Linear-Algebra-Done-Right · Unit 24 · Video 2 · Interactive Practice

Generalized Eigenvectors Span Every Complex Space

IKey Formulas

Formula	Name	Hypotheses
$v \ne 0$ and $(T - \lambda I)^k v = 0$ for some positive integer $k$	$v$ is a generalized eigenvector of $T$ corresponding to $\lambda$	$T \in \mathcal{L}(V)$ , $\lambda$ an eigenvalue of $T$ ; $k = 1$ gives the eigenvectors
$v \ne 0$ is a generalized eigenvector for $\lambda$ $\iff (T - \lambda I)^{n} v = 0$	One power is enough to test	$V$ finite-dimensional, $n = \dim V \ge 1$ ; the null spaces of the powers of $T - \lambda I$ stop growing by the power $n$
$(T - \lambda I)^k v = 0$ , $v \ne 0$ $\Rightarrow$ $\lambda$ is an eigenvalue of $T$	No new "generalized eigenvalues"	$T \in \mathcal{L}(V)$ , $k$ a positive integer; if $T - \lambda I$ were injective, so would be its powers
$V$ has a basis of generalized eigenvectors of $T$ ; proof step: $V = \operatorname{null}(T - \lambda I)^n \oplus \operatorname{range}(T - \lambda I)^n$	Basis of generalized eigenvectors	$\mathbf{F} = \mathbf{C}$ , $V$ finite-dimensional, $n = \dim V$ , $\lambda$ an eigenvalue of $T$ ; induction on $n$

Key Insight: Eigenvectors can leave part of $V$ unspanned, but allowing powers of $T - \lambda I$ fixes this over $\mathbf{C}$ : the null space of $(T - \lambda I)^n$ consists of generalized eigenvectors (plus $0$ ), and the invariant range of $(T - \lambda I)^n$ has smaller dimension, so induction covers it.

IIApply $T - \lambda I$ Until It Vanishes

For $T(z_1, z_2, z_3) = (4z_2,\ 0,\ 5z_3)$ , a nonzero $v$ is a generalized eigenvector for $\lambda$ exactly when $(T - \lambda I)^3 v = 0$ .

λ z₁ 1 z₂ 1 z₃ 0

IIISplit, Then Induct

The inductive step splits $\mathbf{C}^3$ as $\operatorname{null}(T - \lambda I)^3 \oplus \operatorname{range}(T - \lambda I)^3$ and finds generalized eigenvectors in each piece.

Eigenvalue used Step 1

IVOver $\mathbf{R}$ the First Step Can Fail

The proof needs an eigenvalue: the rotation of $\mathbf{R}^2$ has none, while $N(x, y) = (y,\ 0)$ has a basis of generalized eigenvectors.

Operator on R²

VQuiz Questions

Problem 1 · Generalized but Not an Eigenvector

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with $T(z_1, z_2, z_3) = (3z_1 + z_2,\ 3z_2,\ -2z_3)$ , whose matrix $\begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \end{pmatrix}$ is upper triangular, so its eigenvalues are $3$ and $-2$ . Which vector is a generalized eigenvector of $T$ for $\lambda = 3$ but not an eigenvector of $T$ ?

✅ Correct!

(T - 3I)(0, 1, 0) = (1, 0, 0) \ne 0

, so it is not an eigenvector, but

(T - 3I)^2(0, 1, 0) = (T - 3I)(1, 0, 0) = 0

: a generalized eigenvector with

k = 2

❌ That one is an eigenvector.

(T - 3I)(1, 0, 0) = (0, 0, 0)

already at

k = 1

, so

(1, 0, 0)

is an eigenvector for

3

; it is a generalized eigenvector, but the question excludes eigenvectors.

❌ Wrong eigenvalue.

T(0, 0, 1) = (0, 0, -2)

, so

(0, 0, 1)

is an eigenvector for

-2

. For

\lambda = 3

(T - 3I)^k (0, 0, 1) = \big(0, 0, (-5)^k\big) \ne 0

for every

k

❌ Mixing eigenvalues spoils it.

(T - 3I)^3(0, 1, 1) = (0, 0, -125) \ne 0

: the

(0, 0, 1)

part belongs to

-2

, and no power of

T - 3I

removes it, so

(0, 1, 1)

is not a generalized eigenvector for

3

Show solution

Here $(T - 3I)(z_1, z_2, z_3) = (z_2,\ 0,\ -5z_3)$ and $n = \dim \mathbf{C}^3 = 3$ .

(T - 3I)(1,0,0) = (0,0,0),\qquad (T - 3I)(0,1,0) = (1,0,0),\qquad (T - 3I)^2(0,1,0) = (0,0,0)

(T - 3I)^3(0,0,1) = (0,0,-125),\qquad (T - 3I)^3(0,1,1) = (0,0,-125)

By the power test, the generalized eigenvectors for $3$ are the nonzero vectors with $(T - 3I)^3 v = 0$ , that is, the nonzero $(z_1, z_2, 0)$ . Among the options, $(1, 0, 0)$ and $(0, 1, 0)$ qualify, and only $(0, 1, 0)$ fails the eigenvector condition $(T - 3I)v = 0$ . Answer: $(0, 1, 0)$ .

Problem 2 · Which Power Kills It?

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ , so $1$ is its only eigenvalue, and $v = (2, -1, 1)$ . Find the smallest positive integer $k$ with $(T - I)^k v = 0$ , if there is one.

✅ Correct!

(T - I)^2 v = (1, 0, 0) \ne 0

but

(T - I)^3 v = 0

, so

v

is a generalized eigenvector for

1

that needs exactly the power

\dim \mathbf{C}^3 = 3

❌ Not an eigenvector.

(T - I)v = (-1, 1, 0) \ne 0

, so

k = 1

fails:

v

is not an eigenvector of

T

❌ One more power.

(T - I)^2 v = (T - I)(-1, 1, 0) = (1, 0, 0) \ne 0

, so

k = 2

fails too.

❌ You stopped too early. The test power is

\dim V = 3

, not

2

(T - I)^3 v = (T - I)(1, 0, 0) = (0, 0, 0)

, so

v

is a generalized eigenvector.

Show solution

$T - I$ has matrix $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ , so $(T - I)(z_1, z_2, z_3) = (z_2,\ z_3,\ 0)$ . Apply it one power at a time:

(T - I)(2, -1, 1) = (-1, 1, 0),\qquad (T - I)^2 v = (1, 0, 0),\qquad (T - I)^3 v = (0, 0, 0).

The first zero appears at $k = 3$ . This is the largest power the test ever needs, since $n = \dim \mathbf{C}^3 = 3$ : had $(T - I)^3 v$ been nonzero, no higher power could have sent $v$ to $0$ .

Problem 3 · Assemble the Basis

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & -1 \end{pmatrix}$ , eigenvalues $2$ and $-1$ . Its eigenvectors are the nonzero multiples of $(1, 0, 0)$ and of $(1, -3, 9)$ . Which list is a basis of $\mathbf{C}^3$ consisting of generalized eigenvectors of $T$ ?

✅ Correct!

\operatorname{null}(T - 2I)^3 = \{(z_1, z_2, 0)\}

supplies

(1, 0, 0)

and

(0, 1, 0)

, the eigenvector

(1, -3, 9)

is a generalized eigenvector for

-1

, and the three vectors are linearly independent.

❌ The standard basis fails here.

(T - 2I)^3(0, 0, 1) = (-3, 9, -27) \ne 0

and

(T + I)^3(0, 0, 1) = (6, 9, 0) \ne 0

, so

(0, 0, 1)

is not a generalized eigenvector for either eigenvalue.

❌ Eigenvectors alone run out. These two vectors are eigenvectors, hence generalized eigenvectors, but two vectors cannot span

\mathbf{C}^3

: this is exactly why

T

is not diagonalizable.

❌ The theorem says otherwise. Over

\mathbf{C}

every operator on a finite-dimensional space has a basis of generalized eigenvectors; the missing direction is supplied by

(0, 1, 0)

, which satisfies

(T - 2I)^2(0, 1, 0) = 0

Show solution

Use the power test with $n = 3$ . For $\lambda = 2$ , $(T - 2I)(z) = (z_2,\ z_3,\ -3z_3)$ and

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

which is $0$ exactly when $z_3 = 0$ : the generalized eigenvectors for $2$ are the nonzero $(z_1, z_2, 0)$ , a plane, while the eigenvectors for $2$ fill only the line through $(1, 0, 0)$ .

For $\lambda = -1$ , $(T + I)^3(z_1, z_2, z_3) = (27z_1 + 27z_2 + 6z_3,\ 27z_2 + 9z_3,\ 0)$ , which vanishes exactly on the multiples of $(1, -3, 9)$ .

Take $(1, 0, 0)$ and $(0, 1, 0)$ for $2$ and $(1, -3, 9)$ for $-1$ . They are linearly independent: in $a(1,0,0) + b(0,1,0) + c(1,-3,9) = 0$ the third coordinate gives $c = 0$ , then $a = b = 0$ . Three independent vectors in $\mathbf{C}^3$ form a basis. Option (b) fails because $(0, 0, 1)$ is not a generalized eigenvector at all.

Problem 4 · Run the Inductive Step

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $\begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 3 \end{pmatrix}$ , so $T(z_1, z_2, z_3) = (z_2 + z_3,\ z_3,\ 3z_3)$ , eigenvalues $0$ and $3$ . Run the proof's inductive step with $\lambda = 3$ : $\mathbf{C}^3 = \operatorname{null}(T - 3I)^3 \oplus \operatorname{range}(T - 3I)^3$ , where $\operatorname{null}(T - 3I)^3 = \operatorname{span}\big((4, 3, 9)\big)$ .

What is range (T − 3I)³?

With S = T restricted to that range, its nonzero vectors are generalized eigenvectors of T for…

✅ Correct!

S(z_1, z_2, 0) = (z_2, 0, 0)

and

S^2 = 0

, so the range supplies

(1, 0, 0)

(0, 1, 0)

for

0

; with

(4, 3, 9)

for

3

they form a basis of generalized eigenvectors of

T

❌ That is the null space.

\operatorname{span}\big((4, 3, 9)\big)

\operatorname{null}(T - 3I)^3

; the range is the other summand, of dimension

3 - 1 = 2

❌ Check the third coordinate.

T - 3I

sends

(z_1, z_2, z_3)

(-3z_1 + z_2 + z_3,\ -3z_2 + z_3,\ 0)

, so every output of

(T - 3I)^3

has third coordinate

0

;

(0, 0, 1)

is not in the range.

❌ Not surjective.

3

is an eigenvalue, so

T - 3I

is not injective and neither is

(T - 3I)^3

; since

\dim \operatorname{null}(T - 3I)^3 = 1

, the range has dimension

2

❌ The range carries the other eigenvalues. A generalized eigenvector for

3

lies in

\operatorname{null}(T - 3I)^3

by the power test, and that null space meets the range only in

0

because the sum is direct.

❌ That is what the induction is for.

S

is an operator on a complex space of dimension

2 < 3

, so the induction hypothesis gives a basis of generalized eigenvectors of

S

, and these are generalized eigenvectors of

T

❌ Those are only the eigenvectors of $S$ .

S(0, 1, 0) = (1, 0, 0)

and

S^2(0, 1, 0) = 0

, so

(0, 1, 0)

is a generalized eigenvector for

0

with

k = 2

; in fact

S^2 = 0

on the whole range.

Show solution

The range. $(T - 3I)(z_1, z_2, z_3) = (-3z_1 + z_2 + z_3,\ -3z_2 + z_3,\ 0)$ , and

(T - 3I)^3 = \begin{pmatrix} -27 & 27 & 3 \\ 0 & -27 & 9 \\ 0 & 0 & 0 \end{pmatrix}.

Its columns lie in $\{(z_1, z_2, 0)\}$ and the first two, $(-27, 0, 0)$ and $(27, -27, 0)$ , span it. So $\operatorname{range}(T - 3I)^3 = \{(z_1, z_2, 0)\}$ , of dimension $2$ , and $1 + 2 = 3$ as the direct sum requires. Check the null space: $(T - 3I)(4, 3, 9) = (-12 + 3 + 9,\ -9 + 9,\ 0) = (0, 0, 0)$ .

The restriction. The range is invariant under $T$ , and $S(z_1, z_2, 0) = T(z_1, z_2, 0) = (z_2, 0, 0)$ . Then $S^2(z_1, z_2, 0) = S(z_2, 0, 0) = (0, 0, 0)$ , so every nonzero vector of the range is a generalized eigenvector of $S$ for $0$ , hence of $T$ for $0$ (since $S$ and $T$ agree on the range).

The basis. $(1, 0, 0), (0, 1, 0)$ from the range and $(4, 3, 9)$ from the null space form a basis of $\mathbf{C}^3$ consisting of generalized eigenvectors of $T$ .

Solved: 0 / 4

Generalized Eigenvectors Span Every Complex Space

IKey Formulas

IIApply T−λIT - \lambda IT−λI Until It Vanishes

IIISplit, Then Induct

IVOver R\mathbf{R}R the First Step Can Fail

VQuiz Questions

IIApply $T - \lambda I$ Until It Vanishes

IVOver $\mathbf{R}$ the First Step Can Fail