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\begin{center}
Math 3013.62979\\[0pt]Solutions to Final Exam\\[0pt]2:00pm -- 4:00pm, December
8, 2020
\end{center}

1. Definitions

2. 

\begin{itemize}
\item[(a)] No solution (There is a pivot in the last column of the third row
which implies the contradictory equation $0=1$.)
\end{itemize}



\begin{itemize}
\item[(b)] There are infinitely many solutions. Since columns 3 and 4 don't
have pivots, there will be two free parameters in the general solution.
\end{itemize}



\begin{itemize}
\item[(c)] There is a unique solution (since each column of the agumented
matrix in RREF has a pivot).
\end{itemize}



3.%
\[
\mathbf{A}^{-1}=\left[
\begin{array}
[c]{|ccc}%
2 & 1 & 0\\
1 & 0 & 0\\
0 & -2 & 1
\end{array}
\right]
\]


4. 
\begin{itemize}
\item Since $W$ is closed under both scalar multiplication and vector
addition, $W$ is a subspace.
\end{itemize}

5. Consider the vectors $\left\{  [1,-1,-1,1],[2,-1,-2,0],[1,0,-1,-1],\left[
3,-2,-3,1\right]  \right\}  \in\mathbb{R}^{4}$

(a) (10 pts) Determine if these vectors are linearly independent.

\begin{itemize}
\item We form a $4\times4$ matrix $\mathbf{A}$ using the given vectors as
rows.
\[
\mathbf{A}=\left[
\begin{array}
[c]{cccc}%
1 & -1 & -1 & 1\\
2 & -1 & -2 & 0\\
1 & 0 & -1 & -1\\
3 & -2 & -3 & 1
\end{array}
\right]  \quad\rightarrow\quad\left[
\begin{array}
[c]{cccc}%
1 & -1 & -1 & 1\\
0 & 1 & 0 & 2\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\right]
\]
Since the REF of $\mathbf{A}$ has two zero rows, the original set of vectors
are not linearly independent.
\end{itemize}

(b) (5 pts) What is the dimension of the subspace generated by these vectors
(i.e. the subspace spanned by these vectors)?

\begin{itemize}
\item The two non-zero rows of the REF of $\mathbf{A}$ provide a basis for the
$RowSp\left(  \mathbf{A}\right)  =span\left(
[1,-1,-1,1],[2,-1,-2,0],[1,0,-1,-1],\left[  3,-2,-3,1\right]  \right)  .$

Since we have two basis vectors, the dimension of the subspace is $2$.
\end{itemize}



6. 

\begin{itemize}
\item
\[
\mathbf{A}_{T}=\left[
\begin{array}
[c]{ccc}%
\uparrow & \uparrow & \uparrow\\
T\left(  \left[  1,0,0\right]  \right)  & T\left(  \left[  0,1,0\right]
\right)  & T\left(  \left[  0,0,1\right]  \right) \\
\downarrow & \downarrow & \downarrow
\end{array}
\right]  =\left[
\begin{array}
[c]{ccc}%
-1 & 1 & 0\\
1 & -1 & 0
\end{array}
\right]
\]

\end{itemize}

(b) (10 pts) Find a basis for the kernel of $T$ (i.e. the set of vectors
$\mathbf{x}$ such that $T\left(  \mathbf{x}\right)  =\mathbf{0}$).

\begin{itemize}
\item
\[
\text{basis for }Ker\left(  T\right)  =\left\{  \left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  \ ,\ \left[
\begin{array}
[c]{c}%
0\\
0\\
1
\end{array}
\right]  \right\}
\]

\end{itemize}


\newpage
7. 

(a) \ (15 pts) Find the eigenvalues and the eigenvectors of the following
matrix : $\mathbf{A=}\left[
\begin{array}
[c]{ccc}%
2 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 2
\end{array}
\right]  $

(b) (5 pts) What are the algebraic multiplicities and geometric multiplicities
of the eigenvalues of $\mathbf{A}$?

\begin{itemize}
\item We have%
\[%
\begin{array}
[c]{cccc}%
\text{eigenvalue} & \text{basis for eigenspace} & \text{alg. mult.} &
\text{geom. mult.}\\
2 & \left\{  \left[
\begin{array}
[c]{c}%
1\\
0\\
0
\end{array}
\right]  \right\}  & 2 & 1\\
0 & \left\{  \left[
\begin{array}
[c]{c}%
-\frac{1}{2}\\
1\\
0
\end{array}
\right]  \right\}  & 1 & 1
\end{array}
\]

\end{itemize}

(c) (5 pts) Is this matrix diagonalizable?

\begin{itemize}
\item No. \ We need 3 linearly independent eigenvectors in order to
diagonalize a $3\times3$ matrix. \ But we only found two linearly independent
eigenvectors; and so $\mathbf{A}$ is not diagonalizable.
\end{itemize}



8. (10 pts) Let $\mathbf{A}$ be the matrix $\left[
\begin{array}
[c]{cc}%
3 & 1\\
1 & 3
\end{array}
\right]  $ . Find a $2\times2$ matrix $\mathbf{C}$ and a diagonal matrix
$\mathbf{D}$ such that $\mathbf{C}^{-1}\mathbf{AC}=\mathbf{D}$.

\begin{itemize}
\item
\[
\mathbf{D}=\left[
\begin{array}
[c]{cc}%
2 & 0\\
0 & 4
\end{array}
\right]  \qquad,\qquad\mathbf{C}=\left[
\begin{array}
[c]{cc}%
-1 & 1\\
1 & 1
\end{array}
\right]
\]

\end{itemize}

9. (10 pts) Find an orthonormal basis for the subspace $W$ generated by the
vectors $\mathbf{v}_{1}=\left[  1,-1,0\right]  $ and $\mathbf{v}_{2}=\left[
1,0,1\right]  $

\begin{itemize}
\item
\begin{align*}
\mathbf{n}_{1} &  =\frac{\mathbf{o}_{1}}{\sqrt{\mathbf{o}_{1}\cdot
\mathbf{o}_{1}}}=\frac{1}{\sqrt{2}}\left[  1,-1,0\right]  =\left[  \frac
{1}{\sqrt{2}}\ ,\ -\frac{1}{\sqrt{2}}\ ,\ 0\right]  \\
\mathbf{n}_{2} &  =\frac{\mathbf{o}_{2}}{\sqrt{\mathbf{o}_{2}\cdot
\mathbf{o}_{2}}}=\frac{1}{\sqrt{\left(  \frac{1}{4}\right)  ^{2}+\left(
\frac{1}{4}\right)  ^{2}+\left(  1\right)  ^{2}}}\left[
\begin{array}
[c]{ccc}%
\frac{1}{2} & \frac{1}{2} & 1
\end{array}
\right]  =\left[  \frac{1}{\sqrt{6}}\ ,\ \frac{1}{\sqrt{6}}\ ,\ \frac{2}%
{\sqrt{6}}\right]
\end{align*}

\end{itemize}



10. 
\begin{itemize}
\item
\[
\quad\left[  \mathbf{A}\mid\mathbf{b}\right]  =\quad\left[
\begin{array}
[c]{cccc}%
1 & 1 & -1 & 2\\
1 & 0 & 1 & 1\\
0 & 1 & 1 & 0
\end{array}
\right]  \rightarrow\left[  \mathbf{I}\mid\mathbf{v}_{B}\right]  =\left[
\begin{array}
[c]{cccc}%
1 & 0 & 0 & \frac{4}{3}\\
0 & 1 & 0 & \frac{1}{3}\\
0 & 0 & 1 & -\frac{1}{3}%
\end{array}
\right]
\]
and%
\begin{align*}
\mathbf{v}_{W} &  =\frac{4}{3}\left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  +\frac{1}{3}\left[
\begin{array}
[c]{c}%
1\\
0\\
1
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
\frac{5}{3}\\
\frac{4}{3}\\
\frac{1}{3}%
\end{array}
\right]  \\
\mathbf{v}_{\perp} &  =-\frac{1}{3}\left[
\begin{array}
[c]{c}%
-1\\
1\\
1
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
\frac{1}{3}\\
-\frac{1}{3}\\
-\frac{1}{3}%
\end{array}
\right]  :
\end{align*}

\end{itemize}


\end{document}
