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\headers{Math343}{Homework 12}{Due Nov 12, 2025}

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\noindent
\textbf{Directions:} You may work to solve these problems in groups, but all written work must be your own.  Show all work; no credit for solutions without work..

\begin{enumerate}

\item {} [5.3.4] Given the factorization $A=PDP^{-1}$ below, find a formula for $A^k$ where $k$ is a nonnegative integer.
\[ \mtrix{2}{-6 & 8 \\ -4 & 6} = \mtrix{2}{1 & 2 \\ 1 & 1}\mtrix{2}{2 & 0 \\ 0 & -2}\mtrix{2}{-1 & 2 \\ 1 & -1} \]

\item {} [5.3.\{7-20\}] Diagonalize the following matrices if possible.  That is, for each diagonalizable matrix $A$ below, construct an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$.  (There is no need to compute $P^{-1}$ explicitly.)  For each matrix $A$ below that is not diagonalizable, explain why not.
\begin{multicols}{2}
\begin{enumerate}

\item $\mtrix{2}{1 & 0 \\ 6 & -1}$
\item $\mtrix{2}{3 & -1 \\ 1 & 5}$
\item $\mtrix{3}{-1 & 4 & -2\\ -3 & 4 & 0 \\ -3 & 1 & 3}$
\item $\mtrix{3}{4 & 0 & 2\\ 2 & 3 & 4\\ 0 & 0 & 3}$
\item $\mtrix{4}{2 & 0 & 0 & 0\\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 2}$

\end{enumerate}
\end{multicols}

\item {} [5.3.\{21-28\}] True/False.  In the following, $A$, $P$, and $D$ are $(n\times n)$ matrices.  Justify your answers.
\begin{enumerate}
	\item $A$ is diagonalizable if $A=PDP^{-1}$ for some matrix $D$ and some invertible matrix $P$.
	\item If $\RR^n$ has a basis of eigenvectors of $A$, then $A$ is diagonalizable.
	\item $A$ is diagonalizable if and only if $A$ has $n$ eigenvalues, counting multiplicities.
	\item If $A$ is diagonalizable, then $A$ is invertible.
	\item $A$ is diagonalizable if $A$ has $n$ eigenvectors.
	\item If $A$ is diagonalizable, then $A$ has $n$ distinct eigenvalues.
	\item If $AP=PD$, with $D$ diagonal, then the nonzero columns of $P$ must be eigenvectors of $A$.
	\item If $A$ is invertible, then $A$ is diagonalizable.
\end{enumerate}

\item {} [5.3.31] A is a $4\times 4$ matrix with three eigenvalues.  One eigenspace is one-dimensional, and one of the other eigenspaces is two-dimensional.  Is it possible that $A$ is not diagonalizable?  Justify your answer.

\item {} Let $\vv v_1 = \MTWBO{3}{1}$ and $\vv v_2 = \MTWBO{2}{7}$, and let $\lambda_1$ and $\lambda_2$ be scalars.  Construct a $(2\times 2)$-matrix $A$ having eigenvectors $\vv v_1$ and $\vv v_2$ with respective eigenvalues $\lambda_1$ and $\lambda_2$.  (Here, the entries of $A$ will depend on $\lambda_1$ and $\lambda_2$.)


\end{enumerate}
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