Eigenvalues and Eigenvectors

The directions a linear map leaves in place, and the factors by which it scales them. They are found as the roots of $det (A - λ I) = 0$ and the null spaces of $A - λ I$ ; whether enough of them exist to describe the whole space is a separate question with a precise answer.

Definition

Let $A$ be an $n \times n$ matrix over a field $F$ . A scalar $λ \in F$ is an eigenvalue of $A$ when there is a nonzero $v$ with

A v = λ v .

Such a $v$ is an eigenvector for $λ$ . The requirement $v \neq 0$ is essential: $A 0 = λ 0$ holds for every $λ$ , so admitting the zero vector would make every scalar an eigenvalue.

Finding them. Rewriting $A v = λ v$ as $(A - λ I) v = 0$ says $v$ lies in the kernel of $A - λ I$ . A nonzero kernel element exists exactly when $A - λ I$ is singular, so the eigenvalues are the roots of the characteristic polynomial

p (λ) = det (A - λ I) .

For each root, the corresponding eigenvectors are the nonzero elements of $\ker (A - λ I)$ , called the eigenspace $E_{λ}$ . A subspace, since it is a kernel, once the zero vector is included.

For a $2 \times 2$ matrix the characteristic polynomial is $λ^{2} - (tr A) λ + det A$ , where $tr A$ is the sum of the diagonal entries.

Multiplicities. The algebraic multiplicity of $λ$ is its multiplicity as a root of $p$ . The geometric multiplicity is $\dim E_{λ}$ . Always

1 \leq geometric \leq algebraic,

and the inequality can be strict. A matrix with geometric less than algebraic for some eigenvalue is defective, and a defective matrix has no basis of eigenvectors.

Two invariants. Over a field where $p$ splits, counting roots with algebraic multiplicity:

tr A = \sum_{i} λ_{i}, det A = \prod_{i} λ_{i} .

Both follow from comparing coefficients of the characteristic polynomial, and both serve as cheap checks on a computed answer.

Triangular matrices. The eigenvalues of a triangular matrix are its diagonal entries, since $det (A - λ I)$ is then the product of the diagonal differences.

Assumptions and scope

Only square matrices have eigenvalues. The equation $A v = λ v$ requires $A v$ and $v$ to live in the same space.
Eigenvectors are nonzero by definition; eigenvalues may be zero. Zero is an eigenvalue exactly when the matrix is singular, and its eigenspace is the kernel.
The field matters. A real matrix may have no real eigenvalues while having complex ones, and for a real matrix the complex eigenvalues arrive in conjugate pairs with conjugate eigenvectors.
Geometric multiplicity is at least 1 and at most algebraic. When it is strictly less for some eigenvalue the matrix is defective and no basis of eigenvectors exists.
The characteristic polynomial is a computational route, not the definition. For large matrices it is numerically poor, and practical algorithms compute eigenvalues without forming it.
Eigenvalues are unchanged by a change of basis: similar matrices have the same characteristic polynomial, which is why the eigenvalues belong to the transformation rather than to its matrix.

Worked material

Example

Four matrices, four behaviours

Triangular: read the diagonal. $D = (\begin{matrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 4 & 5 & - 1 \end{matrix})$ is lower triangular, so its eigenvalues are $2$ , $3$ and $- 1$ . Checking one: $det (D - 2 I) = 0$ . The invariants agree, trace $2 + 3 - 1 = 4$ and $det D = 2 \cdot 3 \cdot (- 1) = - 6$ , matching the sum and product of the eigenvalues.

Defective: a shear. $B = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$ is triangular, so $λ = 3$ with algebraic multiplicity 2. But

B - 3 I = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

has rank 1, so its kernel has dimension 1: $E_{3} = span {(1, 0)}$ , geometric multiplicity 1. Checking: $B (1, 0)^{T} = (3, 0)^{T} = 3 (1, 0)^{T}$ .

Geometric $1 <$ algebraic $2$ , so $B$ is defective: there is no basis of $R^{2}$ made of its eigenvectors, and no change of basis makes it diagonal. Geometrically it fixes the horizontal direction and shears everything else, never settling any second direction.

Complex: a rotation. $C = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ rotates the plane by a quarter turn. Trace $0$ , determinant $1$ , so $p (λ) = λ^{2} + 1$ , whose roots are $\pm i$ , no real eigenvalues at all, which the picture predicts: a quarter turn leaves no real direction fixed.

Over $C$ it is diagonalisable. For $λ = i$ , solving $(C - i I) v = 0$ gives $v = (1, - i)$ , and

C (\begin{matrix} 1 \\ - i \end{matrix}) = (\begin{matrix} i \\ 1 \end{matrix}) = i (\begin{matrix} 1 \\ - i \end{matrix})

since $i (- i) = 1$ . The other eigenvalue $- i$ has eigenvector $(1, i)$ , the entrywise conjugate. That pairing is general: a real matrix has real characteristic polynomial, so its non-real roots come in conjugate pairs, with conjugate eigenvectors.

Singular: zero as an eigenvalue. $S = (\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix})$ has $det S = 0$ , so $0$ is an eigenvalue with $E_{0} = \ker S = span {(2, - 1)}$ ; indeed $S (2, - 1)^{T} = (0, 0)^{T}$ . The trace is 5, so the other eigenvalue is 5, with eigenvector $(1, 2)$ : $S (1, 2)^{T} = (5, 10)^{T} = 5 (1, 2)^{T}$ .

Eigenvalues can be read off structure (triangular), can exist without enough eigenvectors (defective), can require a larger field (rotation), and can be zero (singular). Only the second is an obstruction that no change of setting removes.

Non-example

Five ways an eigenvalue computation goes wrong

Subtracting $λ$ from every entry. Writing $A - λ$ rather than $A - λ I$ produces a matrix unrelated to the problem, and a characteristic polynomial of the wrong degree. Only the diagonal is affected.

Accepting the zero vector as an eigenvector. After reducing $(A - λ I) v = 0$ , reporting $v = 0$ as the answer. That solution always exists and says nothing; the eigenspace is the kernel, and the eigenvectors are its nonzero elements. If reduction produced no free variable, the eigenvalue or the arithmetic is wrong, $A - λ I$ is singular by construction.

Reporting one vector as "the" eigenvector. Every nonzero multiple of an eigenvector is an eigenvector, and an eigenspace of dimension 2 has a whole plane of them. The answer is a basis for $E_{λ}$ , not a representative.

Concluding non-diagonalisability from a repeated root. Algebraic multiplicity above 1 does not by itself mean defective: the identity matrix has $λ = 1$ with algebraic multiplicity $n$ and geometric multiplicity $n$ , and is already diagonal. The verdict needs the geometric multiplicity, which means solving for the eigenspace rather than stopping at the polynomial.

Concluding a real matrix has no eigenvalues. A rotation has no real eigenvalues, which is a statement about the field rather than about the matrix. Over $C$ every $n \times n$ matrix has $n$ eigenvalues with multiplicity, because a degree- $n$ polynomial always splits there. Saying "no eigenvalues" without naming the field leaves out the part that decides it.

The first is mechanical. The next two mistake one solution for the solution set. The last two draw a conclusion from insufficient evidence. A repeated root without the eigenspace, or an absence of roots without the field. Each produces a confident answer that is wrong in a different way, and only the first is visible in the arithmetic.

Common errors

Common misconception

The zero vector is an eigenvector for every eigenvalue, since $A 0 = λ 0$ holds; and correspondingly any scalar satisfying that equation is an eigenvalue.

Eigenvalues and Eigenvectors

Assumptions and scope

Worked material

Four matrices, four behaviours

Five ways an eigenvalue computation goes wrong

Common errors

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