Algebraic and Geometric Multiplicity, linear algebra I lesson

The $2 \times 2$ matrix $2I$ and the matrix $J$ with rows $(2, 1)$ and $(0, 2)$ have the same characteristic polynomial $(t - 2)^2$ . For $2I$ every nonzero vector of the plane is an eigenvector, while for $J$ the eigenvectors fill only one line. So the number of times an eigenvalue occurs as a root and the number of independent eigenvectors it supplies are different quantities. This lesson names them algebraic and geometric multiplicity, proves that the second never exceeds the first, and names the matrices where it falls short. Throughout, a tuple such as $(1, 0)$ stands for a column vector, and matrices and eigenvalues are real.

Key terms

Repeated eigenvalue

Definition Let $\lambda$ be an eigenvalue of a real $n \times n$ matrix $A$ , with characteristic polynomial $\chi_A(t) = \det(tI - A)$ . The eigenvalue $\lambda$ is repeated if the factor $t - \lambda$ occurs at least twice in $\chi_A$ , that is, if $\chi_A(t) = (t - \lambda)^2 q(t)$ for some polynomial $q$ . Otherwise $\lambda$ is simple. For a $2 \times 2$ matrix, $\chi_A(t) = t^2 - (\operatorname{tr} A)\,t + \det A$ has a repeated root exactly when its discriminant $(\operatorname{tr} A)^2 - 4\det A$ is zero, and the repeated eigenvalue is then $\tfrac{1}{2}\operatorname{tr} A$ . Being repeated is a statement about the factorization of a polynomial. It does not say how many independent eigenvectors belong to $\lambda$ .

In plain words A repeated eigenvalue is a root that the characteristic polynomial counts more than once. The count comes from algebra and does not yet tell which directions the matrix keeps.

Example Both $2I$ and

J = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}

have $\chi(t) = (t - 2)^2$ , so $2$ is a repeated eigenvalue of each. The matrix $2I$ doubles every vector, so every nonzero vector is an eigenvector. The matrix $J$ sends $(x, y)$ to $(2x + y, 2y)$ , which equals $2(x, y)$ only when $y = 0$ . Its eigenvectors are the nonzero vectors on the horizontal axis and nothing else.

Algebraic multiplicity

Definition Let $\lambda$ be an eigenvalue of a real $n \times n$ matrix $A$ . Its algebraic multiplicity $\operatorname{am}(\lambda)$ is the number of times the factor $t - \lambda$ occurs in $\chi_A$ , that is, the largest $k$ with

\chi_A(t) = (t - \lambda)^k q(t)

for some polynomial $q$ . For this largest $k$ the remaining factor satisfies $q(\lambda) \ne 0$ . So $\lambda$ is simple when $\operatorname{am}(\lambda) = 1$ and repeated when $\operatorname{am}(\lambda) \ge 2$ . Because $\chi_A$ has degree $n$ , the algebraic multiplicities of the real eigenvalues add up to at most $n$ , with equality exactly when $\chi_A$ factors completely into real linear factors. For a triangular matrix, $tI - A$ is triangular too, so $\chi_A(t)$ is the product of the factors $t - a_{ii}$ over the diagonal entries.

In plain words The algebraic multiplicity counts how often an eigenvalue occurs as a root of the characteristic polynomial.

Example The diagonal matrix with entries $2, 2, 3$ has $\chi(t) = (t - 2)^2(t - 3)$ , so $\operatorname{am}(2) = 2$ and $\operatorname{am}(3) = 1$ , which add up to $3$ . The triangular matrix

H = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix}

has $\chi_H(t) = (t - 4)^3$ , so $\operatorname{am}(4) = 3$ . The matrix with rows $(0, -1, 0)$ , $(1, 0, 0)$ and $(0, 0, 2)$ has $\chi(t) = (t^2 + 1)(t - 2)$ . Its only real eigenvalue is $2$ , with algebraic multiplicity $1$ , and the sum $1$ is less than $3$ because $t^2 + 1$ has no real root.

Geometric multiplicity

Definition Let $\lambda$ be an eigenvalue of a real $n \times n$ matrix $A$ , with eigenspace $E_\lambda = \operatorname{Null}(A - \lambda I)$ . Its geometric multiplicity is

\begin{aligned} \operatorname{gm}(\lambda) &= \dim E_\lambda \\[0.4em] &= n - \operatorname{rank}(A - \lambda I), \end{aligned}

the number of free variables when $(A - \lambda I)\mathbf{v} = \mathbf{0}$ is solved by elimination. It is the largest number of linearly independent eigenvectors for $\lambda$ . Because $\lambda$ is an eigenvalue, $E_\lambda$ contains a nonzero vector, so $\operatorname{gm}(\lambda) \ge 1$ . The zero vector lies in $E_\lambda$ but adds no direction.

In plain words The geometric multiplicity counts how many independent directions an eigenvalue supplies. Count basis vectors or free variables, never individual eigenvectors, since a nonzero eigenspace contains infinitely many of them.

Example For $2I$ , the matrix $2I - 2I$ is zero, every vector solves the system, and $\operatorname{gm}(2) = 2$ . For $J$ ,

J - 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},

whose only equation is $y = 0$ . The variable $x$ is free, $E_2 = \operatorname{span}\{\mathbf{e}_1\}$ and $\operatorname{gm}(2) = 1$ , although $\operatorname{am}(2) = 2$ . For the diagonal matrix with entries $2, 2, 3$ , the eigenspaces are $E_2 = \operatorname{span}\{\mathbf{e}_1, \mathbf{e}_2\}$ and $E_3 = \operatorname{span}\{\mathbf{e}_3\}$ , of dimensions $2$ and $1$ , which together give three independent eigenvectors. Now replace the entry $1$ of $J$ by a parameter $s$ . The characteristic polynomial stays $(t - 2)^2$ for every $s$ , but the first row of the shifted matrix, now $(0, s)$ , gives the equation $sy = 0$ . For $s = 0$ it holds for every vector and $\operatorname{gm}(2) = 2$ , while for every $s \ne 0$ , however small, it forces $y = 0$ and $\operatorname{gm}(2) = 1$ .

Bound on geometric multiplicity

Definition For every eigenvalue $\lambda$ of a real $n \times n$ matrix $A$ ,

1 \le \operatorname{gm}(\lambda) \le \operatorname{am}(\lambda).

The lower bound holds because an eigenvalue has an eigenvector. For the upper bound, let $g = \operatorname{gm}(\lambda)$ , choose a basis $\mathbf{p}_1, \dots, \mathbf{p}_g$ of $E_\lambda$ , extend it to a basis $\mathbf{p}_1, \dots, \mathbf{p}_n$ of $\mathbb{R}^n$ , and let $P$ be the invertible matrix with these columns. For $j \le g$ we have $P^{-1}AP\,\mathbf{e}_j = P^{-1}A\mathbf{p}_j = \lambda P^{-1}\mathbf{p}_j = \lambda\mathbf{e}_j$ , so in block form

P^{-1}AP = \begin{bmatrix} \lambda I_g & C \\ 0 & B \end{bmatrix}

for some $g \times (n - g)$ block $C$ and $(n - g) \times (n - g)$ block $B$ . When $g = n$ there are no such blocks, so $A = \lambda I$ and $\operatorname{am}(\lambda) = n = g$ . Suppose from now on that $g < n$ . A change of basis does not change the characteristic polynomial, because $tI - P^{-1}AP = P^{-1}(tI - A)P$ and $\det P^{-1} \det P = 1$ . Expanding $g$ times along the first column, where the only nonzero entry is $t - \lambda$ , gives

\begin{aligned} \chi_A(t) &= \det(tI - P^{-1}AP) \\[0.4em] &= (t - \lambda)^g \chi_B(t). \end{aligned}

So $t - \lambda$ occurs at least $g$ times in $\chi_A$ , and $\operatorname{am}(\lambda) \ge g$ . It can occur more often, because $\chi_B(\lambda)$ may also be zero.

In plain words An eigenvalue can supply at most as many independent eigenvectors as the number of times it occurs as a root. It always supplies at least one, and it can supply fewer than its root count.

Example The matrix $A$ with rows $(4, 1)$ and $(-1, 2)$ has $\chi_A(t) = (t - 4)(t - 2) + 1 = (t - 3)^2$ , and $E_3 = \operatorname{span}\{(1, -1)\}$ , so $\operatorname{gm}(3) = 1 < 2 = \operatorname{am}(3)$ . Following the proof, extend $(1, -1)$ by $\mathbf{e}_2$ , so

\begin{gathered} P = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}, \\[0.4em] P^{-1}AP = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}. \end{gathered}

Here $g = 1$ , $C = [\,1\,]$ and $B = [\,3\,]$ , and $\chi_A(t) = (t - 3)\chi_B(t)$ with $\chi_B(t) = t - 3$ , which supplies the second factor without a second eigenvector.

Defective matrix

Definition A real $n \times n$ matrix $A$ is defective if at least one of its eigenvalues $\lambda$ has $\operatorname{gm}(\lambda) < \operatorname{am}(\lambda)$ . A defective matrix has fewer than $n$ linearly independent eigenvectors, so no basis of $\mathbb{R}^n$ consists of its eigenvectors. Indeed, in any independent list of eigenvectors the ones that belong to $\lambda$ lie in $E_\lambda$ , so there are at most $\operatorname{gm}(\lambda)$ of them, and the list has length at most

\sum_{\lambda} \operatorname{gm}(\lambda) < \sum_{\lambda} \operatorname{am}(\lambda) \le n.

Since $\operatorname{gm}(\lambda) \ge 1$ , a defective matrix always has a repeated eigenvalue, and a matrix with $n$ distinct real eigenvalues is never defective. Neither statement reverses, as the diagonal matrix with entries $2, 2, 3$ in the example below shows. When $\chi_A$ factors completely into real linear factors, a matrix that is not defective has $n$ independent eigenvectors, because its geometric multiplicities then add up to $\sum_{\lambda} \operatorname{am}(\lambda) = n$ and bases of the separate eigenspaces, listed together, stay independent. The quarter turn with rows $(0, -1)$ and $(1, 0)$ has no real eigenvector for a different reason, since $t^2 + 1$ has no real root. Once complex eigenvalues are allowed, the same comparison of the two multiplicities applies to them as well, and a matrix is also called defective when a nonreal eigenvalue falls short. The quarter turn is not defective, because the two complex roots of $t^2 + 1$ are distinct, so each of them is simple.

In plain words A defective matrix runs short of eigenvectors. Some repeated eigenvalue supplies fewer independent directions than its root count, and the missing directions cannot be made up by the other eigenvalues.

Example The matrix $J$ is defective, because $\operatorname{gm}(2) = 1 < 2 = \operatorname{am}(2)$ , and its eigenvectors all lie on the horizontal axis. For $H$ above, the matrix $H - 4I$ has a single nonzero entry, a $1$ in row $2$ and column $3$ , so its only equation is $z = 0$ . Then $E_4 = \operatorname{span}\{\mathbf{e}_1, \mathbf{e}_2\}$ and $\operatorname{gm}(4) = 2 < 3 = \operatorname{am}(4)$ , so $H$ is defective, and every eigenvector of $H$ lies in the plane $z = 0$ . The matrix $4I$ of size $3 \times 3$ has the same characteristic polynomial $(t - 4)^3$ , but its eigenspace is all of $\mathbb{R}^3$ , so it is not defective. The diagonal matrix with entries $2, 2, 3$ has a repeated eigenvalue and is not defective either.

Check yourself

Find the eigenvalues of the matrix with rows $(3, 0, 2)$ , $(0, 3, 0)$ and $(0, 0, 5)$ , with their algebraic and geometric multiplicities. Is the matrix defective?

Show the answer

Call the matrix $A$ . It is triangular, so $\chi_A(t) = (t - 3)^2(t - 5)$ , with $\operatorname{am}(3) = 2$ and $\operatorname{am}(5) = 1$ . The matrix $A - 3I$ has rows $(0, 0, 2)$ , $(0, 0, 0)$ and $(0, 0, 2)$ , so its only condition is $z = 0$ . With $x$ and $y$ free, $E_3 = \operatorname{span}\{\mathbf{e}_1, \mathbf{e}_2\}$ and $\operatorname{gm}(3) = 2$ . The matrix $A - 5I$ has rows $(-2, 0, 2)$ , $(0, -2, 0)$ and $(0, 0, 0)$ , which give $x = z$ and $y = 0$ , so $E_5 = \operatorname{span}\{(1, 0, 1)\}$ and $\operatorname{gm}(5) = 1$ . Check that $A(1, 0, 1) = (5, 0, 5)$ . For each eigenvalue the two multiplicities agree, so the matrix is not defective, and $\mathbf{e}_1$ , $\mathbf{e}_2$ , $(1, 0, 1)$ are three independent eigenvectors.

For the matrix with rows $(3, -1)$ and $(1, 1)$ , a student finds the repeated eigenvalue $2$ , writes down the eigenvectors $(1, 1)$ and $(2, 2)$ , and concludes that the geometric multiplicity is $2$ . What went wrong?

Show the answer

Call the matrix $A$ . Its characteristic polynomial is $(t - 3)(t - 1) + 1 = t^2 - 4t + 4 = (t - 2)^2$ , so $\operatorname{am}(2) = 2$ is correct. But $(2, 2) = 2(1, 1)$ , so the two vectors are dependent and span one line. The matrix $A - 2I$ has rows $(1, -1)$ and $(1, -1)$ , of rank $1$ , so

\operatorname{gm}(2) = 2 - 1 = 1.

Every eigenvector is a nonzero multiple of $(1, 1)$ . Since $\operatorname{gm}(2) < \operatorname{am}(2)$ , the matrix is defective. The geometric multiplicity counts independent eigenvectors, so listing multiples of one eigenvector adds nothing.

For which real numbers $a$ and $b$ is the matrix with rows $(2, a, b)$ , $(0, 2, 0)$ and $(0, 0, 3)$ defective?

Show the answer

The matrix is triangular, so $\chi(t) = (t - 2)^2(t - 3)$ for all $a$ and $b$ . The simple eigenvalue $3$ has $1 \le \operatorname{gm}(3) \le \operatorname{am}(3) = 1$ , so only the eigenvalue $2$ can cause trouble. Subtracting $2I$ gives the matrix with rows $(0, a, b)$ , $(0, 0, 0)$ and $(0, 0, 1)$ .

If $a \ne 0$ , the rows $(0, a, b)$ and $(0, 0, 1)$ are independent, the rank is $2$ , and $\operatorname{gm}(2) = 3 - 2 = 1 < 2$ . The matrix is defective, with $E_2 = \operatorname{span}\{\mathbf{e}_1\}$ .

If $a = 0$ , the row $(0, 0, b)$ is $b$ times $(0, 0, 1)$ , the rank is $1$ , and $\operatorname{gm}(2) = 2$ , with $E_2 = \operatorname{span}\{\mathbf{e}_1, \mathbf{e}_2\}$ . The matrix is not defective.

So the matrix is defective exactly when $a \ne 0$ , whatever the value of $b$ .

Algebraic and Geometric Multiplicity

Key terms

Repeated eigenvalue

Algebraic multiplicity

Geometric multiplicity

Bound on geometric multiplicity

Defective matrix

Check yourself

How this lesson connects

Builds on

Used later in