ for that factor; the eigenvalues appear only after a field extension.

Summary of Notation

Further Reading

• Roger A. Horn and Charles R. Johnson, Matrix Analysis (Cambridge University Press, 2nd ed. 2013), Chapter 1 and §6.1, for the spectral projections, the diagonalisation and the distinction between the minimal and the characteristic polynomial.
• Felix R. Gantmacher, The Theory of Matrices, Vol. 1 (Chelsea, 1959), Chapter V, for the Jordan form and the generalized eigenspaces that underlie §The Generalized Eigenvalue Decomposition.
• Nicolas Bourbaki, Algèbre, Chapitres II and VII, for the Peirce decomposition and the spectral idempotents in an associative algebra.
• Richard S. Pierce, Associative Algebras (Springer, 1982), for the Peirce decomposition by a complete orthogonal family of idempotents.
• Serge Lang, Algebra (Springer, 3rd ed. 2002), Chapters V and XIV, for the minimal polynomial, the Chinese remainder theorem and the diagonalisability criterion.
• Paul R. Halmos, Finite-Dimensional Vector Spaces (Springer, 1974), for the spectral theorem in the finite-dimensional setting and the simultaneous diagonalisation of commuting operators.