Linear-Algebra-Done-Right · Unit 26 · Video 3 · Interactive Practice

Proving Jordan Form via an Invariant Complement

IKey Formulas

Formula	Name	Hypotheses
$U = \operatorname{span}(u, Tu, \dots, T^{m-1}u)$ , $\ \dim U = m$	The span of one string	$T \in \mathcal{L}(V)$ nilpotent, $m$ the smallest positive integer with $T^m = 0$ , $T^{m-1}u \neq 0$
$W = \{v \in V : \varphi(T^k v) = 0 \text{ for } k = 0, 1, \dots, m-1\}$	The invariant complement	$\varphi \in V'$ with $\varphi(T^{m-1}u) \neq 0$
$Sv = \bigl(\varphi(v), \varphi(Tv), \dots, \varphi(T^{m-1}v)\bigr)$ , $\ W = \operatorname{null} S$ , $\ \dim W \ge \dim V - m$	Counting dimensions	$S \in \mathcal{L}(V, \mathbf{F}^m)$ , $V$ finite-dimensional
$T\|_{G(\lambda_k, T)} = \lambda_k I + (T - \lambda_k I)\|_{G(\lambda_k, T)}$	From nilpotent to every operator	$\mathbf{F} = \mathbf{C}$ , $\lambda_1, \dots, \lambda_r$ the distinct eigenvalues of $T$

Key Insight: An arbitrary complement of $U$ need not be invariant; the $m$ conditions $\varphi(T^k v) = 0$ produce one that is, and $V = U \oplus W$ lets induction on $\dim V$ finish the nilpotent case.

IIWhich Functionals Give a Complement?

For $T(x, y, z) = (y, 0, 0)$ and $u = e_2$ , the subspace $W$ defined by $\varphi$ complements $U$ exactly when $\varphi(Tu) \neq 0$ .

c₁ 1 c₂ 0 c₃ 0

IIIWhy $U \cap W = \{0\}$

With $m = 3$ , the numbers $\varphi(T^k v)$ for $v \in U$ form a triangular system, so they cannot all vanish unless $v = 0$ .

c₀ 0 c₁ 2 c₂ −1

💡 Here $\varphi(u) = 2$ , $\varphi(Tu) = -1$ , $\varphi(T^2 u) = 1$ ; any values work as long as $\varphi(T^2 u) \neq 0$ .

IVQuiz Questions

Problem 1 · Dimensions in the Inductive Step

Given: $T \in \mathcal{L}(\mathbf{F}^7)$ nilpotent with $T^4 = 0$ and $T^3 u \neq 0$ , $U = \operatorname{span}(u, Tu, T^2u, T^3u)$ , and $W$ defined from a functional $\varphi$ with $\varphi(T^3 u) \neq 0$ — find $\dim U$ and what the map $S$ guarantees about $\dim W$ .

$\dim U$

From $W = \operatorname{null} S$ with $S \colon \mathbf{F}^7 \to \mathbf{F}^4$

✅ Correct!

u, Tu, T^2u, T^3u

is linearly independent, so

\dim U = 4

; and

\dim W = 7 - \dim \operatorname{range} S \ge 7 - 4 = 3

❌ The string has $m = 4$ vectors.

u, Tu, T^2u, T^3u

runs up to

T^{m-1}u

with

m = 4

❌ Not quite. Count the vectors

u, Tu, \dots, T^{m-1}u

; they are linearly independent.

❌ $\operatorname{range} S \subseteq \mathbf{F}^4$ has dimension at most $4$ . The fundamental theorem gives

\dim W = 7 - \dim \operatorname{range} S \ge 3

, not

\ge 4

❌ The inequality goes the other way. A smaller range means a larger null space, so the bound on

\dim W

is from below.

❌ Not quite. Apply the fundamental theorem of linear maps to

S

, whose range lies in

\mathbf{F}^4

Show solution

Here $m = 4$ . The list $u, Tu, T^2u, T^3u$ is linearly independent, so $\dim U = 4$ .

$S \in \mathcal{L}(\mathbf{F}^7, \mathbf{F}^4)$ , $Sv = \bigl(\varphi(v), \varphi(Tv), \varphi(T^2v), \varphi(T^3v)\bigr)$ , has $\operatorname{null} S = W$ . By the fundamental theorem of linear maps,

\dim W = 7 - \dim \operatorname{range} S \ge 7 - 4 = 3.

Together with $U \cap W = \{0\}$ , $\dim(U \oplus W) \ge 4 + 3 = 7$ , so $V = U \oplus W$ and in fact $\dim W = 3$ .

Problem 2 · Compute the Complement

Given: $T \in \mathcal{L}(\mathbf{F}^3)$ defined by $T(x, y, z) = (z, 0, 0)$ , $u = e_3$ (so $m = 2$ and $U = \operatorname{span}(e_3, e_1)$ ), and $\varphi(x, y, z) = x + y$ — find $W = \{v : \varphi(v) = \varphi(Tv) = 0\}$ .

✅ Correct!

\varphi(Tv) = z = 0

and

\varphi(v) = x + y = 0

; the line

\operatorname{span}((1, -1, 0))

is mapped to

0

T

, so it is invariant, and it meets

U

only in

0

❌ That uses only the condition $\varphi(v) = 0$ .

W

also needs

\varphi(Tv) = 0

, and

\varphi(T(0, 0, 1)) = \varphi(1, 0, 0) = 1

❌ Check the sign.

\varphi(1, 1, 0) = 2 \neq 0

❌ $\varphi(e_2) = 1 \neq 0$ .

e_2

would be the answer for

\varphi(x, y, z) = x

; here

\varphi

also reads

y

Show solution

$T^2 = 0 \neq T$ and $Te_3 = e_1 \neq 0$ , so $m = 2$ and $\varphi(Tu) = \varphi(e_1) = 1 \neq 0$ , as required. For $v = (x, y, z)$ :

\varphi(v) = x + y, \qquad \varphi(Tv) = \varphi(z, 0, 0) = z.

So $W = \{(x, y, z) : x + y = 0,\ z = 0\} = \operatorname{span}\bigl((1, -1, 0)\bigr)$ . Indeed $T(1, -1, 0) = 0 \in W$ , and $(1, -1, 0) \notin \operatorname{span}(e_3, e_1)$ , so $\mathbf{F}^3 = U \oplus W$ .

Problem 3 · The Key Step for $U \cap W$

Given: a nonzero $v = c_0 u + c_1 Tu + \cdots + c_{m-1}T^{m-1}u$ in $U$ , with $j$ the smallest index such that $c_j \neq 0$ — find $T^{m-j-1}v$ .

✅ Correct! Then

\varphi(T^{m-j-1}v) = c_j\,\varphi(T^{m-1}u) \neq 0

with

0 \le m-j-1 \le m-1

, so

v \notin W

❌ The term $c_{m-1}T^{m-1}u$ is sent to $c_{m-1}T^{2m-j-2}u$ . Unless

j = m-1

that power is at least

m

, so the term vanishes.

❌ The power of $T$ adds up.

T^{m-j-1}(c_j T^j u) = c_j T^{m-1}u

❌ The $c_j$ term survives. It lands exactly on

T^{m-1}u

, which is nonzero; only the later terms reach

T^m = 0

Show solution

Since $c_0 = \cdots = c_{j-1} = 0$ , $v = c_j T^j u + c_{j+1}T^{j+1}u + \cdots + c_{m-1}T^{m-1}u$ . Apply $T^{m-j-1}$ :

T^{m-j-1}v = c_j T^{m-1}u + c_{j+1}T^{m}u + \cdots + c_{m-1}T^{2m-j-2}u = c_j T^{m-1}u,

because every power from $T^m$ on is $0$ . Applying $\varphi$ gives $c_j\varphi(T^{m-1}u) \neq 0$ , which contradicts $v \in W$ since $m - j - 1 \in \{0, \dots, m-1\}$ . So $U \cap W = \{0\}$ .

Problem 4 · One Block or Several?

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ defined by $T(z_1, z_2, z_3) = (2z_1 + z_2 + z_3,\ 2z_2 + z_3,\ 2z_3)$ — find the matrix of $T$ with respect to a Jordan basis.

✅ Correct!

N = T - 2I

has

N^2 e_3 = e_1 \neq 0

, so

N^2e_3, Ne_3, e_3

is a single string filling

\mathbf{C}^3

: one

3 \times 3

block.

❌ $T \neq 2I$ . A diagonal Jordan matrix would make

T - 2I = 0

❌ That would make $(T - 2I)^2 = 0$ . But

(T - 2I)^2 e_3 = e_1 \neq 0

, so the longest string has length

3

❌ That is the standard matrix of $T$ . A Jordan matrix has

0

's everywhere except the diagonal and the line directly above it.

Show solution

The only eigenvalue is $2$ , so $G(2, T) = \mathbf{C}^3$ and $N = T - 2I$ is nilpotent: $N(z_1, z_2, z_3) = (z_2 + z_3,\ z_3,\ 0)$ .

Ne_3 = (1, 1, 0), \qquad N^2 e_3 = N(1, 1, 0) = (1, 0, 0) \neq 0, \qquad N^3 = 0.

So $N^2e_3, Ne_3, e_3 = (1, 0, 0), (1, 1, 0), (0, 0, 1)$ is a basis made of one string, and in it $\mathcal{M}(N)$ is the $3 \times 3$ block with $1$ 's above the diagonal. Adding $2I$ puts $2$ 's on the diagonal:

\mathcal{M}(T) = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}.

Solved: 0 / 4

Proving Jordan Form via an Invariant Complement

IKey Formulas

IIWhich Functionals Give a Complement?

IIIWhy U∩W={0}U \cap W = \{0\}U∩W={0}

IVQuiz Questions

IIIWhy $U \cap W = \{0\}$