HW 7 - Generalized Eigenvectors, Operator Decomposition

8.A: Generalized Eigenvectors

2

Question

Define $T \in L (C^{2})$ by:

T (w, z) = (- z, w)

Find the generalized eigenspaces corresponding to the distinct eigenvalues of $T$

Proof
Notice that $T$ has:

M (T) = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]

Here $λ = \pm i$ and thus the eigenvectors are $v_{1} = (i, 1), v_{2} = (- i, 1)$ :

T v_{1} = T (i, 1) = (- 1, i) = i (i, 1), T v_{2} = T (- i, 1) = (- 1, - i) = - i (- i, 1)

E (i, T) = {c (i, 1) : c \in C}, E (- i, T) = {c (- i, 1) : c \in C}

However, notice that since $\dim (V) = 2$ here that:

G (i, T) = null (T - i I)^{\dim (V)}

T^{2} (w, z) = T (- z, w) = (- w, - z) = - 1 (w, z) = - I (w, z) \Leftrightarrow T^{2} = - I

\begin{aligned} (T - i I)^{\dim (V)} & = (T - i I)^{2} \\ = T^{2} - 2 i T - I \\ = - I - 2 i T - I \\ = - 2 I - 2 i T \\ = - 2 (i T + I) \end{aligned}

So since $G$ is just the eigenvalues of the above operator:

\begin{aligned} G (i, T) & = null (T - i I)^{\dim (V)} \\ = null (i T + I) \end{aligned}

\begin{aligned} (i T + I) (w, z) & = i T (w, z) + (w, z) \\ = i (- z, w) + (w, z) \\ = (w - i z, z + i w) \end{aligned}

Finding the nullspace says:

w - i z = 0 \Rightarrow w = i z \Rightarrow z + i w = 0 \Rightarrow z + i (i z) = z - z = 0

Thus $z \in C$ then choose $w = i z$ . Thus:

G (i, T) = {(i z, z) : z \in C}

Similarly for $G (- i, T)$ we have:

(T + i I)^{\dim (V)} = T^{2} + 2 i T - I = - 2 I + 2 i T = 2 (i T - I)

\begin{aligned} (i T - I) (w, z) & = i T (w, z) - (w, z) \\ = i (- z, w) - (w, z) \\ = (- w - z i, - z + w i) \end{aligned}

Finding the nullspace:

- w - z i = 0 \Rightarrow w = - z i, - z + w i = 0 = - z + (- z i) i = - z + z = 0

Thus let $z \in C$ while have $w = - z i$ . Thus:

G (- i, T) = {(- i z, z)}

☐

3

Question

Suppose $T \in L (V)$ is invertible. Then $G (λ, T) = G (\frac{1}{λ}, T^{- 1})$ for every $λ \in F$ with $λ \neq 0$ .

Proof
First we should show that:

null (T - λ I)^{n} = null {(T^{- 1} - \frac{1}{λ} I)}^{n}

for all $n \in Z^{+}$ . We'll do this via induction. For the base case we know that:

\begin{aligned} v \in E (λ, T) & \Leftrightarrow T v = λ v \\ \Leftrightarrow T^{- 1} T v = T^{- 1} λ v \\ \Leftrightarrow v = λ T^{- 1} v \\ \Leftrightarrow \frac{v}{λ} = T^{- 1} v & λ \neq 0 \\ \Leftrightarrow v \in E (\frac{1}{λ}, T^{- 1}) \end{aligned}

For the inductive step, suppose it holds for all $n < k$ . We'll show $k + 1$ case holds:

\begin{aligned} v \in null (T - λ I)^{k} & \Leftrightarrow (T - λ I)^{k} v = \vec{0} \\ \Leftrightarrow (T - λ I)^{k - 1} ((T - λ I) v) = \vec{0} \\ \Leftrightarrow (T - λ I) v \in null (T - λ I)^{k - 1} \\ \Leftrightarrow (T - λ I) v \in null {(T^{- 1} - \frac{1}{λ} I)}^{k - 1} & Inductive Hypothesis \\ \Leftrightarrow {(T^{- 1} - \frac{1}{λ} I)}^{k - 1} (T - λ I) v = \vec{0} \\ \Leftrightarrow (T - λ I) {(T^{- 1} - \frac{1}{λ} I)}^{k - 1} v = \vec{0} & p (T) q (T) = q (T) p (T) \\ \Leftrightarrow {(T^{- 1} - \frac{1}{λ} I)}^{k - 1} v \in null (T - λ I) \\ \Leftrightarrow {(T^{- 1} - \frac{1}{λ} I)}^{k - 1} v \in null (T^{- 1} - \frac{1}{λ} I) & Inductive Hypothesis (Base Case) \\ \Leftrightarrow {(T^{- 1} - \frac{1}{λ} I)}^{k} v = \vec{0} \\ \Leftrightarrow v \in null {(T^{- 1} - \frac{1}{λ} I)}^{k} \end{aligned}

Thus $null (T - λ I)^{k} \subseteq null {(T^{- 1} - \frac{1}{λ} I)}^{k}$ and $\supseteq$ too. Thus the sets equal.

The theorem holds by applying the lemma with $n = \dim (V)$ :

G (λ, T) = null (T - λ I)^{\dim (V)} = null {(T^{- 1} - \frac{1}{λ} I)}^{\dim (V)} = G (\frac{1}{λ}, T^{- 1})

☐

4

Question

Suppose $T \in L (V)$ and $α, β \in F$ where $α \neq β$ . Then:

G (α, T) \cap G (β, T) = {\vec{0}}

Proof
Suppose $v \in G (α, T) \cap G (β, T)$ and assume for contradiction that $v \neq \vec{0}$ . Then $v$ is a generalized eigenvector corresponding to two distinct generalized eigenvalues $α, β$ via $T$ . But since $α, β$ are distinct eigenvalues and $v, v$ (here we treat $v$ as 'different' generalized eigenvectors to create the contradiction) then that implies $v, v$ is linearly independent which is a contradiction. So then $v = \vec{0}$ completing the proof.
☐

5

Question

Suppose $T \in L (V)$ , where $m \in Z^{+}$ , and $v \in V$ is such that $T^{m - 1} v \neq \vec{0}$ but $T^{m} v = \vec{0}$ . Prove that:

v, T v, . . ., T^{m - 1} v

is linearly independent.

Proof
Consider:

\sum_{i = 0}^{m - 1} α_{i} T^{i} v = \vec{0}

Take $T^{m - 1}$ on both sides:

\begin{aligned} T^{m - 1} (\sum_{i = 0}^{m - 1} α_{i} T^{i} v) & = T^{m - 1} \vec{0} \\ \sum_{i = 0}^{m - 1} α_{i} T^{i + m - 1} v & = \vec{0} \\ α_{0} T^{m - 1} v + \dots + a_{m - 1} T^{2 m - 2} v & = \vec{0} \\ α_{0} T^{m - 1} v & = \vec{0} \end{aligned}

Since $T^{m - 1} v \neq \vec{0}$ then we require $α_{0} = 0$ .

Repeat this with $T^{m - 2}$ on both sides:

\begin{aligned} T^{m - 2} (\sum_{i = 0}^{m - 1} α_{i} T^{i} v) & = T^{m - 2} \vec{0} \\ \sum_{i = 0}^{m - 1} α_{i} T^{i + m - 2} v & = \vec{0} \\ {α_{0} T^{m - 2} v}^{\to 0} + α_{1} T^{m - 2} v \dots + a_{m - 1} T^{2 m - 2} v & = \vec{0} \\ α_{1} T^{m - 2} v & = \vec{0} \end{aligned}

Again $T^{m - 2} v \neq \vec{0}$ so then $α_{1} = 0$ .

Repeat this process using $T^{m - 3}, . . ., T, I$ . This gets that all $α_{i} = 0$ , so then the set is LI.
☐

6

Question

Suppose $T \in L (C^{3})$ is defined by $T (z_{1}, z_{2}, z_{3}) = (z_{2}, z_{3}, 0)$ . Prove that $T$ has no square root. More precisely, prove that there does not exist $S \in L (C^{3})$ such that $S^{2} = T$ .

Proof
Assume for contradiction that there is some such square root $S \in L (C^{3})$ . Notice first that $T$ has only $λ = 0$ as it's eigenvalue since:

T (\vec{z}) = λ \vec{z} \Leftrightarrow (z_{2}, z_{3}, 0) = (λ z_{1}, λ z_{2}, λ z_{3})

Notice for this to work we require that either $λ = 0 \Leftrightarrow \vec{z} = (1, 0, 0)$ works and is the only eigenvector (if we let $λ \in C ∖ {0}$ then we require that $z_{3} = 0 \Rightarrow z_{2} = 0 \Rightarrow z_{1} = 0$ implying $\vec{z} = \vec{0}$ which isn't an eigenvector). As such, we know that:

C^{3} = G (0, T) = null (T - 0 I)^{\dim (V)} = null (T^{3}) = null (S^{6})

But wait! We know $\dim (V) = 3$ in this case. That means that:

\begin{aligned} null (S^{3}) = null (S^{4}) = \dots = null (S^{6}) & = null (T^{3}) \end{aligned}

Thus:

\begin{aligned} V & = null (T^{3}) \\ = null (S^{3}) \\ = null (S T) \\ \subseteq null (S^{2} T) & Sequence of increasing nullspaces \\ = null (T^{2}) \end{aligned}

But we can see that $null (T^{2}) = {T^{2} \vec{z} = \vec{0}} = {(z, 0, 0) : z \in C}$ and clearly $V ⊈ null (T^{2})$ here. Thus $S$ must not exist.
☐

7

Question

Suppose $N \in L (V)$ is nilpotent. Prove that $0$ is the only eigenvalue of $N$ .

Proof
Consider $λ$ is an eigenvalue of $N$ , with corresponding eigenvector $v \neq \vec{0}$ . Since $N$ is nilpotent, then $N^{\dim (V)} = {\vec{0}}_{N}$ . As such, then:

N^{\dim (V)} v = {\vec{0}}_{N} v = \vec{0}

While since $v$ is an eigenvector with $λ$ :

N^{\dim (V)} v = (\underset{\dim (V)}{\underset{⏟}{N \circ \dots \circ N}}) v = (\underset{\dim (V) - 1}{\underset{⏟}{N \circ \dots \circ N}}) λ v = \dots = λ^{\dim (V)} v

λ^{\dim (V)} v = \vec{0}

Since $v \neq \vec{0}$ then $λ^{\dim (V)} = 0 \Leftrightarrow λ = 0$ .
☐

8

Question

Prove or give a counterexample: The set of nilpotent operators on $V$ is a subspace of $L (V)$ .

Proof
This statement is false. Consider $S, T \in L (C^{2})$ , where:

S (z_{1}, z_{2}) = (0, z_{1}), T (z_{1}, z_{2}) = (z_{2}, 0)

Notice that $S^{2} = T^{2} = \vec{0}$ so then both are nilpotent. However:

(S + T) = S (z_{1}, z_{2}) = T (z_{1}, z_{2}) = (z_{2}, z_{1})

Notice that $(S + T)^{2} (z_{1}, z_{2}) = (z_{1}, z_{2})$ is the identity so $(S + T)^{2} = I$ and thus can never be nilpotent!
☐

9

Question

Suppose $S, T \in L (V)$ and $S T$ is nilpotent. Prove that $T S$ is nilpotent.

Proof
Since $S T$ is nilpotent then $\exists n \in Z^{+}$ where $(S T)^{n} = \vec{0}$ . Namely the nilpotent raised to the dimension is still the zero operator though:

(S T)^{\dim (V)} = \vec{0} \Leftrightarrow null ((S T)^{\dim (V)}) = V

But notice:

\begin{aligned} null ((T S)^{\dim (V)}) & = null (T S)^{\dim (V) + 1} & Null spaces stop growing \\ = null ((T S)^{\dim (V)} T S) \\ = null (T \circ \underset{\dim (V)}{\underset{⏟}{S \circ \dots \circ T}} \circ S) \\ = null (T (S T)^{\dim (V)} S) \\ = null (T \vec{0} S) & (S T)^{\dim (V)} = \vec{0} \\ = V \end{aligned}

Thus then $(T S)^{\dim (V)} = \vec{0}$ as a result so then $T S$ is nilpotent.
☐

12

Question

Suppose $N \in L (V)$ and there exists a basis of $V$ with respect to which $N$ has an upper-triangular matrix with only 0's on the diagonal. Prove that $N$ is nilpotent.

Proof
$M (N, β)$ (whatever $β$ may be) is UT and has only 0's on the diagonal. This is the form of the matrix of a nilpotent operator, so $N$ is nilpotent.

In more detail, say $β = {v_{1}, . . ., v_{n}}$ is such a basis. We know:

M (N, β) = [\begin{matrix} 0 & * \\ ⋱ \\ 0 & 0 \end{matrix}]

That implies $N (v_{1}) = \vec{0}$ looking at the first column. Further it shows that $N v_{2} \in span (v_{1})$ . This implies that:

N v_{2} = α v_{1}

for some $α \in F$ . But notice if we apply $N$ to both sides:

N^{2} v_{2} = N (α v_{1}) = α N v_{1} = α \vec{0} = \vec{0}

A similar argument will show that $N^{i} v_{i} = \vec{0}$ for all $1 \leq i \leq n$ . Clearly if we have $N^{n}$ then:

N^{n} v_{i} = \vec{0}

for all $i$ , so then $n = \dim (V)$ is the positive integer to choose to show $N$ is nilpotent.
☐

13

Question

Suppose $V$ is an inner product space and $N \in L (V)$ is normal and nilpotent. Then $N = \vec{0}$ .

Proof
Since $N$ is nilpotent then $\exists n \in Z^{+}$ where $N^{n} = \vec{0}$ . Recall from HW 3 - Self-Adjoint and Normal Operators#17 that for all $k \in Z^{+}$ then:

null (T^{k}) = null (T)

since $T$ is normal.

As such, let $v \in V$ be arbitrary. Since $N^{\dim (V)} = \vec{0}$ then $null (N^{\dim (V)}) = null (N)$ So since $v \in null (N^{\dim (V)}) = null (N)$ then $N v = \vec{0}$ . As $v$ was arbitrary then $N$ must be the zero map.

☐

15

Question

Suppose $N \in L (V)$ is such that $null (N^{\dim (V) - 1}) \neq null (N^{\dim (V)})$ . Prove that $N$ is nilpotent and that:

\dim (null (N^{j})) = j

for all $j \in N$ where $0 \leq j \leq \dim (V)$ .

Proof
Focus on the second part. Notice that we know that since:

null (N^{\dim (V)}) = null (N^{\dim (V) + 1}) = \dots

then since $null (N^{\dim (V) - 1}) \neq null (N^{\dim (V)})$ then $null (N^{\dim (V) - 1}) \neq null (N^{\dim (V) + i})$ for all $i \geq 0$ . Further, since:

null (T^{0}) \subseteq null (T^{1}) \subseteq \dots \subseteq null (N^{\dim (V) - 1})

Then because once we have $=$ in our subset chain that implies equality for the following nullspaces, then if we had equality for any two adjacent nullspaces, that implies equality all the way down the line, which would conflict with the fact that $null (N^{\dim (V) - 1}) \neq null (N^{\dim (V)})$ . Thus:

\dim (null (T^{0})) \leq \dim (null (T^{1})) \leq \dots \leq \dim (null (N^{\dim (V) - 1})) \leq \dim (null (N^{\dim (V)}))

becomes:

\dim (null (T^{0})) < \dim (null (T^{1})) < \dots < \dim (null (N^{\dim (V) - 1})) < \dim (null (N^{\dim (V)}))

Notice that $\dim (null (T^{0})) = 0$ . We know that $\dim (null (N^{\dim (V)}))$ is at most $\dim (V)$ . Because we are ordering $\dim (V)$ distinct dimension values from $0, 1, . . ., \dim (V)$ then by the well ordering principle we must have each $\dim (null (N^{j})) = j$ for all $0 \leq j \leq \dim (V)$ .

Thus since $\dim (null (N^{\dim (V)})) = \dim (V)$ as a result, then $null (N^{\dim (V)}) = V$ showing that $N$ is nilpotent.
☐

16 (latex)

Question

Suppose $T \in L (V)$ . Show that:

V = range (T^{0}) \supseteq range (T) \supseteq \dots \supseteq range (T^{k}) \supseteq range (T^{k + 1}) \supseteq \dots

Proof
Clearly we know that $range (T^{0}) = range (I) = V$ . We just need to show that $range (T^{k}) \supseteq range (T^{k + 1})$ . Let $v \in range (T^{k + 1})$ be arbitrary. Thus $\exists w$ such that:

T^{k + 1} w = v

Notice then:

T^{k + 1} w = T^{k} (T w) = v

So there exists vector $T w$ such that shows that $v \in range (T^{k})$ , showing $\supseteq$ .
☐

17

Question

Suppose $T \in L (V)$ and $m \in Z^{+}$ such that:

range (T^{m}) = range (T^{m + 1})

Prove that $range (T^{k}) = range (T^{m})$ for all $k > m$ .

Proof
Let $k \in Z^{+}$ such that $k > m$ . Using HW 7 - Generalized Eigenvectors, Operator Decomposition#16 we know that:

V = range (T^{0}) \supseteq range (T) \supseteq \dots \supseteq range (T^{m}) \supseteq range (T^{m + 1}) \supseteq \dots

So we already know that $range (T^{k}) \supseteq range (T^{m})$ since $k > m$ . As such we only have to show that $range (T^{k}) \subseteq range (T^{m})$ .

Let $v \in range (T^{k})$ . Then $\exists w$ where $T^{k} w = v$ . Since $k > m$ then we can deconstruct as:

T^{k} w = T^{m} (T^{k - m} w) = v

Thus $\exists T^{k - m} w$ such that shows that $v \in range (T^{m})$ , completing the proof.
☐

18

Question

Suppose $T \in L (V)$ . Let $n = \dim (V)$ . Prove that:

range (T^{n}) = range (T^{n + 1}) = \dots

Proof
All we have to show is that $range (T^{n}) = range (T^{n + 1})$ since if we set $m = \dim (V)$ with HW 7 - Generalized Eigenvectors, Operator Decomposition#17, then we get equality for the rest of the terms.

Assume for contradiction that instead $range (T^{n}) \neq range (T^{n + 1})$ . Then using HW 7 - Generalized Eigenvectors, Operator Decomposition#16 then:

V = \underset{\dim (V)}{\underset{⏟}{range (T^{0})}} \supset \underset{\dim (V) - 1}{\underset{⏟}{range (T)}} \supset \dots \supset \underset{0}{\underset{⏟}{range (T^{n})}} \supset range (T^{n + 1})

As each of the $\supset$ 's are strict inclusions going the other way, then the dimension must decrease by at least 1 per step. This implies that at least $\dim (range (T^{n + 1})) < 0$ which is impossible. Thus, then our assumption is wrong, so $range (T^{n}) = range (T^{n})$ .
☐

8.B: Decomposition of an Operator

1

Question

Suppose $V$ is a complex vector space, $N \in L (V)$ , and $0$ is the only eigenvalue of $N$ . Then $N$ is nilpotent.

Proof
Since 0 is the only eigenvalue of $N$ , then:

V = G (0, N)

So any $v \in G (0, N)$ is arbitrary. As a result then $(N - 0 I)^{\dim (V)} v = N^{\dim (V)} v = \vec{0}$ by definition (since $G (0, N) = null (N)^{\dim (V)}$ ). Notice that since $v$ is arbitrary then $N^{\dim (V)} = \vec{0}$ which is the zero-transformation, so choose $\dim (V)$ as the power to raise $N$ to show that it is nilpotent.
☐

2

Question

Give an example of an operator $T$ on a finite-dimensional real vector space such that 0 is the only eigenvalue of $T$ but $T$ is not nilpotent.

Proof
The power of complex vector spaces in the process of #1 is that it allows any $T$ to be turned into an UT matrix, and thus if all $λ = 0$ then we get an UT matrix with only 0's on the diagonal; exactly the form of a nilpotent operator. But in $R$ instead, such a guaruntee isn't granted, so let's construct a $M (T)$ that isn't UT and then make $T$ from that. Consider the matrix related to the standard basis: using $V = R^{3}$

M (T) = [\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 1 \\ 0 & - 1 & 0 \end{matrix}]

Then notice that $T (x, y, z) = (- y, x + z, - y)$ . Further if we try to find all our $λ$ 's:

(- y, x + z, - y) = (λ x, λ y, λ z)

Thus $λ (x) = - y = λ z$ so maybe $x = z$ so $x + z = 2 x = λ y = λ (- λ x) = - λ^{2} x$ . Then that implies that $- λ^{2} = 2$ but $λ \in R$ so there's no solution here. Thus $x \neq z$ . Then $λ = 0$ and then that implies $x = - z, y = 0$ so having $λ = 0$ has the only eigenvector $(1, 0, - 1)$ .

Furthermore, notice that $T$ is not nilpotent since:

M (T^{3}) = {[\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 1 \\ 0 & - 1 & 0 \end{matrix}]}^{3} = - 2 [\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 1 \\ 0 & - 1 & 0 \end{matrix}] = - 2 M (T)

And $M (T^{2}) \neq \vec{0}$ . Thus not matter how high the power gets, we just never have $M (T) = \vec{0}$ showing $T$ is not nilpotent. It's especially apparent here since we raised it to the third power which is $\dim (V)$ in this case, and if it was nilpotent we must have gottent the zero map.
☐

3

Question

Suppose $T \in L (V)$ . Suppose $S \in L (V)$ is invertible. Prove that $T$ and $S^{- 1} T S$ have the same eigenvalues with the same multiplicities.

Proof
Notice also we cannot assume $V$ is a complex vector space here.

For this purpose, say some $λ_{i}$ is an eigenvalue of $T$ , with corresponding multiplicity $d_{i}$ . Then $G (λ_{i}, T) = span (v_{1}, . . ., v_{d_{i}})$ is the span of all generalized eigenvectors $v_{1}, . . ., v_{d_{i}}$ with eigenvalue $λ_{i}$ via $T$ . Each $v_{j} \in null (T - λ_{i} I)^{\dim (V)}$ .

But notice that:

S^{- 1} T S - λ_{i} I = S^{- 1} T S - λ_{i} S^{- 1} S = S^{- 1} (T - λ_{i} I) S

(S^{- 1} T S - λ_{i} I)^{\dim (V)} = (S^{- 1} \circ (T - λ_{i}) \circ S)^{\dim (V)} = S^{- 1} (T - λ_{i})^{\dim (V)} S

So:

G (λ_{i}, S^{- 1} T S) = {v \in V : S^{- 1} (T - λ_{i} I)^{\dim (V)} S v = \vec{0}}

We want to show that there's an isomorphism between $G (λ_{i}, T) \sim G (λ_{i,} S^{- 1} T S)$ , namely that there's a vector:

U_{j} \in G (λ_{i}, S^{- 1} T S) = null (S^{- 1} T S - λ_{i} I)^{\dim (V)}

Constructed choosing some $v_{j}$ , which is equivalent to, as our analysis above shows:

S^{- 1} (T - λ_{i} I)^{\dim (V)} S u_{j} = \vec{0}

Notice that $S$ is invertible, so it must be both injective and surjective. As such, $\exists u_{j}$ where $S u_{j} = v_{j}$ . Notice then:

\begin{aligned} S^{- 1} (T - λ_{i} I)^{\dim (V)} S u_{j} & = S^{- 1} (T - λ_{i} I)^{\dim (V)} v_{j} \\ = S^{- 1} \vec{0} \\ = \vec{0} \end{aligned}

Thus for each $v_{j}$ we can construct a $u_{j}$ , showing both spaces have the same dimension. As result, they must have the same multiplicities, and correspondingly the same eigenvalues.
☐

4

Question

Suppose $V$ is an $n$ -dimensional complex vector space and $T$ is an operator on $V$ such that $null (T^{n - 2}) \neq null (T^{n - 1})$ . Prove that $T$ has at most two distinct eigenvalues.

Proof
Recall that $null (T^{n}) = G (0, T)$ by the definitions. Further, we know that since:

{\vec{0}} = null (T^{0}) \subseteq null (T^{1}) \subseteq \dots \subseteq null (T^{n}) = null (T^{n + 1}) = \dots = G (0, T)

and since $null (T^{n - 2}) \neq null (T^{n - 1})$ , notice then that if any $null (T^{i}) = null (T^{i + 1})$ (where $i \in {0, 1, . . ., n - 2}$ then that implies equality later down the line, which would contradict $null (T^{n - 2}) \neq null (T^{n - 1})$ . Thus, we have instead:

{\vec{0}} = null (T^{0}) \subset null (T^{1}) \subset \dots \subset null (T^{n - 2}) \neq null (T^{n - 1}) \subseteq null (T^{n}) = \dots

So namely:

0 < \dim (null (T)) < \dots < \dim (null (T^{n - 2})) < \dim (null (T^{n - 1})) \leq \dim (null (T^{n})) = \dim (null (T^{n + 1})) = \dots

Since all the dimensions are integer values less than $\dim (V) = n$ , then:

n - 1 \leq \dim (null (T^{n - 1})) \leq \dim (G (0, T))

Thus clearly $G (0, T) \geq n - 1$ . Thus since $G (0, T) \leq n$ then $G (0, T)$ can only be of dimension $n$ or $n - 1$ . If it's $n$ then it's the only eigenvalue since then:

V = G (0, T)

If instead we have $G (0, T) = n - 1$ then it's only possible that we have $G (λ \neq 0, T)$ be of dimension 1:

V = G (0, T) \oplus G (λ, T) \Leftrightarrow \dim (V) = \dim (G (0, T)) + \dim (G (λ, T)) \Leftrightarrow \dim (G (λ, T)) = n - (n - 1) = 1

Thus it's impossible to add any more different distinct eigenvalues, hence only a maximum of two eigenvalues are allowed (namely 0 and some $λ \in F$ ).
☐

5

Question

Suppose $V$ is a complex vector space and $T \in L (V)$ . Prove that $V$ has a basis consisting of eigenvectors of $T$ iff every generalized eigenvector of $T$ is an eigenvector of $T$ . (For $F = C$ the exercise above adds an equivalence to the list in the conditions equivalent to diagonalizability).

Proof
( $\to$ ): Suppose $V$ has a basis consisting of eigenvectors of $T$ . Denote the basis $β = {e_{1}, . . ., e_{n}}$ . As a result, by the conditions equivalent to diagonalizability, then:

V = ⨁_{i = 1}^{m} E (λ_{i}, T)

Since $V$ is a complex vector space, then we must have:

V = ⨁_{i = 1}^{m} G (λ_{i}, T)

So consider some $v \in G (λ_{i}, T)$ . Let's induct over $m$ to show that $E (λ_{i}, T) = G (λ_{i}, T)$ . For $m = 1$ as the base case it's trivial since $V = E (λ_{1}, T) = G (λ_{1}, T)$ . Notice for $m = 2$ since:

E (λ_{1}, T) \oplus E (λ_{2}, T) = G (λ_{1}, T) \oplus G (λ_{2}, T)

Then if $w \in G (λ_{1}, T)$ then clearly $w \notin G (λ_{2}, T)$ . As a result, by contradiction if we had $w \in E (λ_{2}, T)$ that would imply that $w \in G (λ_{2}, T)$ which is a contradiction. Thus, then $w \in E (λ_{1}, T)$ . The other way is shown directly since $E (λ_{i}, T) \subseteq G (λ_{i}, T)$ . Thus $E (λ_{1}, T) = G (λ_{1}, T)$ . Showing the same for $λ_{2}$ is done the same way. Repeat this type of process up until $m$ .

Therefore since $E (λ_{i}, T) = G (λ_{i}, T)$ then clearly $v \in E (λ_{i}, T)$ so since $v$ was any generalized eigenvector and it always was an eigenvector then that applies for all the generalized eigenvectors.

( $\leftarrow$ ): Suppose all generalized eigenvectors of $T$ are eigenvectors of $T$ . Thus any $v \in G (λ, T) \leftrightarrow v \in E (λ, T)$ (since all eigenvectors are generalized eigenvectors) so $G (λ, T) = E (λ, T)$ . You then know that:

V = ⨁_{i = 1}^{m} G (λ_{i}, T) = ⨁_{i = 1}^{m} E (λ_{i}, T)

Therefore via the conditions equivalent to diagonalizability, there is a basis consisting of eigenvectors of $T$ .
☐