CommeUnJeu · L2 MP

Reduction: eigen-elements and diagonalisation

⌚ ~83 min ▢ 10 blocks ✓ 18 exercises ➣ Prerequisites : Determinants, Change of basis$\virgule$ equivalence$\virgule$ similarity, Polynomials, Further linear algebra

Given an endomorphism $u$ of a vector space $E$, reduction asks for a basis of $E$ in which the matrix of $u$ is as simple as possible --- ideally diagonal, so that $u$ merely scales each basis vector. This chapter builds the two tools that decide whether such a basis exists. The first is geometric: the eigen-elements of $u$ --- the directions $u$ only stretches. The second is algebraic: the characteristic polynomial, which turns the question « is $\lambda$ an eigenvalue? » into « is $\lambda$ a root? ».
The headline results are three. An endomorphism is diagonalisable exactly when $E$ is the direct sum of its eigenspaces --- equivalently, when its characteristic polynomial is split and each eigenspace has the expected dimension. It is trigonalisable (a triangular matrix) exactly when its characteristic polynomial is split. And a nilpotent endomorphism is precisely one that is trigonalisable with $0$ as its only eigenvalue.

Conventions

Throughout this chapter, $\mathbb{K}$ is a subfield of $\mathbb{C}$ and $E$ is a finite-dimensional, non-zero $\mathbb{K}$-vector space, with $n = \dim E \ge 1$. We write $u \in \mathcal{L}(E)$ for an endomorphism, $M \in \mathcal{M}_n(\mathbb{K})$ for a square matrix, $\operatorname{Id}_E$ for the identity and $I_n$ for the identity matrix. Determinant, trace, rank, similar matrices and the matrix of an endomorphism in a basis are those of Determinants and Change of basis; polynomials, roots, multiplicity and split polynomials those of Polynomials; direct sums, stable subspaces and the induced endomorphism those of Further linear algebra. The characteristic polynomial is defined as $\det(XI_n - M)$; its degree, monicity and coefficients are established in § 2.

I Eigen-elements

I.1 Eigenvalues$\virgule$ eigenvectors$\virgule$ eigenspaces

The reduction problem starts with a single observation: some vectors are merely stretched by $u$. A non-zero vector $x$ with $u(x) = \lambda x$ keeps its direction --- $u$ acts on the line it spans as the homothety of ratio $\lambda$. Such directions are the raw material of reduction: a basis made entirely of them turns the matrix of $u$ diagonal. This subsection names them and gathers their first properties.

Definition — Eigenvalue and eigenvector

A scalar $\lambda \in \mathbb{K}$ is an eigenvalue of $u$ when there exists a vector $x \neq 0_E$ such that $u(x) = \lambda x$. Such a vector $x$ is then an eigenvector of $u$ associated with $\lambda$. An eigenvector is, by definition, never the zero vector.

Example — Eigen-elements of three familiar endomorphisms

A homothety $\lambda_0 \operatorname{Id}_E$ admits every non-zero vector as an eigenvector, all for the single eigenvalue $\lambda_0$. A projector $p$ has eigenvalues among $0$ and $1$: every vector of $\operatorname{Im} p$ satisfies $p(x) = x$ and every vector of $\operatorname{Ker} p$ satisfies $p(x) = 0$ --- both $0$ and $1$ occur as soon as $\operatorname{Ker} p$ and $\operatorname{Im} p$ are both non-zero. A vector symmetry $s$ has eigenvalues among $-1$ and $1$, both occurring when its fixed and anti-fixed subspaces are both non-zero.

Definition — Eigenspace

Let $\lambda \in \mathbb{K}$. The eigenspace of $u$ associated with $\lambda$ is $$ E_\lambda(u) = \operatorname{Ker}(u - \lambda \operatorname{Id}_E). $$ It is a subspace of $E$, being the kernel of an endomorphism. The scalar $\lambda$ is an eigenvalue of $u$ if and only if $E_\lambda(u) \neq \{0_E\}$, and $E_\lambda(u)$ is then exactly the set of eigenvectors associated with $\lambda$, together with $0_E$.

Example — An eigenspace as a kernel

For $M = \begin{psmallmatrix} 1 & 2 \\ 2 & 1 \end{psmallmatrix}$, take $\lambda = 3$: the eigenspace is $E_3 = \operatorname{Ker}(M - 3I_2) = \operatorname{Ker}\begin{psmallmatrix} -2 & 2 \\ 2 & -2 \end{psmallmatrix}$. The system $-2x + 2y = 0$ has solution line $\operatorname{Vect}(1, 1)$, so $E_3 = \operatorname{Vect}(1, 1)$, a line; $3$ is an eigenvalue since $E_3 \neq \{0\}$. For $\lambda = 0$, $\operatorname{Ker} M = \{0\}$ (as $M$ is invertible), so $0$ is not an eigenvalue.

Definition — Spectrum

The spectrum of $u$, written $\operatorname{Sp}(u)$, is the set of eigenvalues of $u$ in $\mathbb{K}$. For a matrix $M \in \mathcal{M}_n(\mathbb{K})$ one defines likewise $\operatorname{Sp}(M)$, the set of $\lambda \in \mathbb{K}$ for which $MX = \lambda X$ has a non-zero column solution $X$ --- such an $X$ being a matrix eigenvector, tied to the eigenvectors of $u$ by the matrix-translation Proposition below.

Example — Eigenspaces and spectrum of a diagonal matrix

Let $u$ be the endomorphism of $\mathbb{R}^3$ with matrix $D = \operatorname{diag}(5, 5, -2)$ in the canonical basis $(e_1, e_2, e_3)$. Then $u(e_1) = 5e_1$, $u(e_2) = 5e_2$ and $u(e_3) = -2e_3$, so the spectrum is $\operatorname{Sp}(u) = \{5, -2\}$. The eigenspaces are $E_5(u) = \operatorname{Vect}(e_1, e_2)$, a plane, and $E_{-2}(u) = \operatorname{Vect}(e_3)$, a line; here $E_5(u) \oplus E_{-2}(u) = \mathbb{R}^3$.

The notion of spectral value is out of the program scope. In infinite dimension one distinguishes eigenvalues from a wider set of « spectral values »; here, in finite dimension, the spectrum is simply the set of eigenvalues and nothing more is needed.

Proposition — Matrix translation

Let $\mathcal{B}$ be a basis of $E$ and $M = \operatorname{Mat}_\mathcal{B}(u)$. A vector $x$, with coordinate column $X$ in $\mathcal{B}$, is an eigenvector of $u$ associated with $\lambda$ if and only if $X \neq 0$ and $MX = \lambda X$. Consequently $\operatorname{Sp}(u) = \operatorname{Sp}(M)$.