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\noindent
\parbox{2.25in}{\bf Math 4010-101}
\hfill {\Large \bf Homework \#5} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Mar. $23^\mathrm{rd}$, 2022}

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%\noindent {\bf\large Name:} \quad \underline{\large\qquad This is my Name \qquad}

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\noindent
Please remember when submitting any work via email or in person to\dots\\
\force \hfill \fbox{\fbox{\bf\large\color{red}PUT YOUR NAME ON YOUR WORK!}} \hfill \force


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\begin{description}
\item[\#1. Invariant, Cyclic, Minimal] I'll go over some background. This may look intimidating, but the problems below are actually pretty straightforward.

Let $T:V \to V$ be a linear operator on a finite dimensional vector spaces over some field $\mathbb{F}$. Let $f(t)=a_nt^n+\cdots+a_1t+a_0$ be a polynomial with coefficients in $\mathbb{F}$ (i.e., $f(t) \in \mathbb{F}[t]$). Recall that $f(T):V \to V$ is a linear operator where $f(T)(\B v)=a_nT^n(\B v)+\cdots+a_1T(\B v)+a_0\B v$ for all $\B v \in V$ and $T^k(\B v)=\underbrace{T(T(\cdots T(}_{k-\mathrm{times}}\B v)\cdots))$. 

There is a monic (i.e., leading coefficient is 1) polynomial, $m(t)$, such that $m(T)=0$ (the zero operator: $(m(T))(\B v)=\B 0$ for all $\B v \in V$) and given any other polynomial $g(t)$ such that $g(T)=0$, we have that $g(t)$ is a multiple of $m(t)$. This polynomial is called the {\it minimal polynomial} of $T$. 

{\it Note:} One can show that the set of all polynomials $g(t)$ that annihilate $T$ (i.e., $g(T)=0$) form something an {\it ideal} in the polynomial ring $\mathbb{F}[t]$. It turns out (since $\mathbb{F}[t]$ is a principal ideal domain) that this ideal has a generator. In fact, it has a {\it unique} monic (= leading coefficient is 1) generator. This unique monic generator is nothing more than our minimal polynomial.

Let $W$ be a subspace of $V$. We say $W$ is $T$-invariant if $T(\B w) \in W$ for all $\B w \in W$ (i.e., $T(W) \subseteq W$).

If $W$ is $T$-invariant, we can restrict $T$'s  domain to $W$ and since $T(W)\subseteq W$, we can also cut its codomain down to $W$. Thus we get a map $S=T\Big|_W:W \to W$. In other words, $S$ is a linear operator on $W$ (instead of $V$) and it matches $T$ on its domain (i.e., $S(\B w)=T(\B w)$ for all $\B w \in W$).

\begin{enumerate}[(a)]
\item Suppose $W$ is $T$-invariant. Show that $W$ is also $f(T)$-invariant.

\item Let $\B v \in V$ and let $W = \mathrm{span}\{\B v, T(\B v), T^2(\B v),\dots \}$ (i.e., $W$ is the $T$-cyclic subspace generated by $\B v$). Obviously $W$ is a subspace because it is defined as a span. Show that $W$ is $T$-invariant.

\item Again, let $W = \mathrm{span}\{\B v, T(\B v), T^2(\B v),\dots \}$. Suppose that $T\Big|_W=S$ (the restriction of $T$ to $W$) has a minimal polynomial $m(t)=t^3-2t+5$. Prove that $\alpha = \{\B v,T(\B v), T^2(\B v)\}$ is a basis for $W$ and find $[S]_\alpha$.

{\it Hint:} Showing $\alpha$ is a basis -- elements of $W$ are of the form $g(T)(\B v)$ for some polynomial $g(t)$. Consider dividing $g(t)$ by $m(t)$: there exist unique polynomials $q(t)$ and $r(t)$ such that $g(t)=q(t)m(t)+r(t)$ where either $r(t)=0$ or the degree of $r(t)$ is less than the degree of $m(t)$.

{\it Note:} $[S]_\alpha$ is an example of something called a {\it companion matrix}. Companion matrices are the building blocks for rational canonical form.
\end{enumerate}

\item[\#2. Eigenproblem] Let $T:V \to V$ be a linear operator on a finite dimensional vector spaces over some field $\mathbb{F}$.

\begin{enumerate}[(a)]
\item Explain why $\lambda=0$ is an eigenvalue if and only if $T$ is not invertible. Then suppose $T$ is invertible and show that if $\lambda$ is an eigenvalue of $T$, then $\lambda^{-1}$ is an eigenvalue of $T^{-1}$.

\item {}{\bf [Extra credit:]} Recall that $T^*:V^* \to V^*$ is the transpose map defined by $T^*(f)=f\circ T$. Show that $T$ and $T^*$ have the same eigenvalues.
\end{enumerate}

\item[\#3. Jordan Forms] If $A$ is a $2\times 2$ matrix (working over an algebraically closed field like the complex numbers), then $A$ must either have one or two distinct eigenvalues. The possible Jordan forms of $A$ look like: $J_1=\begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}$, $J_2=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}$, or $J_3=\begin{bmatrix} \lambda & 0 \\ 0 & \mu \end{bmatrix}$. These go with characteristic polynomials $f_1(t)=(t-\lambda)^2$, $f_2(t)=(t-\lambda)^2$, and $f_3t(t)=(t-\lambda)(t-\mu)$ and minimal polynomials $m_1(t)=t-\lambda$, $m_2(t)=(t-\lambda)^2$, and $m_3(t)=(t-\lambda)(t-\mu)$. The first and third are diagonalizable while the second is not.

\begin{enumerate}[(a)]
\item Suppose $\lambda$, $\mu$, and $\eta$ are 3 distinct scalars. If $A$ is $3\times 3$ it can have eigenvalues: $\lambda$, $\mu$, and $\eta$ (all distinct); $\lambda$, $\mu$ with $\lambda$ repeated twice; OR just $\lambda$ repeated 3 times. 

Write down all possible $3 \times 3$ Jordan forms (up to reorganizing). State each form's characteristic and minimal polynomial and whether the corresponding matrix is diagonalizable.
\item A square matrix $A$ is {\it nilpotent} if $A^k=0$ for some positive exponent $k>0$. Suppose $A$ is diagonalizable, show that $A=0$. 

{\it Hint:} Eigenvalues of nilpotents are? Minimal polynomial of a diagonalizable matrix?

\item We have a matrix $B$ whose eigenvalues are $1$ and $2$. Moreover: $\dim(\ker(B-I))=2$, $\dim(\ker(B-I)^2)=3$, $\dim(\ker(B-I)^3)=3$; $\dim(\ker(B-2I))=2$, $\dim(\ker(B-2I)^2)=4$, $\dim(\ker(B-2I)^3)=4$. Write down the Jordan form of $B$ along with $B$'s characteristic and minimal polynomials. Is $B$ diagonalizable? Is $B$ invertible? What is $\det(B)$? What is $\mathrm{trace}(B)$?
\end{enumerate}

\item[\#4. Computational Difficulties] You probably should do these computations in Maple (or similar software). Feel free to submit a Maple file (or pdf print of such).
\begin{enumerate}[(a)]
\item Let $A = \begin{bmatrix} 2 & -1 & -1 & -1 & 0 & -1/2 & -1 \\ 2 & 4 & -1 & 3 & -1 & -1/2 & -1 \\ -6 & -4 & 5 & -5 & 1 & 1 & 3 \\ -4 & -3 & 3 & -3 & 2 & 2 & 3\\ -6 & -4 & 3 & -5 & 3 & 3/2 & 3 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 8 & 5 & -5 & 7 & -2 & -5/2 & -3 \end{bmatrix}$.

Copy/paste into Maple:

{\color{red}
\verb#with(LinearAlgebra):#\\
\verb#A := <<2,2,-6,-4,-6,0,8>|<-1,4,-4,-3,-4,0,5>|<-1,-1,5,3,3,0,-5>|<-1,3,-5,-3,-5,0,7>|#\\
\verb#       <0,-1,1,2,3,0,-2>|<-1/2,-1/2,1,2,3/2,2,-5/2>|<-1,-1,3,3,3,0,-3>>;#
}

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Find the Jordan form of $A$ as well as an invertible matrix $P$ (i.e., a {\it transition matrix}) such that $P^{-1}AP$  is the Jordan form of $A$. You may use Maple to do your computations, but I want to see your work building up a basis of generalized eigenvectors (i.e. don't just use the ``JordanForm" command). 

\item Let $\displaystyle B = \begin{bmatrix} -17 & -9 & -3 \\ 36 & 19 & 6 \\ 18 & 9 & 4 \end{bmatrix}$.

Copy/paste into Maple:

{\color{red}
\verb#B := <<-17,36,18>|<-9,19,9>|<-3,6,4>>;#
}

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As before, find the Jordan form and a corresponding transition matrix for $B$ (show your work). Then use these matrices to find $\sqrt{B}$ (i.e., $(\sqrt{B})^2=B$) and $e^B$. [{\bf Extra credit:} Find $\cos(B)$.]

\end{enumerate}

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