The Lagrange–Sylvester Eigenprojector Formula

Introduction

The eigenvalue decomposition of The Eigenvalue Decomposition of an Element produces idempotents $e_i$ whose existence is guaranteed by the Chinese remainder theorem. This article gives them explicitly, as polynomials in the element, and gives the corresponding formula for every polynomial or analytic function of the element. The first formula is Lagrange interpolation at the eigenvalues; the second is Sylvester's formula. Together they reduce the computation of $f(a)$ to the values of $f$ at the eigenvalues and the eigenprojectors.

The article assumes the algebra generated by an element of The Algebra Generated by an Element and the Chinese Remainder Decomposition, the separability and splitting conditions of Separable Elements and Split Minimal Polynomials, and the eigenvalue decomposition of The Eigenvalue Decomposition of an Element. The repeated-root case uses the generalized eigenprojectors of the same article and anticipates the Jordan–Chevalley decomposition of Nilpotent Elements and the Jordan–Chevalley Decomposition. The base is a field $k$ and $A$ is a finite-dimensional unital associative $k$-algebra. The analytic case is stated over $\mathbb{C}$.

Lagrange Interpolation

The interpolation problem

Let $\lambda_1, \dots, \lambda_r$ be pairwise distinct elements of $k$, and let $v_1, \dots, v_r \in k$. The interpolation problem asks for a polynomial $h \in k[x]$ of degree less than $r$ with

$$ h(\lambda_i) = v_i, \qquad i = 1, \dots, r . $$

The solution exists and is unique. Uniqueness: the difference of two solutions is a polynomial of degree less than $r$ with $r$ distinct roots, hence zero. Existence: the Lagrange basis polynomials

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

satisfy $h_i(\lambda_j) = \delta_{ij}$, so

$$ h(x) = \sum_{i=1}^{r} v_i h_i(x) $$

is a solution. The denominators $\lambda_i - \lambda_j$ are nonzero because the nodes are distinct; this is the only place where the distinctness is used.

The Lagrange basis is a basis of the $r$-dimensional space of polynomials of degree less than $r$. The coefficients $v_i$ are the coordinates of $h$ in this basis, and the evaluation map $h \mapsto (h(\lambda_1), \dots, h(\lambda_r))$ is an isomorphism from that space onto $k^r$.

Interpolation in the algebra

Suppose now that $a$ has the eigenvalue decomposition

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

of The Eigenvalue Decomposition of an Element, with the $\lambda_i$ distinct. Evaluating the Lagrange basis at $a$ gives elements $h_i(a) \in k[a]$. Since $h_i(\lambda_j) = \delta_{ij}$ and $e_j$ is the component on which $a$ acts by $\lambda_j$,

$$ h_i(a) = \sum_{j=1}^{r} h_i(\lambda_j) e_j = e_i . $$

This is the eigenprojector formula: the Lagrange basis polynomial at the node $\lambda_i$, evaluated at $a$, is the eigenprojector $e_i$. It expresses the idempotents of the Chinese remainder decomposition as explicit polynomials in $a$, with coefficients in $k$.

For a polynomial $f$ of degree less than $r$, the same computation gives

$$ f(a) = \sum_{i=1}^{r} f(\lambda_i) e_i . $$

Indeed, $f$ is its own interpolation polynomial at the $r$ nodes, because both have degree less than $r$ and agree at the nodes. For a polynomial $f$ of arbitrary degree, the formula still holds, but the proof needs the minimal polynomial; it is the content of §Sylvester's Formula.

The Chinese remainder interpretation

The map $k[x] \to k^r$, $f \mapsto (f(\lambda_1), \dots, f(\lambda_r))$, factors through the Chinese remainder isomorphism $k[x]/(m_a) \cong k^r$ when $m_a = \prod_i (x-\lambda_i)$. The Lagrange basis $h_i$ is the inverse image of the standard basis vector $(0, \dots, 1, \dots, 0)$; the idempotent $e_i$ is the image of $h_i$ under the evaluation map $f \mapsto f(a)$, and it corresponds to that basis vector under the Chinese remainder isomorphism. The eigenprojector formula is therefore the explicit form of the Chinese remainder decomposition, and the algebra $k[a]$ is the image of the interpolation map.

The Eigenprojector Formula

The formula

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

$$ \boxed{\;e_i = \prod_{j \neq i} \frac{a - \lambda_j 1_A}{\lambda_i - \lambda_j}\;} . $$

Proof. The polynomial on the right is $h_i(a)$, and $h_i(\lambda_j) = \delta_{ij}$. Evaluating $h_i$ on the decomposition gives

$$ h_i(a) = \sum_{j=1}^{r} h_i(\lambda_j) e_j = e_i . $$

The formula is the definition of the Lagrange basis transported from $k$ to $k[a]$ by the algebra homomorphism $f \mapsto f(a)$.

The minimal-polynomial form

Let $m = m_a = \prod_{i=1}^{r}(x-\lambda_i)$ and let

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

be the product of all the factors except the $i$-th. Then $m_i(\lambda_j) = 0$ for $j \neq i$, while

$$ m_i(\lambda_i) = \prod_{j \neq i}(\lambda_i - \lambda_j) \neq 0 . $$

Hence

$$ e_i = \frac{m_i(a)}{m_i(\lambda_i)} . $$

Since $m'(x) = \sum_i m_i(x)$ and all the terms except $m_i$ vanish at $\lambda_i$, one has $m'(\lambda_i) = m_i(\lambda_i)$. Therefore

$$ e_i = \frac{m_i(a)}{m_a'(\lambda_i)} . $$

This is the Sylvester form of the eigenprojector formula. It replaces the product over $j \neq i$ by one division by the derivative at the eigenvalue. The two forms are the same polynomial in $a$; the derivative form is shorter to write and is the one usually remembered.

A two-by-two example

For $a = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ and $m = (x-1)(x-2)$:

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

The derivative form gives the same result: $m_1 = x-2$, $m_1(1) = -1$, so $e_1 = -(a-2I)$; $m_2 = x-1$, $m_2(2) = 1$, so $e_2 = a-I$. The two projections sum to $I$ and are orthogonal.

The companion matrix of $x^3 - x$

For $C = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$ and $m = x(x-1)(x+1)$, the eigenprojector at $0$ is

$$ e_0 = \frac{(C-I)(C+I)}{(0-1)(0+1)} = \frac{C^2 - I}{-1} = I - C^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}. $$

Equivalently, with $m_0 = (x-1)(x+1) = x^2-1$ and $m_0(0) = -1$, the derivative form gives $e_0 = m_0(C)/m_0(0) = I - C^2$.

With $m_1 = x(x+1)$ and $m_1(1) = 2$,

$$ e_1 = \frac{C^2 + C}{2} = \begin{pmatrix} 0 & 0 & 0 \\ \tfrac12 & \tfrac12 & \tfrac12 \\ \tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix}. $$

With $m_{-1} = x(x-1)$ and $m_{-1}(-1) = 2$,

$$ e_{-1} = \frac{C^2 - C}{2} = \begin{pmatrix} 0 & 0 & 0 \\ -\tfrac12 & \tfrac12 & -\tfrac12 \\ \tfrac12 & -\tfrac12 & \tfrac12 \end{pmatrix}. $$

These are the idempotents of The Algebra Generated by an Element and the Chinese Remainder Decomposition, §Worked Examples, now obtained by the general formula rather than by inspection.

Change of basis

The formula is polynomial in $a$, so it is invariant under every algebra homomorphism: if $\varphi : A \to B$ is a unital algebra homomorphism and $a$ has the eigenvalue decomposition with eigenvalues $\lambda_i$ in $k$, then $\varphi(a)$ has the eigenvalue decomposition with the same eigenvalues and eigenprojectors $\varphi(e_i)$. The formula commutes with $\varphi$ because it is a polynomial identity. In matrix terms, conjugating $a$ by an invertible matrix conjugates each $e_i$ by the same matrix; the eigenvalues and the coefficients of the Lagrange polynomials do not change.

Sylvester's Formula

The formula

Theorem (Sylvester). Let $a = \sum_{i=1}^{r} \lambda_i e_i$ be an eigenvalue decomposition, and let $f \in k[x]$ be any polynomial. Then

$$ \boxed{\;f(a) = \sum_{i=1}^{r} f(\lambda_i) e_i\;} . $$

Proof. Let $h$ be the interpolation polynomial of $f$ at the nodes $\lambda_1, \dots, \lambda_r$, of degree less than $r$. The difference $f - h$ vanishes at every $\lambda_i$, so it is divisible by $x-\lambda_i$ for every $i$; since the $\lambda_i$ are distinct, it is divisible by $m_a = \prod_i (x-\lambda_i)$. Evaluating at $a$ gives $f(a) = h(a)$, and $h(a) = \sum_i h(\lambda_i)e_i = \sum_i f(\lambda_i)e_i$ by the eigenprojector formula.

The theorem says that the homomorphism $k[x] \to k[a]$, $f \mapsto f(a)$, factors through the interpolation map at the eigenvalues. The kernel is $(m_a)$, and the quotient $k[x]/(m_a)$ is identified with the functions on the $r$ eigenvalues. The formula is the explicit form of this identification.

A two-by-two example

For $a = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ and $f(x) = x^3$:

$$ f(a) = 1^3 e_1 + 2^3 e_2 = e_1 + 8e_2 = \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 8 \\ 0 & 8 \end{pmatrix} = \begin{pmatrix} 1 & 7 \\ 0 & 8 \end{pmatrix}. $$

A direct computation gives $a^2 = \begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix}$ and $a^3 = \begin{pmatrix} 1 & 7 \\ 0 & 8 \end{pmatrix}$, in agreement.

The companion matrix

For $C$ the companion matrix of $x^3-x$ and $f(x) = x^5 + 2x^3 - 1$, the values at the eigenvalues are

$$ f(0) = -1, \qquad f(1) = 1 + 2 - 1 = 2, \qquad f(-1) = -1 - 2 - 1 = -4 . $$

Hence

$$ f(C) = -e_0 + 2e_1 - 4e_{-1}, $$

which is a linear combination of the three idempotents. The formula avoids computing $C^5$; the result is a matrix in $\mathbb{Q}[C]$, as it must be, because $f(C)$ is a polynomial in $C$.

Analytic functions

Over $\mathbb{C}$, the formula extends to functions holomorphic on a neighbourhood of the spectrum. If $f$ is such a function, define $f(a)$ by the Cauchy integral

$$ f(a) = \frac{1}{2\pi i}\oint_{\Gamma} f(z)(z-a)^{-1}\,dz, $$

where $\Gamma$ is a contour enclosing the spectrum and avoiding it. The resolvent expansion of §The Resolvent and the Contour Integral gives

$$ f(a) = \sum_{i=1}^{r} f(\lambda_i)e_i . $$

Equivalently, one may define $f(a)$ by taking the interpolation polynomial of $f$ at the eigenvalues, which agrees with the holomorphic functional calculus because $f$ and its interpolation polynomial have the same values at the spectrum. The formula is the finite-dimensional form of the holomorphic functional calculus, and it is the reason the spectrum is the right notion of "the set of values that matter" for a matrix or an element of a finite-dimensional algebra.

The Resolvent and the Contour Integral

The resolvent identity

Let $z \in k$ be different from every eigenvalue. The resolvent of $a$ at $z$ is $(z 1_A - a)^{-1}$. From the eigenvalue decomposition,

$$ (z 1_A - a)^{-1} = \sum_{i=1}^{r} \frac{e_i}{z - \lambda_i} . $$

Indeed, multiplying the right-hand side by $z 1_A - a = \sum_i (z-\lambda_i)e_i$ gives $\sum_i e_i = 1_A$, and the inverse is unique. The identity is the partial fraction decomposition of the rational function $(z-x)^{-1}$ modulo $m_a$; at each node $\lambda_i$, both sides have the residue $e_i$.

The contour formula

Over $\mathbb{C}$, the eigenprojector is the residue of the resolvent at the eigenvalue:

$$ e_i = \frac{1}{2\pi i}\oint_{\Gamma_i} (z 1_A - a)^{-1}\,dz, $$

where $\Gamma_i$ is a small positively oriented circle around $\lambda_i$ enclosing no other eigenvalue. The formula follows from the resolvent identity: the integral of $e_j/(z-\lambda_j)$ around $\Gamma_i$ is $e_i$ if $i=j$ and zero otherwise, by Cauchy's theorem. This is Sylvester's contour formula. It defines the spectral projection without choosing a basis and without knowing the minimal polynomial explicitly; it is the form used in the holomorphic functional calculus and in the theory of operator algebras.

For a matrix, the contour formula is the standard residue formula for the spectral projection. For a general finite-dimensional algebra, the same computation is carried out in the algebra of operators on $A$, and the result lies in $k[a]$ because the resolvent does.

The resolvent in the repeated case

If $m_a$ has repeated roots, the resolvent has higher-order poles. With the generalized eigenprojectors $\varepsilon_i$ and the nilpotent parts $n_i = (a-\lambda_i)\varepsilon_i$ of §The Generalized Eigenvalue Decomposition of The Eigenvalue Decomposition of an Element,

$$ (z 1_A - a)^{-1} = \sum_{i=1}^{r} \sum_{k=0}^{e_i-1} \frac{n_i^k}{(z-\lambda_i)^{k+1}}, $$

where $e_i$ is the multiplicity of $\lambda_i$ in $m_a$. Indeed, on the $i$-th block, $a = \lambda_i + n_i$ with $n_i^{e_i}=0$, and the geometric series for $(z-\lambda_i-n_i)^{-1}$ terminates. The coefficient of $(z-\lambda_i)^{-1}$ is $\varepsilon_i$; the higher coefficients are the nilpotent parts. The contour integral of the resolvent around $\lambda_i$ therefore gives $\varepsilon_i$, not an idempotent that separates the roots inside a repeated block: the block is local, and its only idempotent is $\varepsilon_i$.

The Repeated-Root Case

The failure of the simple formula

Suppose $m_a = \prod_{i=1}^{r}(x-\lambda_i)^{e_i}$ with some $e_i > 1$. The factors are still pairwise coprime, so the generalized eigenprojectors $\varepsilon_i$ exist. The simple formula

$$ e_i = \frac{m_i(a)}{m_i(\lambda_i)}, \qquad m_i = \frac{m_a}{x-\lambda_i}, $$

cannot apply, because $m_i(\lambda_i) = 0$ when $e_i > 1$: the factor $x-\lambda_i$ still occurs in $m_i$ with exponent $e_i-1$. Equivalently, $m_a'(\lambda_i) = 0$. The derivative form of the eigenprojector formula is therefore a criterion for simple roots, not a formula that survives their failure.

The correct replacement is Hermite interpolation. Let $q_i = m_a/(x-\lambda_i)^{e_i}$. The polynomial $q_i$ is a unit modulo $(x-\lambda_i)^{e_i}$, because $q_i(\lambda_i) \neq 0$. Let $u_i$ be its inverse modulo $(x-\lambda_i)^{e_i}$,

$$ q_i u_i \equiv 1 \pmod{(x-\lambda_i)^{e_i}} . $$

Then the generalized eigenprojector is

$$ \varepsilon_i = q_i(a)u_i(a). $$

It satisfies $\varepsilon_i \equiv 1 \pmod{(x-\lambda_i)^{e_i}}$ and $\varepsilon_i \equiv 0 \pmod{(x-\lambda_j)^{e_j}}$ for $j \neq i$, so it is the CRT idempotent of the factor $(x-\lambda_i)^{e_i}$. In the simple-root case $e_i=1$, $u_i$ is the constant $1/q_i(\lambda_i)$, and the formula reduces to $\varepsilon_i = q_i(a)/q_i(\lambda_i) = m_i(a)/m_i(\lambda_i)$.

Hermite interpolation

Theorem. Let $m = \prod_{i=1}^{r}(x-\lambda_i)^{e_i}$ with the $\lambda_i$ distinct, and let $f \in k[x]$. There is a unique polynomial $p$ of degree less than $\deg m$ such that

$$ p^{(k)}(\lambda_i) = f^{(k)}(\lambda_i), \qquad 0 \le k < e_i, \quad 1 \le i \le r . $$

For this polynomial, $f(a) = p(a)$.

Proof. The conditions are $\deg m$ linear conditions on the coefficients of $p$, and the corresponding homogeneous system has only the zero solution: a polynomial of degree less than $\deg m$ with a zero of order at least $e_i$ at each $\lambda_i$ is divisible by $m$, hence zero. Existence follows by dimension, or by the explicit Hermite basis. The difference $f-p$ has a zero of order at least $e_i$ at $\lambda_i$, so it is divisible by $(x-\lambda_i)^{e_i}$ for every $i$; the factors are coprime, so $m \mid f-p$, and evaluating at $a$ gives $f(a)=p(a)$.

The Hermite polynomial is the Taylor expansion of $f$ at each eigenvalue, truncated at order $e_i-1$ and patched by the CRT idempotents. In the simple-root case $e_i=1$, it is the Lagrange interpolation polynomial.

The generalized Sylvester formula

On the $i$-th block, $a = \lambda_i \varepsilon_i + n_i$ with $n_i = (a-\lambda_i)\varepsilon_i$ and $n_i^{e_i}=0$. Taylor's formula gives

$$ f(a)\varepsilon_i = \sum_{k=0}^{e_i-1} \frac{f^{(k)}(\lambda_i)}{k!} n_i^k . $$

Summing over $i$,

$$ \boxed{\;f(a) = \sum_{i=1}^{r} \sum_{k=0}^{e_i-1} \frac{f^{(k)}(\lambda_i)}{k!} (a-\lambda_i)^k \varepsilon_i\;} . $$

This is the generalized Sylvester formula, also called the Hermite–Sylvester formula. It reduces to Sylvester's formula when all $e_i=1$, because the $k=0$ term is $f(\lambda_i)\varepsilon_i$ and the higher terms are absent. It is the explicit form of the Jordan–Chevalley decomposition of $f(a)$: the $k=0$ terms give the semisimple part, and the terms with $k \ge 1$ give the nilpotent part.

A Jordan block

For $a = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $f(x) = x^3$, the minimal polynomial is $(x-1)^2$, so $e_1 = 2$ and the Hermite polynomial is

$$ p(x) = f(1) + f'(1)(x-1) = 1 + 3(x-1) = 3x - 2 . $$

Then

$$ f(a) = p(a) = 3a - 2I = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = a^3 . $$

The generalized Sylvester formula gives the same result with $\varepsilon_1 = 1$ and $n_1 = a-I$:

$$ f(a) = f(1)\cdot 1 + f'(1)\cdot (a-I) = 1 + 3\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}. $$

The resolvent is

$$ (zI - a)^{-1} = \begin{pmatrix} \frac{1}{z-1} & \frac{1}{(z-1)^2} \\ 0 & \frac{1}{z-1} \end{pmatrix}, $$

with a double pole at the eigenvalue and residue $I$, the generalized eigenprojector.

A block with two eigenvalues

For $a = \operatorname{diag}(J_2(1), 2) = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$, the minimal polynomial is $(x-1)^2(x-2)$. The generalized eigenprojectors are

$$ \varepsilon_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad \varepsilon_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, $$

and $a = 1\cdot\varepsilon_1 + 2\cdot\varepsilon_2 + n$ with $n = E_{12}$. The generalized Sylvester formula for $f(x)=x^3$ is

$$ f(a) = f(1)\varepsilon_1 + f'(1)n + f(2)\varepsilon_2 = \varepsilon_1 + 3E_{12} + 8\varepsilon_2 = \begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{pmatrix}, $$

which is $a^3$. The formula separates the double eigenvalue at $1$, with its derivative term, from the simple eigenvalue at $2$.

Why the Formula Needs Distinct Roots

The Lagrange and Sylvester formulas divide by $\lambda_i - \lambda_j$ and by $m_a'(\lambda_i)$. Both quantities are nonzero exactly when the roots are distinct. When a root is repeated, the denominators vanish, the idempotent formula has no meaning, and the correct statement is not that the idempotents fail to exist but that the simple formula for them fails. The CRT idempotents $\varepsilon_i$ of the repeated factors exist and are computable by the inverse-modulo formula of §The failure of the simple formula; they project onto the generalized eigenspaces, not onto the individual roots inside a repeated block, because inside such a block there is no nontrivial idempotent at all.

The distinction is the same one that separates the eigenvalue decomposition from the generalized eigenvalue decomposition in The Eigenvalue Decomposition of an Element: the formula with distinct roots produces a diagonal decomposition, while the Hermite formula produces a decomposition with nilpotent parts. The two formulas agree in the semisimple case and diverge exactly when the derivative vanishes.

Summary

Lagrange interpolation at distinct nodes $\lambda_1, \dots, \lambda_r$ gives the basis polynomials $h_i(x) = \prod_{j \neq i}(x-\lambda_j)/(\lambda_i-\lambda_j)$, with $h_i(\lambda_j)=\delta_{ij}$. Evaluating at an element $a$ with eigenvalue decomposition $a = \sum_i \lambda_i e_i$ gives the eigenprojectors $e_i = h_i(a)$, and the minimal-polynomial form $e_i = m_i(a)/m_i(\lambda_i) = m_i(a)/m_a'(\lambda_i)$ with $m_i = m_a/(x-\lambda_i)$.

Sylvester's formula is $f(a) = \sum_i f(\lambda_i)e_i$ for every polynomial $f$; it follows because $f$ and its interpolation polynomial differ by a multiple of $m_a$. The resolvent is $(z-a)^{-1} = \sum_i e_i/(z-\lambda_i)$, and over $\mathbb{C}$ the eigenprojector is the contour integral of the resolvent around the eigenvalue. The formula is invariant under algebra homomorphisms and under conjugation.

When the minimal polynomial has repeated roots, the simple formula fails because $m_a'(\lambda_i)=0$. The correct replacement is Hermite interpolation, with the generalized eigenprojectors $\varepsilon_i = q_i(a)u_i(a)$ where $q_i = m_a/(x-\lambda_i)^{e_i}$ and $u_i$ is the inverse of $q_i$ modulo $(x-\lambda_i)^{e_i}$. The generalized Sylvester formula is $f(a) = \sum_i \sum_{k=0}^{e_i-1} f^{(k)}(\lambda_i)(a-\lambda_i)^k\varepsilon_i/k!$, and the resolvent has poles of order $e_i$. The two formulas agree in the semisimple case and diverge exactly when a derivative vanishes.

Summary of Notation

Further Reading

• J. J. Sylvester, "On the equation to the secular inequalities in the planetary theory", Philosophical Magazine 16 (1883), 267–269, for the formula for a function of a matrix in terms of its eigenvalues and eigenprojectors.
• Roger A. Horn and Charles R. Johnson, Matrix Analysis (Cambridge University Press, 2nd ed. 2013), §6.1, for Sylvester's formula, the resolvent and the spectral projections.
• Felix R. Gantmacher, The Theory of Matrices, Vol. 1 (Chelsea, 1959), Chapter V, for the Hermite interpolation form and the Jordan decomposition.
• Nicholas J. Higham, Functions of Matrices: Theory and Computation (SIAM, 2008), Chapters 1 and 4, for the holomorphic functional calculus, the contour integral and the Hermite–Sylvester formula.
• Serge Lang, Algebra (Springer, 3rd ed. 2002), Chapter IV, for Lagrange interpolation and the Chinese remainder theorem.
• Nelson Dunford and Jacob T. Schwartz, Linear Operators, Part I (Interscience, 1958), Chapter VII, for the resolvent and the spectral projections in the general setting.