Linear-Algebra-Done-Right · Unit 24 · Video 3 · Interactive Practice

Linear Independence of Generalized Eigenvectors

IKey Formulas

Formula	Name	Hypotheses
$(T - \lambda I)^k v = 0$ for some $k \ge 1$ $\iff$ $(T - \lambda I)^n v = 0$	Generalized eigenvector for $\lambda$	$v \neq 0$ , $T \in \mathcal{L}(V)$ , $n = \dim V$
$(T - \lambda I)^n = \sum_{k=0}^{n} \binom{n}{k} (\alpha - \lambda)^{n-k} (T - \alpha I)^k$	Binomial step	Any scalars $\alpha, \lambda$ : write $T - \lambda I = (T - \alpha I) + (\alpha - \lambda) I$ ; the two summands commute
$v$ a generalized eigenvector for $\alpha$ and for $\lambda$ $\Rightarrow$ $\alpha = \lambda$	One vector, one eigenvalue (Theorem 1)	$V$ finite-dimensional, $T \in \mathcal{L}(V)$
$v_1, \dots, v_m$ is linearly independent	Independence across eigenvalues (Theorem 2)	Each $v_k$ a generalized eigenvector of $T$ for $\lambda_k$ , with $\lambda_1, \dots, \lambda_m$ distinct

Key Insight: If $v$ is a generalized eigenvector for $\alpha$ and $m$ is the smallest power with $(T - \alpha I)^m v = 0$ , then applying $(T - \alpha I)^{m-1}$ to the binomial expansion kills every term with $k \ge 1$ , so $(T - \alpha I)^{m-1} (T - \lambda I)^n v = (\alpha - \lambda)^n (T - \alpha I)^{m-1} v$ , which is nonzero whenever $\lambda \neq \alpha$ .

IIOne Vector, One Ladder to Zero

For $T(z_1, z_2, z_3) = (4z_2, 0, 5z_3)$ on $\mathbf{C}^3$ , no nonzero $v$ is sent to $0$ by powers of both $T$ and $T - 5I$ .

z₁ 2 z₂ 1 z₃ 0

IIIThe Binomial Step, Term by Term

For the same $T$ and $v$ with $T^m v = 0$ ( $m$ smallest), $T^{m-1}$ erases every term of $(T - \lambda I)^3 v$ except $(0 - \lambda)^3\, T^{m-1} v$ .

Vector v Test value λ 2

$k$	$\binom{3}{k}$	$(0 - \lambda)^{3-k}$	$T^k v$	$\binom{3}{k} (0 - \lambda)^{3-k}\, T^k v$	$T^{m-1+k} v$

IVShrinking a Dependence Relation

Applying $(T - 5I)^3$ to $a_1 v_1 + a_2 v_2$ erases the $v_2$ term and leaves $a_1$ times a nonzero generalized eigenvector for $0$ .

v₁ = (z₁, z₂, 0): z₁ 1 z₂ 1 a₁ 2 a₂ −1

💡 With $m$ eigenvalues the same move turns a shortest dependent list of length $m$ into a dependent list of length $m - 1$ ; here $m = 2$ , so the shorter list would be the single vector $(T - 5I)^3 v_1$ , which is nonzero and so cannot be dependent.

VQuiz Questions

Problem 1 · Which Power Kills w?

Given: the operator on $\mathbf{F}^4$ with the block upper-triangular matrix $A$ below (eigenvalues $5$ and $-1$ ) and $w = (-2, -1, 1, 1)$ — find the smallest $j \ge 1$ with $(A + I)^j w = 0$ and the smallest $k \ge 1$ with $(A - 5I)^k w = 0$ .

A = \left[\begin{array}{cc|cc} 5 & 1 & 6 & 6 \\ 0 & 5 & 0 & 6 \\ \hline 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{array}\right]

Smallest power of A + I

Smallest power of A − 5I

✅ Correct!

(A + I)w = (-1, 0, 1, 0)

and

(A + I)^2 w = 0

, so

w

is a generalized eigenvector for

-1

with

j = 2

; by Theorem 1 it is not one for

5

, so no power of

A - 5I

sends it to

0

❌ One step short.

(A + I)w = (-1, 0, 1, 0) \neq 0

; that vector is an eigenvector for

-1

, and one more application of

A + I

sends it to

0

❌ Not the smallest. The power

4 = \dim \mathbf{F}^4

does send

w

0

, but already

(A + I)^2 w = 0

❌ The ladder does reach zero.

(A + I)w = (-1, 0, 1, 0)

and

(A + I)(-1, 0, 1, 0) = (-6 + 6, 0, 0, 0) = 0

❌ That count belongs to $-1$ .

(A - 5I)^2 w = (-60, -36, 24, 36) \neq 0

; the number

2

is the power of

A + I

that kills

w

❌ The power $n$ only kills generalized eigenvectors for $5$ .

(A - 5I)^4 w = (-1728, -1296, 432, 1296) \neq 0

, and if no power up to

n = 4

works, no higher power does either.

❌ $w$ is not an eigenvector for $5$ .

(A - 5I)w = (11, 6, -5, -6) \neq 0

Show solution

Subtract $-1$ from the diagonal:

A + I = \begin{bmatrix} 6 & 1 & 6 & 6 \\ 0 & 6 & 0 & 6 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \qquad (A + I)w = \begin{bmatrix} -12 - 1 + 6 + 6 \\ -6 + 6 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}.

(A + I)\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -6 + 6 \\ 0 \\ 0 \\ 0 \end{bmatrix} = 0.

So $j = 2$ , and $w$ is a generalized eigenvector of $A$ for $-1$ . A generalized eigenvector corresponds to only one eigenvalue, so $w$ is not a generalized eigenvector for $5$ : no power of $A - 5I$ sends $w$ to $0$ . As a check, $(A - 5I)^4 w = (-1728, -1296, 432, 1296) \neq 0$ , and the power $4 = \dim \mathbf{F}^4$ works whenever any power does.

Problem 2 · The Expansion Before Anything Cancels

Given: $T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))$ defined by $T(a + bx + cx^2) = (5a + b) + 5bx - cx^2$ , with eigenvalues $5$ and $-1$ . The polynomial $x$ is a generalized eigenvector for $\alpha = 5$ . Testing $\lambda = -1$ , write $T + I = (T - 5I) + 6I$ — find $(T + I)^3 x$ by the binomial theorem.

✅ Correct!

(T + I)^3 x = 6^3 x + \binom{3}{1} 6^2 (T - 5I)x = 216x + 108

; the terms with

k \ge 2

vanish because

(T - 5I)^2 x = 0

❌ The $k = 1$ term is still there.

\binom{3}{1} 6^2 (T - 5I)x = 108 \cdot 1

is nonzero; it disappears only after applying

(T - 5I)^{m-1} = T - 5I

, which is a later step of the proof.

❌ Missing a binomial coefficient. The

k = 1

term carries

b_1 = \binom{3}{1} = 3

, so it is

3 \cdot 36 \cdot 1 = 108

, not

36

❌ Sign of $\alpha - \lambda$ .

T + I = (T - 5I) + (5 - (-1))I = (T - 5I) + 6I

, so the scalar is

+6

; using

-6

gives

(-6)^3 x + 3 \cdot 36 = -216x + 108

Show solution

First the ladder for $5$ : $(T - 5I)(a + bx + cx^2) = b - 6cx^2$ , so $(T - 5I)x = 1$ and $(T - 5I)^2 x = (T - 5I)1 = 0$ . Hence $m = 2$ .

Since $T - 5I$ commutes with $6I$ , the binomial theorem gives

(T + I)^3 x = \sum_{k=0}^{3} \binom{3}{k} 6^{3-k} (T - 5I)^k x = 216x + 3 \cdot 36 \cdot 1 + 0 + 0 = 216x + 108.

Check directly: $(T + I)(a + bx + cx^2) = (6a + b) + 6bx$ , so $x \mapsto 1 + 6x \mapsto 12 + 36x \mapsto 108 + 216x$ .

The proof's next step applies $(T - 5I)^{m-1} = T - 5I$ : it sends $216x + 108$ to $216 = 6^3 (T - 5I)x \neq 0$ , so $(T + I)^3 x \neq 0$ and $x$ is not a generalized eigenvector for $-1$ .

Problem 3 · The Shrinking Step on a 4 × 4 Matrix

Given: the matrix $A$ of Problem 1, $u = (1, 1, 0, 0)$ (a generalized eigenvector for $5$ ) and $w$ (a generalized eigenvector for $-1$ ). To show $u, w$ is linearly independent, the proof applies $(A + I)^4$ to $a u + b w = 0$ — find $(A + I)^4 u$ , then the smallest $m \ge 1$ with $(A - 5I)^m (A + I)^4 u = 0$ .

Step 1 · expand

Step 2 · smallest power of A − 5I

✅ Correct!

(A + I)^4 u = (2160, 1296, 0, 0)

is nonzero and is again a generalized eigenvector for

5

(with

m = 2

), so

a (A + I)^4 u = 0

forces

a = 0

, and then

b w = 0

forces

b = 0

❌ A term was dropped.

(A - 5I)u = (1, 0, 0, 0) \neq 0

, so the

k = 1

term

\binom{4}{1} 6^3 (A - 5I)u = (864, 0, 0, 0)

must be added to

6^4 u

❌ $(A + I)^4$ kills only the $-1$ vectors. It sends

w

0

, but

u

is a generalized eigenvector for

5

, and by Theorem 1 not one for

-1

, so

(A + I)^4 u \neq 0

❌ Sign of $\alpha - \lambda$ .

A + I = (A - 5I) + 6I

, not

(A - 5I) - 6I

; with

+6

the first coordinate is

1296 + 864

, not

1296 - 864

❌ Not an eigenvector.

(A - 5I)(2160, 1296, 0, 0) = (1296, 0, 0, 0) \neq 0

; one more application of

A - 5I

is needed.

❌ The new vector stays in the group for $5$ . Powers of

A - 5I

and

A + I

commute, so

(A - 5I)^2 (A + I)^4 u = (A + I)^4 (A - 5I)^2 u = 0

❌ Not the smallest. The power

4

works, but

(A - 5I)^2

already sends

(2160, 1296, 0, 0)

0

Show solution

On $u$ : $(A - 5I)u = (1, 0, 0, 0)$ and $(A - 5I)^2 u = 0$ . Write $A + I = (A - 5I) + 6I$ and expand:

(A + I)^4 u = \sum_{k=0}^{4} \binom{4}{k} 6^{4-k} (A - 5I)^k u = 1296\,(1, 1, 0, 0) + 4 \cdot 216\,(1, 0, 0, 0) = (2160, 1296, 0, 0).

Then $(A - 5I)(2160, 1296, 0, 0) = (1296, 0, 0, 0) \neq 0$ and $(A - 5I)(1296, 0, 0, 0) = 0$ , so $m = 2$ .

In the proof: applying $(A + I)^4$ to $au + bw = 0$ removes $bw$ and leaves $a (A + I)^4 u = 0$ . The vector $(A + I)^4 u$ is nonzero (Theorem 1) and is a generalized eigenvector for $5$ , a list of length $1$ that is independent, so $a = 0$ ; then $bw = 0$ with $w \neq 0$ gives $b = 0$ .

Problem 4 · Inside the Minimal-Length Argument

Given: in the proof of independence, $m$ is the smallest length of a linearly dependent list $v_1, \dots, v_m$ of generalized eigenvectors for distinct $\lambda_1, \dots, \lambda_m$ , and $a_1 v_1 + \cdots + a_m v_m = 0$ with not all $a_k = 0$ . Applying $(T - \lambda_m I)^n$ , $n = \dim V$ , gives $(\ast)$ : $a_1 (T - \lambda_m I)^n v_1 + \cdots + a_{m-1} (T - \lambda_m I)^n v_{m-1} = 0$ , with new vectors $(T - \lambda_m I)^n v_k$ for $k \lt m$ . Identify the reason behind each step.

Why are the coefficients nonzero?

Why is each new vector a generalized eigenvector?

✅ Correct! Minimality makes every coefficient nonzero, and the new vectors are nonzero generalized eigenvectors for

\lambda_1, \dots, \lambda_{m-1}

, so equation

(\ast)

is a dependence of length

m - 1

❌ Nonzero vectors can carry zero coefficients. A relation

0 \cdot v_1 + a_2 v_2 + \cdots + a_m v_m = 0

with nonzero vectors is still possible a priori; what rules it out is minimality, since dropping

v_1

would give a shorter dependent list.

❌ That fact is used later. Theorem 1 shows

(T - \lambda_m I)^n v_k \neq 0

; the coefficients are nonzero because

m

is the smallest length.

❌ Distinctness doesn't touch the coefficients. A sublist of the

v_k

still has distinct eigenvalues; the point is that such a sublist would be a shorter dependent list, against the choice of

m

❌ That explains why $v_m$ drops out. It says nothing about

v_k

for

k \lt m

; for those one needs nonzero (Theorem 1) and

(T - \lambda_k I)^n (T - \lambda_m I)^n v_k = (T - \lambda_m I)^n (T - \lambda_k I)^n v_k = 0

❌ The coefficient plays no role here. Whether

(T - \lambda_m I)^n v_k

is a generalized eigenvector depends only on the vector;

a_k \neq 0

is used to make

(\ast)

a genuine dependence.

❌ It is not invertible.

(T - \lambda_m I)^n

sends the nonzero vector

v_m

0

; nonzeroness of

(T - \lambda_m I)^n v_k

comes from Theorem 1 instead.

Show solution

Coefficients. If some $a_k = 0$ , delete $v_k$ : the remaining $m - 1$ vectors still satisfy a nontrivial relation, so they form a shorter dependent list of generalized eigenvectors for distinct eigenvalues, contradicting the choice of $m$ .

New vectors. Apply $(T - \lambda_m I)^n$ , $n = \dim V$ , to the relation. The term $a_m v_m$ vanishes and

a_1 (T - \lambda_m I)^n v_1 + \cdots + a_{m-1} (T - \lambda_m I)^n v_{m-1} = 0. \qquad (\ast)

For $k \lt m$ , $(T - \lambda_m I)^n v_k \neq 0$ , otherwise $v_k$ would be a generalized eigenvector for both $\lambda_k$ and $\lambda_m$ . Both powers are polynomials in $T$ , so they commute:

(T - \lambda_k I)^n (T - \lambda_m I)^n v_k = (T - \lambda_m I)^n (T - \lambda_k I)^n v_k = 0.

So $(\ast)$ , with all $a_k \neq 0$ , is a dependence among $m - 1$ generalized eigenvectors for the distinct $\lambda_1, \dots, \lambda_{m-1}$ , contradicting the minimality of $m$ .

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