8.A: Generalized Eigenvectors

2

\begin{proof}
Notice that $T$ has:

Here $\lambda =\pm i$ and thus the eigenvectors are $v_1=(i,1),v_2=(-i,1)$:

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

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

Finding the nullspace says:

Thus $z\in \mathbb{C}$ then choose $w=iz$. Thus:

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

Finding the nullspace:

Thus let $z\in \mathbb{C}$ while have $w=-zi$. Thus:

\end{proof}

3

\begin{proof}
First we should show that:

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

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

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

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

\end{proof}

4

\begin{proof}
Suppose $v\in G(\alpha ,T)\cap G(\beta ,T)$ and assume for contradiction that $v\ne \overrightarrow{0}$. Then $v$ is a generalized eigenvector corresponding to two distinct generalized eigenvalues $\alpha ,\beta$ via $T$. But since $\alpha ,\beta$ 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 $v=\overrightarrow{0}$ completing the proof.
\end{proof}

5

\begin{proof}
Consider:

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

Since $T^{m-1}v\ne \overrightarrow{0}$ then we require ${\alpha }_0=0$.

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

Again $T^{m-2}v\ne \overrightarrow{0}$ so then ${\alpha }_1=0$.

Repeat this process using $T^{m-3},...,T,I$. This gets that all ${\alpha }_i=0$, so then the set is LI.
\end{proof}

6

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

Notice for this to work we require that either $\lambda =0\iff \overrightarrow{z}=(1,0,0)$ works and is the only eigenvector (if we let $\lambda \in \mathbb{C}\setminus \{0\}$ then we require that $z_3=0\Rightarrow z_2=0\Rightarrow z_1=0$ implying $\overrightarrow{z}=\overrightarrow{0}$ which isn't an eigenvector). As such, we know that:

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

Thus:

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

7

\begin{proof}
Consider $\lambda$ is an eigenvalue of $N$, with corresponding eigenvector $v\ne \overrightarrow{0}$. Since $N$ is nilpotent, then $N^{\dim (V)}={\overrightarrow{0}}_N$. As such, then:

While since $v$ is an eigenvector with $\lambda$:

Since $v\ne \overrightarrow{0}$ then ${\lambda }^{\dim (V)}=0\iff \lambda =0$.
\end{proof}

8

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

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

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!
\end{proof}

9

\begin{proof}
Since $ST$ is nilpotent then $\mathrm{\exists }n\in {\mathbb{Z}}^{+}$ where $(ST{)}^n=\overrightarrow{0}$. Namely the nilpotent raised to the dimension is still the zero operator though:

But notice:

Thus then $(TS{)}^{\dim (V)}=\overrightarrow{0}$ as a result so then $TS$ is nilpotent.
\end{proof}

12

\begin{proof}
$\mathcal{M}(N,\beta )$ (whatever $\beta$ 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 $\beta =\{v_1,...,v_n\}$ is such a basis. We know:

That implies $N(v_1)=\overrightarrow{0}$ looking at the first column. Further it shows that $N{v}_2\in \text{span}(v_1)$. This implies that:

for some $\alpha \in \mathbb{F}$. But notice if we apply $N$ to both sides:

A similar argument will show that $N^i{v}_i=\overrightarrow{0}$ for all $1\le i\le n$. Clearly if we have $N^n$ then:

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

13

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

since $T$ is normal.

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

\end{proof}

15

\begin{proof}
Focus on the second part. Notice that we know that since:

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

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 $\text{null}(N^{\dim (V)-1})\ne \text{null}(N^{\dim (V)})$. Thus:

becomes:

Notice that $\dim (\text{null}(T^0))=0$. We know that $\dim (\text{null}(N^{\dim (V)}))$ 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 $\dim (\text{null}(N^j))=j$ for all $0\le j\le \dim (V)$.

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

16 (latex)

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

Notice then:

So there exists vector $Tw$ such that shows that $v\in \text{range}(T^k)$, showing $\supseteq$.
\end{proof}

17

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

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

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

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

18

\begin{proof}
All we have to show is that $\text{range}(T^n)=\text{range}(T^{n+1})$ 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 $\text{range}(T^n)\ne \text{range}(T^{n+1})$. Then using HW 7 - Generalized Eigenvectors, Operator Decomposition#16 then:

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 $\dim (\text{range}(T^{n+1}))<0$ which is impossible. Thus, then our assumption is wrong, so $\text{range}(T^n)=\text{range}(T^n)$.
\end{proof}

8.B: Decomposition of an Operator

1

\begin{proof}
Since 0 is the only eigenvalue of $N$, then:

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

2

\begin{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 $\lambda =0$ then we get an UT matrix with only 0's on the diagonal; exactly the form of a nilpotent operator. But in $\mathbb{R}$ instead, such a guaruntee isn't granted, so let's construct a $\mathcal{M}(T)$ that isn't UT and then make $T$ from that. Consider the matrix related to the standard basis: using $V={\mathbb{R}}^3$

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

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

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

And $\mathcal{M}(T^2)\ne \overrightarrow{0}$. Thus not matter how high the power gets, we just never have $\mathcal{M}(T)=\overrightarrow{0}$ 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.
\end{proof}

3

\begin{proof}
Notice also we cannot assume $V$ is a complex vector space here.

For this purpose, say some ${\lambda }_i$ is an eigenvalue of $T$, with corresponding multiplicity $d_i$. Then $G({\lambda }_i,T)=\text{span}(v_1,...,v_{d_i})$ is the span of all generalized eigenvectors $v_1,...,v_{d_i}$ with eigenvalue ${\lambda }_i$ via $T$. Each $v_j\in \text{null}(T-{\lambda }_{i}I{)}^{\dim (V)}$.

But notice that:

So:

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

Constructed choosing some $v_j$, which is equivalent to, as our analysis above shows:

Notice that $S$ is invertible, so it must be both injective and surjective. As such, $\mathrm{\exists }{u}_j$ where $S{u}_j=v_j$ . Notice then:

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.
\end{proof}

4

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

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

So namely:

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

Thus clearly $G(0,T)\ge n-1$. Thus since $G(0,T)\le n$ 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:

If instead we have $G(0,T)=n-1$ then it's only possible that we have $G(\lambda \ne 0,T)$ be of dimension 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 $\lambda \in \mathbb{F}$).
\end{proof}

5

\begin{proof}
($\to$): Suppose $V$ has a basis consisting of eigenvectors of $T$. Denote the basis $\beta =\{e_1,...,e_n\}$. As a result, by the conditions equivalent to diagonalizability, then:

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

So consider some $v\in G({\lambda }_i,T)$. Let's induct over $m$ to show that $E({\lambda }_i,T)=G({\lambda }_i,T)$. For $m=1$ as the base case it's trivial since $V=E({\lambda }_1,T)=G({\lambda }_1,T)$. Notice for $m=2$ since:

Then if $w\in G({\lambda }_1,T)$ then clearly $w\notin G({\lambda }_2,T)$. As a result, by contradiction if we had $w\in E({\lambda }_2,T)$ that would imply that $w\in G({\lambda }_2,T)$ which is a contradiction. Thus, then $w\in E({\lambda }_1,T)$. The other way is shown directly since $E({\lambda }_i,T)\subseteq G({\lambda }_i,T)$. Thus $E({\lambda }_1,T)=G({\lambda }_1,T)$. Showing the same for ${\lambda }_2$ is done the same way. Repeat this type of process up until $m$.

Therefore since $E({\lambda }_i,T)=G({\lambda }_i,T)$ then clearly $v\in E({\lambda }_i,T)$ 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 $v\in G(\lambda ,T)\leftrightarrow v\in E(\lambda ,T)$ (since all eigenvectors are generalized eigenvectors) so $G(\lambda ,T)=E(\lambda ,T)$. You then know that:

Therefore via the conditions equivalent to diagonalizability, there is a basis consisting of eigenvectors of $T$.
\end{proof}