8.D: 1,2,3,4,5
9.A: 2,3,4,5,6,7,8,15,19

8.D: Jordon Form

1

\begin{proof}
Since $\mathcal{M}(N)$ is upper triangular, then all the eigenvalues are on the diagonal of the matrix. As such, then $\lambda =0$ is the only eigenvalue for $N$, and furthermore by HW 8 - Operator Decomposition, Characteristic and Minimal Polynomials#11 (TODO) the multiplicity $d_i$ of each eigenvalue is precisely the number of times it appears on the diagonal. Thus $\lambda =0$ is of multiplicity $4$. Thus:

For the minimal polynomial we have the choices of ${\mu }_N(z)=z,z^2,z^3,z^4$, but we know that $z^4$ is the only viable option since:

and only holds true when ${\mu }_N(z)=p_N(z)$ because $N$ is nilpotent where only $N^3\ne \overrightarrow{0}$ while $N^4=\overrightarrow{0}$. The math of the matrices of $\mathcal{M}(N)$ raised to various powers verifies this:

Thus, $\boxed{{\displaystyle {\mu }_N(z)=z^4}}$.
\end{proof}

2

\begin{proof}
Using the same reasoning about eigenvalues mentioned above, clearly:

However we have to go through and see of all $N^i=\overrightarrow{0}$ for $1\le i\le 5$ and see the lowest $i$ that holds up:

Thus this implies that $\boxed{{\displaystyle {\mu }_N(z)=z^3}}$.
\end{proof}

3

\begin{proof}
Let $\beta$ be any Jordan basis for $N$, so:

Now because we are considering the basis $\beta$, then the length of each string of consecutive 1's are each of the $m_j$'s in the equation above. That's because if we consider the rightmost vector in each "cycle" of the basis, then we start at some $v_j$, and because of the form of the 1's on the upper diagonal of $\mathcal{M}(N,\beta )$ then:

Notice then that:

Now if we apply $N$ to both sides again that implies:

Now if $m$ is the longest length of consecutive 1's on the upper diagonal, applying this logic shows that that's equivalent to the largest $m_j$ that we find. Hence $m=\text{max}(m_1,..,m_n)$. In that case, then:

for some selected $k$ where $m=m_k$. It's important to note that by definition here then $N^m=v_{k-m}\ne \overrightarrow{0}$.

Okay. Now because $N$ is nilpotent, then $\lambda =0$ is the only possible eigenvalue (see HW 7 - Generalized Eigenvectors, Operator Decomposition#7). As such, then:

Now ${\mu }_N(z)=z,...,z^{\dim (V)}$ are the options. However, I claim that:

We know that ${\mu }_{N(z)}\ne z^m$ because I can choose vector $v_k$ up above, where by definition:

So that eliminates options $z,...,z^m$. Now all we have to do is check that ${\mu }_N(z)=z^{m+1}$ by checking that for all $v\in V$ that ${\mu }_N(N)v=\overrightarrow{0}$. First, keep in mind that $v=\sum _{i=1}^n\sum _{j=1}^{m_i}{N}^j{v}_i$ by the definition of a basis using $\beta$ as that basis. Therefore:

so since $v$ was arbitrary, then ${\mu }_N(N)=\overrightarrow{0}$ showing that ${\mu }_N(z)=z^{m+1}$ as desired.
\end{proof}

4

\begin{proof}
For clarity, let's consider this case for a basis $\beta =\{v_1,...,v_n\}$ for any arbitrary matrix $A=\mathcal{M}(T,\beta )$. By the way we construct our matrix, we have the relationship:

is equivalent to saying:

so notice that if I swap $v_1\leftrightarrow v_n$, ... for all the vectors in $\beta$, creating a new basis ${\beta }^{\prime }$, then this is equivalent to starting with the bottom equation for our matrix, then working our way up:

which is equivalent to having the matrix:

Notice though that this is just rotating the matrix clockwise!

Therefore, to describe this for any matrix where $v_1,...,v_n$, then we start with a Jordon Form of the matrix:

where each $A_j$ follows:

Thus, using ${\beta }^{\prime }=\{v_n,...,v_1\}$ we get:

where each $A_j$ is just the rotation of the above description of $A_j$ just rotated:

\end{proof}

5

\begin{proof}
We can actually just use HW 8 - Operator Decomposition, Characteristic and Minimal Polynomials#9 to just calculate the form. For clarity:

where each $A_j$ follows:

So using the theorem we previously derived, then:

If $A_j$ is the matrix for the basis ${\beta }_j=\{v_{j+1},...,v_{j+m_j}\}$ then, by the definition that $A_j=\mathcal{M}(S,{\beta }_j)$ for some $S\in \mathcal{L}(V)$:

But notice that:

Therefore we can derive the matrix version for $A_j^2$:

Therefore, now each $A_j^2$ is of the form of the matrix above.
\end{proof}

9.A: Complexification

2

\begin{proof}
(additivity): Let $u_1+i{v}_1,u_2+i{v}_2\in V_C$ (so $u_1,u_2,v_1,v_2\in V$). Then:

(homogeneity): Let $u+iv\in V_C$ and $\lambda \in \mathbb{C}$. Then:

Thus $T_C\in \mathcal{L}(V_C)$.
\end{proof}

3

\begin{proof}
($\to$): Suppose $v_1,...,v_m\in V$ is LI in $V_C$ meaning that $\sum _{j=1}^m{\alpha }_j{v}_j=\overrightarrow{0}\iff \mathrm{\forall }j({\alpha }_j=0)$ where ${\alpha }_j\in \mathbb{C}$. Consider when ${\alpha }_j\in \mathbb{R}$:

Since each ${\alpha }_j\in \mathbb{R}$ equals ${\beta }_j={\alpha }_j+i0\in \mathbb{C}$ then:

so since ${\beta }_j\in \mathbb{C}$ then since we have LI in $V_C$ (see above) then for all $j$ then ${\beta }_j=0={\alpha }_j+0i={\alpha }_j$ showing LI in $V$.

($\leftarrow$): Suppose $v_1,...,v_m$ is LI in $V$, so then $\sum _{j=1}^m{\alpha }_j{v}_j=\overrightarrow{0}\iff \mathrm{\forall }j({\alpha }_j=0)$ for constants ${\alpha }_j\in \mathbb{R}$. Consider the sum:

for ${\alpha }_j\in \mathbb{C}$. Clearly:

where $a_j,b_j\in \mathbb{R}$. Expanding the sum:

We can equate just the real and imaginary components to get:

Thus then for all $j$ then $a_j+i{b}_j={\alpha }_j=0$ showing LI on $V_C$.
\end{proof}

4

\begin{proof}
From HW 9 - Jordon Form, Complexification#3 we have that both $V$ and $V_C$ would have $v_1,...,v_m$ be LI in their respective vector spaces. Since $\dim (V)=\dim (V_C)$ that implies that both are basis of both (as an equivalence), so clearly they must span their respective vector spaces.
\end{proof}

5

\begin{proof}
Let $w=u+iv\in V_C$ be arbitrary, so $u,v\in V$. Notice:

Since $w$ was arbitrary then $(S+T{)}_C=S_C+T_C$ immediately follows.

Similarly, for $\lambda \in \mathbb{R}$:

thus $(\lambda T{)}_C=\lambda T_C$ since $w$ was arbitrary.
\end{proof}

6

\begin{proof}
We'll prove the opposite, that $T_C$ is not invertible iff $T$ is not invertible.

$T=T-0I$ is not invertible iff $0$ is an eigenvalue of $T$. Since $\lambda =0\in \mathbb{R}$ then that's iff $\lambda =0$ is an eigenvalue of $T_C$ iff $T_C-0I=T_C$ is not invertible.
\end{proof}

7

\begin{proof}
($\leftarrow$): $N$ is nilpotent iff $N^{\dim (V)}=\overrightarrow{0}$. Consider an arbitrary $w\in V_C$, so then:

so since $w$ was arbitrary then $N_C^{\dim (V)}=\overrightarrow{0}$ iff $N_C$ is nilpotent.

$(\to )$: $N_C$ is nilpotent iff $N_C^{\dim (V)}=\overrightarrow{0}$. Consider an arbitrary $v\in V$. Consequently $v\in V_C$, so then:

So clearly $N^{\dim (V)}=\overrightarrow{0}$ iff $N$ is nilpotent.
\end{proof}

8

\begin{proof}
Clearly since $\dim ({\mathbb{R}}^3)=3$ then if $\dim G(5,T)=2$ then $\dim G(7,T)=1$ and thus since $T_C$ shares all real eigenvalues of $T$, then $T_C$ would only have $5,7$ has eigenvalues which are all real. A similar arguments works if instead $\dim G(7,T)=2$ and $\dim G(5,T)=1$.

Clearly the only situation we have a different eigenvalue is if $\dim G(5,T)=\dim G(7,T)=1$. Then if we have a new eigenvalue $\lambda$ then $\dim G(\lambda ,T)=1$ since the sum of the dimensions must equal 3 in this case. But notice that $\lambda$ cannot be complex, since for contradiction if it was then that implies that the multiplicity of $\lambda$ for $T_C$ would have to equal that for $\bar{\lambda }$ of $T_C$, which implies that the sum of the dimensions of the generalized eigenspaces would be $\dim G(5,T_C)+\dim G(7,T)+\dim G(\lambda ,T_C)+G(\bar{\lambda },T_C)=2+\dim G(\lambda ,T_C)+G(\bar{\lambda },T_C)\ge 4$ which would imply $\dim (V)\ge 4$ which is a contradiction.

Thus $\lambda \in \mathbb{R}$, so then only $5,7,\lambda$ are the eigenvalues for $T_C$ which are all real, as desired.
\end{proof}

15 (can i make the argument of invariance near the end?)

\begin{proof}
Let $U$ be a subspace of $V$ invariant under $T$. We'll show $\dim (U)$ must be even. To do this, assume for contradiction that it's odd. Then the basis ${\beta }_U=\{u_1,...,u_k\}$ spans $U$, namely:

it's composed of two subspaces $U_o+U_e$. Here $U_o$ is just $\text{span}(u_1)$ and $U_e=\text{span}(u_2,...,u_k)$. Namely:

which comes directly from the fact that we split the basis vectors to create the subspaces $U_e$ and $U_o$. Notice that since $U$ is invariant under $T$ then by HW 5 - Polynomials, Invariants, and Eigenvectors#4 then $U_o,U_e$ must be invariant under $T$.

Notice though that $U_o=\text{span}(u_1)$, so since $U_o$ is invariant under $T$ then $T{u}_1\in U_o=\text{span}(u_1)$ which only happens if $T{u}_1=\lambda u_1$ implying that $u_1$ is an eigenvector of $T$. But that's a contradiction as $T$ has no eigenvalues.

Therefore, we must have $\dim (U)$ is odd, and since $U$ was arbitrary, it holds for all subspaces of $V$ invariant under $T$.
\end{proof}

19 (almost over!)

\begin{proof}
We still have it that:

As such because $\text{null}(T^{n-2})\ne \text{null}(T^{n-1})$ then we must have a strict subset near the end.

But because if we have equality in our subset chain that implies the following must be equal, and that would contradict $\text{null}(T^{n-2})\ne \text{null}(T^{n-1})$, then the ones prior must be strict subsets as well:

Applying the dimensionality argument shows that:

This implies that:

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 $\lambda \in \mathbb{F}$).

This shows that $T$ has at most two distinct eigenvalues. We'll now show that $T_C$ has no non-real eigenvalues. Notice that we have ${\lambda }_1=0$ and ${\lambda }_2\in \mathbb{F}$.

• For both cases, if $\lambda \in \mathbb{C}$ is an eigenvalue for $T_C$ that implies that $\bar{\lambda }$ is an eigenvalue for $T_C$.
• In the former case, clearly $T$ and $T_C$ share real eigenvalues, so then $\lambda$ would be an eigenvalue of
  \end{proof}