Linear-Algebra-Done-Right · Unit 25 · Video 2 · Interactive Practice

Multiplicity Without Determinants: Algebraic vs Geometric

IKey Formulas

Formula	Name	Hypotheses
multiplicity of $\lambda$ $= \dim G(\lambda, T) = \dim \operatorname{null}(T - \lambda I)^{\dim V}$	Multiplicity (algebraic multiplicity)	$V$ finite-dimensional, $\lambda$ an eigenvalue of $T \in \mathcal{L}(V)$
$\dim E(\lambda, T) = \dim \operatorname{null}(T - \lambda I) \le \dim G(\lambda, T)$	Geometric multiplicity, never larger	same
$d_1 + \cdots + d_m = \dim V$	Sum of the multiplicities	$\mathbf{F} = \mathbf{C}$ ; $d_k$ the multiplicity of the distinct eigenvalue $\lambda_k$
$G(\lambda, T) = E(\lambda, T)$	The two multiplicities agree	$V$ a finite-dimensional inner product space, $T$ normal

Key Insight: The two multiplicities differ only in the exponent: the null space of $T - \lambda I$ itself, or of its power $\dim V$ . Over $\mathbf{C}$ , $T$ has a basis of eigenvectors exactly when they agree for every eigenvalue.

IIFrom Eigenspace to Generalized Eigenspace

The geometric multiplicity is the dimension at the power $1$ ; the multiplicity is the dimension at the power $\dim V$ .

Operator λ

IIIDo the Eigenvectors Fill $V$ ?

The multiplicities always add up to $\dim V$ over $\mathbf{C}$ ; the dimensions of the eigenspaces add up to $\dim V$ only sometimes.

Operator

IVQuiz Questions

Problem 1 · Two Multiplicities of One Eigenvalue

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ with matrix $\begin{pmatrix} 2 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$ in the standard basis — find the multiplicity and the geometric multiplicity of the eigenvalue $2$ .

Multiplicity $\dim G(2, T)$

Geometric multiplicity $\dim E(2, T)$

✅ Correct!

T - 2I

has rank

2

, so

\dim E(2, T) = 2

, while

(T - 2I)^4

has rank

1

, so

\dim G(2, T) = 3

❌ That is the geometric multiplicity. The multiplicity uses the power

\dim V = 4

T - 2I

, whose null space is larger.

❌ $-1$ is also an eigenvalue.

e_4

is an eigenvector for

-1

, and

G(-1, T) \cap G(2, T) = \{0\}

, so

\dim G(2, T) \le 3

❌ Not quite. Compute

(T - 2I)^2

and the dimension of its null space; the power stops mattering after that.

❌ Not quite. Find the null space of

T - 2I

: which coordinates must vanish?

Show solution

$(T - 2I)(z_1, z_2, z_3, z_4) = (z_3,\ 0,\ 0,\ -3z_4)$ , so $E(2, T) = \{(z_1, z_2, 0, 0)\}$ and the geometric multiplicity is $2$ .

Squaring: $(T - 2I)^2(z_1, z_2, z_3, z_4) = (T - 2I)(z_3, 0, 0, -3z_4) = (0, 0, 0, 9z_4)$ , and every higher power is $(0, 0, 0, (-3)^k z_4)$ . So

G(2, T) = \operatorname{null}(T - 2I)^4 = \{(z_1, z_2, z_3, 0)\}, \qquad \dim G(2, T) = 3.

The multiplicity is $3$ ; with the multiplicity $1$ of the eigenvalue $-1$ the sum is $4 = \dim \mathbf{C}^4$ .

Problem 2 · A Real Operator

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ , $T(x, y, z) = (-y,\ x,\ 0)$ — find the sum of the multiplicities of all the (real) eigenvalues of $T$ .

✅ Correct! The only real eigenvalue is

0

, with

G(0, T)

the

z

-axis; the sum

1

falls short of

\dim \mathbf{R}^3 = 3

, so the theorem needs

\mathbf{F} = \mathbf{C}

❌ The sum equals $\dim V$ only over $\mathbf{C}$ . On the

xy

-plane

T

is a quarter turn, which contributes no real eigenvalue.

❌ $T$ has a real eigenvalue.

T(0, 0, 1) = (0, 0, 0)

, so

0

is an eigenvalue.

❌ The $xy$ -plane is not in $G(0, T)$ .

T

rotates that plane, so

T^k(x, y, 0) \neq 0

unless

x = y = 0

Show solution

If $Tv = \lambda v$ with $\lambda \in \mathbf{R}$ and $v = (x, y, z) \ne 0$ , then $-y = \lambda x$ , $x = \lambda y$ , $0 = \lambda z$ . The first two give $x = -\lambda^2 x$ , so $x = 0$ , then $y = 0$ , and $z \ne 0$ forces $\lambda = 0$ . So $0$ is the only real eigenvalue.

$T^k(x, y, z)$ is the $k$ -th quarter turn of $(x, y)$ followed by $0$ in the last coordinate, so $T^3(x, y, z) = 0$ exactly when $x = y = 0$ : $G(0, T) = \{(0, 0, z)\}$ has dimension $1$ . The sum of the multiplicities is $1 \neq 3$ .

Problem 3 · Is There an Eigenbasis?

Given: $T \in \mathcal{L}(\mathbf{C}^5)$ has exactly the eigenvalues $0$ and $3$ , $\dim \operatorname{null} T^5 = 3$ and $\dim E(3, T) = 1$ — find the multiplicity of $3$ , and decide whether $\mathbf{C}^5$ has a basis of eigenvectors of $T$ .

Multiplicity of $3$

A basis of eigenvectors?

✅ Correct! The multiplicities add to

5

, so the multiplicity of

3

5 - 3 = 2

; but

\dim E(3, T) = 1 < 2

, so the eigenspaces cannot fill

\mathbf{C}^5

❌ That is the geometric multiplicity. The multiplicity is

\dim G(3, T)

, which can exceed

\dim E(3, T)

❌ The data determine it.

\dim \operatorname{null} T^5 = \dim G(0, T)

, and over

\mathbf{C}

the multiplicities add up to

\dim V

❌ Not quite. Over

\mathbf{C}

\dim G(0, T) + \dim G(3, T) = 5

❌ Not quite. An eigenbasis exists exactly when

\dim E(\lambda, T) = \dim G(\lambda, T)

for every eigenvalue; check

\lambda = 3

Show solution

$\dim V = 5$ , so $\dim G(0, T) = \dim \operatorname{null} T^5 = 3$ . Over $\mathbf{C}$ the multiplicities add up to $\dim V$ :

\dim G(3, T) = 5 - 3 = 2.

$T$ has a basis of eigenvectors exactly when $\dim E(0, T) + \dim E(3, T) = 5$ . Since $\dim E(0, T) \le \dim G(0, T) = 3$ and $\dim E(3, T) = 1$ , the sum is at most $4$ . So there is no basis of eigenvectors, whatever $\dim E(0, T)$ is.

Problem 4 · A Normal Operator

Given: $T$ is a normal operator on $\mathbf{C}^4$ (standard inner product) and $\dim \operatorname{null}(T - 5I) = 2$ — find $\dim \operatorname{null}(T - 5I)^4$ .

✅ Correct!

T - 5I

is normal, and every positive power of a normal operator has the same null space as the operator, so

G(5, T) = E(5, T)

❌ $5$ need not be the only eigenvalue. For a normal operator the multiplicity equals the geometric multiplicity, which is

2

❌ Normality pins it down. That range is right for a general operator, but for a normal one the null spaces of all the powers of

T - 5I

coincide.

❌ Not for a normal operator. If

(T - 5I)^2 v = 0

, then

(T - 5I)v

is in both the range and the null space of

T - 5I

, which are orthogonal, so

(T - 5I)v = 0

Show solution

$N = T - 5I$ is normal: $N^* = T^* - 5I$ , and $N N^* = N^* N$ because $T T^* = T^* T$ .

For a normal $N$ , $\operatorname{null} N = \operatorname{null} N^* = (\operatorname{range} N)^\perp$ . If $N^2 v = 0$ , then $Nv \in \operatorname{range} N \cap \operatorname{null} N$ , so $Nv$ is orthogonal to itself and $Nv = 0$ . Repeating, $N^k v = 0$ implies $Nv = 0$ for every $k \ge 1$ .

\operatorname{null}(T - 5I)^4 = \operatorname{null}(T - 5I), \qquad \dim = 2.

Solved: 0 / 4

Multiplicity Without Determinants: Algebraic vs Geometric

IKey Formulas

IIFrom Eigenspace to Generalized Eigenspace

IIIDo the Eigenvectors Fill VVV?

IVQuiz Questions

IIIDo the Eigenvectors Fill $V$ ?