Theory

Eigenvalues, eigenvectors and diagonalization

How to compute the eigenvalues and eigenspaces of a matrix or of an endomorphism, when a matrix is diagonalizable and how to build the diagonalizing matrix, with applications to powers and to matrices with a parameter.

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Eigenvalues and eigenvectors
The characteristic polynomial

Eigenvalues and eigenvectors

1. Eigenvalue and eigenvector of an endomorphism

Let $V$ be a real vector space and let $f : V \to V$ be a linear map, called an endomorphism of $V$ . A number $λ \in R$ is an eigenvalue of $f$ if there exists a vector $v \in V$ that satisfies the two conditions together

{v \neq = 0 f (v) = λ v

and every vector $v$ with these two properties is called an eigenvector of $f$ associated with the eigenvalue $λ$ . An eigenvector is therefore a nonzero vector that $f$ maps to a multiple of itself, so the line $⟨ v ⟩$ is mapped into itself. The condition $v \neq = 0$ is necessary in the definition, since $f (0) = 0 = λ \cdot 0$ for every $λ$ ; the eigenvalue, on the other hand, can be $0$ , and this happens exactly when $ker f$ contains nonzero vectors.

2. Eigenvalue and eigenvector of a matrix

Let $A \in M_{n} (R)$ . A number $λ \in R$ is an eigenvalue of $A$ if there exists a column vector $v \in R^{n}$ , with $v \neq = 0$ , such that $A v = λ v$ , and $v$ is called an eigenvector of $A$ associated with $λ$ . This is the previous definition applied to the endomorphism $v \mapsto A v$ of $R^{n}$ . To check that a given vector is an eigenvector, one computes $A v$ and checks that it is a multiple of $v$ ; the proportionality coefficient is the eigenvalue.

Example

Let

A = (4213), v = (11), w = (10) .

Since $A v = (5, 5) = 5 v$ , the vector $v$ is an eigenvector of $A$ associated with the eigenvalue $5$ . On the other hand, $A w = (4, 2)$ is not a multiple of $w$ , because the second component of $w$ is zero and that of $A w$ is not, therefore $w$ is not an eigenvector.

A v = 5 v

stays on the line of

v

(dashed): the direction of

v

is invariant;

A w = (4, 2)

leaves the line of

w

, the

x

-axis

The characteristic polynomial

1. Characteristic polynomial

Let $A \in M_{n} (R)$ . The characteristic polynomial of $A$ is

p_{A} (λ) = det (A - λ I_{n}),

a polynomial of degree $n$ in the variable $λ$ , with leading coefficient $(- 1)^{n}$ . It is obtained by subtracting $λ$ from every entry of the main diagonal and computing the determinant with the rules of the topic on determinants. For $A = (4213)$ one gets $p_{A} (λ) = (4 - λ) (3 - λ) - 2 = λ^{2} - 7 λ + 10$ .

2. Eigenvalues as roots of the characteristic polynomial

Let $A \in M_{n} (R)$ and let $λ_{0} \in R$ . Then $λ_{0}$ is an eigenvalue of $A$ if and only if

p_{A} (λ_{0}) = det (A - λ_{0} I_{n}) = 0.

The theorem reduces the search for eigenvalues to solving a polynomial equation of degree $n$ . For the matrix of the previous example $λ^{2} - 7 λ + 10 = (λ - 2) (λ - 5)$ , therefore the eigenvalues are $2$ and $5$ . A matrix of order $n$ has at most $n$ distinct eigenvalues, because a polynomial of degree $n$ has at most $n$ roots.

Proof

The equality $A v = λ_{0} v$ is equivalent to $A v - λ_{0} I_{n} v = 0$ , that is, to $(A - λ_{0} I_{n}) v = 0$ . Therefore $λ_{0}$ is an eigenvalue if and only if the homogeneous linear system $(A - λ_{0} I_{n}) v = 0$ has a solution $v \neq = 0$ . Since the coefficient matrix is square, by the criterion recalled below this happens if and only if $det (A - λ_{0} I_{n}) = 0$ .

Triangular matrices and expansion along a row

If $A$ is triangular, in particular diagonal, then $A - λ I_{n}$ is also triangular, and its determinant is the product of the diagonal entries:

p_{A} (λ) = (a_{11} - λ) (a_{22} - λ) \dots (a_{nn} - λ) .

The eigenvalues of a triangular matrix are therefore its diagonal entries, each counted as many times as it appears. For $(2012)$ the only eigenvalue is $2$ , with $p_{A} (λ) = (2 - λ)^{2}$ . When the matrix is not triangular but has a row or a column in which all the off-diagonal entries are zero, it is convenient to compute $det (A - λ I_{n})$ by Laplace expansion along that row or column, since the polynomial is then obtained already partly factored.

3. Trace, determinant and eigenvalues

Let $A \in M_{n} (R)$ . The constant term of $p_{A}$ is $p_{A} (0) = det A$ and the coefficient of $λ^{n - 1}$ is $(- 1)^{n - 1} tr A$ . For $n = 2$ one therefore has

p_{A} (λ) = λ^{2} - (tr A) λ + det A .

If $p_{A}$ has $n$ real roots $λ_{1}, \dots, λ_{n}$ , counted with their multiplicity, then

tr A det A = λ_{1} + λ_{2} + \dots + λ_{n} = λ_{1} λ_{2} \dots λ_{n} .

The first statement follows by setting $λ = 0$ and observing that in the expansion of $det (A - λ I_{n})$ the powers $λ^{n}$ and $λ^{n - 1}$ come only from the product of the diagonal entries; the second follows by comparing the coefficients with $p_{A} (λ) = (λ_{1} - λ) \dots (λ_{n} - λ)$ . The two equalities serve as a check on the computations: for $(4213)$ one has $tr A = 7 = 2 + 5$ and $det A = 10 = 2 \cdot 5$ . Since $det A = p_{A} (0)$ , the matrix $A$ is invertible if and only if $0$ is not an eigenvalue of $A$ .

The theory continues with 7 more sections (similar matrices and endomorphisms, eigenspaces, algebraic and geometric multiplicity…), plus all worked examples, the solutions and the PDF.

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