Linear-Algebra-Done-Right · Unit 25 · Video 1 · Interactive Practice

Generalized Eigenspaces Split Every Complex Space

IKey Formulas

Formula	Name	Hypotheses
$G(\lambda, T) = \{v \in V : (T - \lambda I)^k v = 0 \text{ for some } k \ge 1\}$	Generalized eigenspace	$T \in \mathcal{L}(V)$ , $\lambda \in \mathbf{F}$
$E(\lambda, T) \subseteq G(\lambda, T) = \operatorname{null}(T - \lambda I)^{\dim V}$	Description of $G(\lambda, T)$	$V$ finite-dimensional, any $\mathbf{F}$
$V = G(\lambda_1, T) \oplus \cdots \oplus G(\lambda_m, T)$	Generalized eigenspace decomposition	$\mathbf{F} = \mathbf{C}$ , $V$ finite-dimensional, $\lambda_1, \dots, \lambda_m$ the distinct eigenvalues of $T$
$T\|_{G(\lambda_k, T)} = \lambda_k I + N_k$ , $\ N_k$ nilpotent	Scalar plus nilpotent on each piece	same as the decomposition

Key Insight: The power $k$ in the definition may depend on $v$ , but the single power $\dim V$ works for every vector at once, so $G(\lambda, T)$ is a null space and hence a subspace.

IIThe Null Spaces of the Powers Stop Growing

The null spaces of $(T - \lambda I)^k$ grow from $E(\lambda, T)$ and reach $G(\lambda, T)$ by the power $k = \dim V$ .

Operator λ Power k 1

IIIOne Vector, Two Pieces

For $S$ on $\mathbf{C}^3$ , every vector is uniquely a vector of $G(3, S)$ plus a vector of $G(1, S)$ .

z₁ 2 z₂ 1 z₃ 4

💡 Here $G(3, S) = \operatorname{span}\big((1,0,0), (0,1,0)\big)$ is a plane but $E(3, S) = \operatorname{span}\big((1,0,0)\big)$ is only a line, while $G(1, S) = E(1, S) = \operatorname{span}\big((1, 2, -4)\big)$ .

IVQuiz Questions

Problem 1 · Eigenspace or Generalized Eigenspace?

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ , $T(z_1, z_2, z_3) = (2z_1 + z_3,\ 2z_2,\ 2z_3)$ — find $G(2, T)$ and $G(5, T)$ .

$G(2, T)$

$G(5, T)$

✅ Correct!

(T - 2I)^2 = 0

, so every vector is in

G(2, T)

, although

E(2, T)

is only a plane; and

5

is not an eigenvalue, so

G(5, T) = \{0\}

❌ That is the eigenspace $E(2, T)$ .

(T - 2I)(z_1, z_2, z_3) = (z_3, 0, 0)

vanishes only when

z_3 = 0

, but

(T - 2I)^2 = 0

, so a higher power kills every vector.

❌ Not quite. Compute

(T - 2I)(z_1, z_2, z_3) = (z_3, 0, 0)

, then apply

T - 2I

again.

❌ Not quite. The definition of

G(\lambda, T)

makes sense for every

\lambda \in \mathbf{F}

; when

\lambda

is not an eigenvalue,

T - \lambda I

is injective, and so is every power of it.

Show solution

Subtract $2I$ :

(T - 2I)(z_1, z_2, z_3) = (z_3, 0, 0), \qquad (T - 2I)^2(z_1, z_2, z_3) = (T - 2I)(z_3, 0, 0) = (0, 0, 0).

So $E(2, T) = \operatorname{null}(T - 2I) = \{(z_1, z_2, 0)\}$ , a plane, while $(T - 2I)^2 = 0$ gives $G(2, T) = \operatorname{null}(T - 2I)^3 = \mathbf{C}^3$ .

Since $\mathbf{C}^3 = G(2, T)$ and the generalized eigenspaces for distinct eigenvalues form a direct sum, $2$ is the only eigenvalue. In particular $T - 5I$ is injective, hence so is $(T - 5I)^3$ , and $G(5, T) = \operatorname{null}(T - 5I)^3 = \{0\}$ .

Problem 2 · One Power for Every Vector

Given: $\dim V = 6$ , $T \in \mathcal{L}(V)$ , $\lambda \in \mathbf{F}$ , and a vector $v \in V$ with $(T - \lambda I)^{9} v = 0$ — find the statement that must hold.

✅ Correct!

v \in \operatorname{null}(T - \lambda I)^9 \subseteq G(\lambda, T) = \operatorname{null}(T - \lambda I)^{\dim V}

, and

\dim V = 6

❌ That would make $v$ an eigenvector or $0$ . A generalized eigenvector need not be an eigenvector: for

N(z_1, z_2) = (z_2, 0)

and

\lambda = 0

, the vector

(0, 1)

satisfies

N^2 v = 0

but

Nv \ne 0

❌ A smaller power is guaranteed. The null spaces of the powers of

T - \lambda I

stop growing by the power

\dim V = 6

❌ Not necessarily. The vector

v

could be

0

; then nothing follows about

\lambda

Show solution

By definition $v \in G(\lambda, T)$ , with $k = 9$ . The description of generalized eigenspaces gives

G(\lambda, T) = \operatorname{null}(T - \lambda I)^{\dim V} = \operatorname{null}(T - \lambda I)^{6},

so $(T - \lambda I)^6 v = 0$ . Option (a) fails for a generalized eigenvector that is not an eigenvector, and option (c) fails for $v = 0$ .

Problem 3 · Split a Vector Between Two Pieces

Given: $R \in \mathcal{L}(\mathbf{C}^3)$ , $R(z_1, z_2, z_3) = (-z_1,\ z_1 + 2z_2,\ z_2 + 2z_3)$ , whose eigenvalues are $-1$ and $2$ — find the component in $G(-1, R)$ of $v = (9, 0, 0)$ in the decomposition $\mathbf{C}^3 = G(-1, R) \oplus G(2, R)$ , and the smallest $k$ with $\big((R - 2I)|_{G(2, R)}\big)^k = 0$ .

Component of $(9, 0, 0)$ in $G(-1, R)$

Smallest $k$ with $\big((R - 2I)|_{G(2, R)}\big)^k = 0$

✅ Correct!

(9, 0, 0) = (9, -3, 1) + (0, 3, -1)

with

(9, -3, 1) \in G(-1, R)

and

(0, 3, -1) \in G(2, R)

; on

G(2, R)

the operator

R - 2I

sends

e_2 \mapsto e_3 \mapsto 0

❌ $(9, 0, 0)$ is not in $G(-1, R)$ .

(R + I)(9, 0, 0) = (0, 9, 0) \neq 0

; the generalized eigenspaces here are not coordinate subspaces.

❌ That is the other component.

(0, 3, -1)

lies in

G(2, R) = \{(0, z_2, z_3)\}

❌ Not quite. First find

G(-1, R) = \operatorname{null}(R + I)

, then subtract the right multiple of its spanning vector from

v

so that what is left has first coordinate

0

❌ Not quite. Restrict to

G(2, R) = \operatorname{span}(e_2, e_3)

and compute

(R - 2I)e_2

and

(R - 2I)e_3

Show solution

$G(-1, R)$ . $(R + I)(z_1, z_2, z_3) = (0,\ z_1 + 3z_2,\ z_2 + 3z_3)$ . Its null space is the line $z_2 = -3z_3$ , $z_1 = 9z_3$ , spanned by $(9, -3, 1)$ ; since $\dim G(-1, R) + \dim G(2, R) = 3$ and $\dim G(2, R) \ge 2$ (below), $G(-1, R)$ is exactly this line.

$G(2, R)$ . $(R - 2I)(z_1, z_2, z_3) = (-3z_1,\ z_1,\ z_2)$ and

(R - 2I)^2(z_1, z_2, z_3) = (9z_1,\ -3z_1,\ z_1), \qquad (R - 2I)^3(z_1, z_2, z_3) = (-27z_1,\ 9z_1,\ -3z_1),

so $G(2, R) = \operatorname{null}(R - 2I)^3 = \{(0, z_2, z_3)\}$ .

Splitting $v$ . Write $(9, 0, 0) = c(9, -3, 1) + (0, z_2, z_3)$ . The first coordinate gives $c = 1$ , so the components are $(9, -3, 1) \in G(-1, R)$ and $(0, 3, -1) \in G(2, R)$ .

The nilpotent part. On $G(2, R)$ , $(R - 2I)e_2 = e_3$ and $(R - 2I)e_3 = 0$ . So the restriction is not $0$ , but its square sends both $e_2$ and $e_3$ to $0$ : the smallest $k$ is $2$ , and $R|_{G(2, R)} = 2I + N$ with $N^2 = 0$ , $N \neq 0$ .

Problem 4 · Why $\mathbf{F} = \mathbf{C}$ ?

Given: $Q \in \mathcal{L}(\mathbf{R}^3)$ , $Q(x, y, z) = (-y,\ x,\ 3z)$ — find the sum $G(\lambda_1, Q) + \cdots + G(\lambda_m, Q)$ over the distinct real eigenvalues of $Q$ .

✅ Correct! The only real eigenvalue is

3

, with

G(3, Q)

the

z

-axis; the plane rotation contributes nothing over

\mathbf{R}

, so the decomposition needs

\mathbf{F} = \mathbf{C}

❌ The decomposition needs $\mathbf{F} = \mathbf{C}$ . On the

xy

-plane

Q

is a quarter turn, and no nonzero vector of that plane is a generalized eigenvector for a real

\lambda

❌ That plane is invariant, but it is no generalized eigenspace.

Q

rotates it by a right angle, so no power of

Q - \lambda I

with real

\lambda

kills a nonzero vector of it.

❌ $Q$ has a real eigenvalue.

Q(0, 0, 1) = (0, 0, 3)

, so

3

is an eigenvalue.

Show solution

$Q(0, 0, 1) = 3(0, 0, 1)$ , so $3$ is an eigenvalue. $(Q - 3I)(x, y, z) = (-y - 3x,\ x - 3y,\ 0)$ ; on the $xy$ -plane this is the matrix $\begin{pmatrix} -3 & -1 \\ 1 & -3 \end{pmatrix}$ , which is invertible (it is $-3I$ plus a quarter turn), and so are its powers. Hence $G(3, Q) = \operatorname{null}(Q - 3I)^3 = \{(0, 0, z)\}$ .

For a real $\lambda \neq 3$ , $Q - \lambda I$ is invertible on the plane (no real eigenvalue there) and multiplies $z$ by $3 - \lambda \neq 0$ , so $G(\lambda, Q) = \{0\}$ . The sum of the generalized eigenspaces is therefore the $z$ -axis, a proper subspace of $\mathbf{R}^3$ . Over $\mathbf{C}$ the eigenvalues $\pm i$ would supply the missing plane.

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