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\runningheader{Math 308L Autumn 2017}
{Final Exam, Page \thepage\ of \numpages}
{December 14, 2017}
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\begin{document}
\begin{center}
Math 308L - Autumn 2017
Final Exam
December 14, 2017
\end{center}
\ifprintanswers
\textbf{\huge KEY}
\else
Name: \hrulefill
Student ID Number: \hrulefill
\fi
\vspace{0.3cm}
\begin{center}
\gradetable[v][questions]
\end{center}
\vspace{0.3cm}
\begin{itemize}
\item
There are 7 problems on this exam. Be sure you have all 7 problems on
your exam.
\item
The final answer must be left in exact form. Box your final answer.
\item
You are allowed the TI-30XIIS calculator. It is possible to complete
the exam without a calculator.
\item
You are allowed a single sheet of 2-sided self-written notes.
\item
You must show your work to receive full credit. A correct answer
with no supporting work will receive a zero.
\item
Use the backsides if you need extra space. Make a note of this if you
do.
\item
Do not cheat. This exam should represent your own work. If you are
caught cheating, I will report you to the Community Standards and
Student Conduct office.
\end{itemize}
\textbf{Conventions}:
\begin{itemize}
\item
I will often denote the zero vector by $0$.
\item
When I define a variable, it is defined for that whole question. The $A$
defined in Question 2 is the same for each part.
\item
I often use $x$ to denote the vector $(x_1,x_2,\ldots,x_n)$. It should be clear from context.
\item
Sometimes I write vectors as a row and sometimes as a column. The
following are the same to me.
\[
(1,2,3) \quad
\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}.
\]
\item
I write the evaluation of linear transformations in a few ways. The
following are the same to me.
\[
T(1,2,3) \quad T((1,2,3)) \quad T \left(
\begin{bmatrix}
1 \\ 2\\ 3
\end{bmatrix}
\right)
\]
\end{itemize}
\newpage
\begin{questions}
\question
Give an example of each of the following. If it is not possible, write
``NOT POSSIBLE''.
\begin{parts}
\part[2]
Give an example of a $2\times 3$ matrix $A$ and a vector $b\in
\mathbb{R}^2$ such that $Ax=b$ has no solutions but $Ax=0$ has
infinitely many solutions.
\begin{solution}
Let
\[
A =
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\]
and $b=(0,1)$.
\end{solution}
\vfill
\part[2]
Give an example of a linear system in 3 variables whose solution space
is the intersection of the $x+y+z=0$ plane and the $xy$-plane.
\begin{solution}
The linear system given by
\[
\begin{aligned}
x+y+z &= 0 \\
z &= 0
\end{aligned}
\]
\end{solution}
\vfill
\part[2]
Give an example of a $2\times 2$ matrix $A$ such that $A^4 = I_2$ but
$A^2 \neq I_2$. If possible, give the matrix $A$ explicitly.
\begin{solution}
Let $A$ be the rotation by $\pi/2$ matrix. This is given by
\[
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}.
\]
\end{solution}
\vfill
\part[2]
Give an example of 2 linear transformations $T:\mathbb{R}^2\to \mathbb{R}^2$
and $S:\mathbb{R}^2\to\mathbb{R}^2$ such that $\range(T)=\ker(S)$.
\begin{solution}
Let $T(x,y)=(x,y)$ and $S(x,y)=(0,0)$.
\end{solution}
\vfill
\part[2]
Give an example of an orthogonal matrix that is not invertible.
\begin{solution}
NOT POSSIBLE. The inverse of an orthogonal matrix is its transpose.
\end{solution}
\vfill
\part[2]
Give an example of an diagonalizable matrix that is not orthogonally
diagonalizable.
\begin{solution}
\[
\begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix}
\]
\end{solution}
\vfill
\end{parts}
\newpage
\question
Let $A$ be defined by
\[
A =
\begin{bmatrix}
1 & 2 & 1\\
2 & 4 & 2\\
0 & 0 & 1
\end{bmatrix}.
\]
\begin{parts}
\part[4]
Find a basis for the solution space $Ax=0$.
\begin{solution}
$\{(2,-1,0)\}$
\end{solution}
\vfill
\part[4]
What is the general solution to $Ax=
\begin{bmatrix}
3 \\ 6 \\ -3
\end{bmatrix}$?
\begin{solution}
$(6,0,-3)+s_1(2,-1,0)$.
\end{solution}
\vfill
\part[4]
Is there a vector $y\in \mathbb{R}^3$ such that $Ax=y$ has no
solutions? If so, give an example. If not, why not?
\begin{solution}
Yes. Many possibilities.
\end{solution}
\vfill
\end{parts}
\newpage
\question
Let $A$ and $B$ be equivalent matrices defined by
\[
A =
\begin{bmatrix}
-3 & 3 & -1 & -9 & 3\\
2 & -2 & 1 & 7 & -1\\
4 & -4 & 5 & 23 & 7
\end{bmatrix}
\sim
\begin{bmatrix}
1 & -1 & 0 & 2 & -2\\
0 & 0 & 1 & 3 & 3\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
= B.
\]
\begin{parts}
\part[4]
Find a basis for the solution space of $Ax=0$.
\begin{solution}
$\{(1,1,0,0,0),(-2,0,-3,1,0),(2,0,-3,0,1)\}$
\end{solution}
\vfill
\part[4]
Let $a_1,a_2,a_3,a_4,a_5$ be the columns of $A$. Define $C=[a_1 \;
a_2\; a_3 \; a_4]$. What is a particular solution to $Cx = a_5$?
\begin{solution}
$(-2, 0, 3, 0)$.
\end{solution}
\vfill
\part[4]
Using the same variables as (b), what is the general solution to
$Cx=3a_4-a_5$?
\begin{solution}
$(8,0,6,0)+s_1(1,1,0,0)+s_2(-2,0,-3,1)$.
\end{solution}
\vfill
\end{parts}
\newpage
\question
Let $S$ be a subspace of $\mathbb{R}^4$ defined by
\[
S=
\spn
\left\{
\begin{bmatrix}
1 \\ 1 \\ 1 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
1 \\ 1 \\ 0 \\ 0
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 0 \\ 1 \\ 1
\end{bmatrix}
\right\}.
\]
\begin{parts}
\part[3]
What is a basis for $(S^\perp)^\perp$?
\begin{solution}
$\{(1,1,0,0),(0,0,1,1)\}$.
\end{solution}
\vfill
\part[3]
What is a basis for $S^\perp$?
\begin{solution}
$\{(1,-1,0,0),(0,0,1,-1)\}$.
\end{solution}
\vfill
\part[3]
Does there exist a rank 2 matrix $A$ such that $\nll(A)=S$? If so, give
an example. If not, why not?
\begin{solution}
If $\nll(A)=S$ then $\row(A)=S^\perp$ so we can take
\[
\begin{bmatrix}
1 & -1 & 0 & 0 \\
0 & 0 & 1 & -1
\end{bmatrix}.
\]
\end{solution}
\vfill
\part[3]
Does there exist a rank 3 matrix $A$ such that $\nll(A)=S$? If so, give
an example. If not, why not?
\begin{solution}
No. By the rank-nullity theorem, $\rank(A)+\nll(A)=4$. Since $\dim
S=2$, the rank of $A$ must be 2.
\end{solution}
\vfill
\end{parts}
\newpage
\question
Let $T:\mathbb{R}^3\to\mathbb{R}^3$ be the linear transform defined by the
following properties:
\begin{itemize}
\item
$T(0,0,1)=(0,0,0)$,
\item
If $v$ is in the $xy$-plane, then $v$ is reflected across the
$x+y=0$ plane.
\end{itemize}
There is a matrix $A$ such that $T(x)=Ax$. The goal of this problem is to
understand $A$.
\begin{parts}
\part[3]
Find a basis $\{u,v,w\}$ where the action of $T$ is
well-understood. Give also $T(u), T(v)$, and $T(w)$.
\begin{solution}
\[
u = (0,0,1), T(u)=(0,0,0)
\]
\[
v = (1,1,0), T(v)=(-1,-1,0)
\]
\[
w = (1,-1,0), T(w)=(1,-1,0)
\]
\end{solution}
\vfill
\part[3]
Find the eigenvalues of $A$ and a basis for each eigenspace of $A$.
(Think geometrically.)
\begin{solution}
Part (a) gives the answer.
$\lambda=0$ is an eigenvalue with eigenspace spanned by $u$.
$\lambda=-1$ is an eigenvalue with eigenspace spanned by $v$.
$\lambda=1$ is an eigenvalue with eigenspace spanned by $w$.
\end{solution}
\vfill
\part[3]
What is $A$? You may express it as product of matrices and their
inverses.
\begin{solution}
Using the theory of diagonalization,
\[
A =
\begin{bmatrix}
0 & 1 & 1 \\
0 & 1 & -1 \\
1 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
0 & 1 & 1 \\
0 & 1 & -1 \\
1 & 0 & 0
\end{bmatrix}^{-1}
\]
\end{solution}
\vfill
\part[3]
What is $A^2$? Give it explicitly as a single matrix. (Think
geometrically.)
\begin{solution}
We can see that $A^2$ is projecting onto the $xy$-plane. So
\[
A^2 =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}.
\]
\end{solution}
\vfill
\end{parts}
\newpage
\question
Let $A$ be the symmetric matrix defined as
\[
A=
\begin{bmatrix}
1 & -1 & -1 \\
-1 & 1 & -1 \\
-1 & -1 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & -1 & -2 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
1 & -1 & -2 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{bmatrix}^{-1}.
\]
\begin{parts}
\part[3]
Find the eigenvalues of $A$ and a basis for each eigenspace of $A$.
\begin{solution}
$\lambda = -1$ is an eigenvalue with $\{(1,1,1)\}$ as a basis for
its eigenspace.
$\lambda = 2$ is an eigenvalue with $\{(-1,0,1),(-2,1,1)\}$ as a
basis for its eigenspace.
\end{solution}
\vfill
\vfill
\vfill
\part[3]
Find a basis for each of the following subspaces.
\begin{itemize}
\item
$\nll(A)$
\vfill
\begin{solution}
Since $0$ is not an eigenvalue, $\nll(A)=\{0\}$ with basis
$\emptyset$.
\end{solution}
\item
$\nll(A-I)$
\begin{solution}
Since $1$ is not an eigenvalue, $\nll(A)=\{0\}$ with basis
$\emptyset$.
\end{solution}
\vfill
\item
$\nll(A-2I)$.
\begin{solution}
We have that $\nll(A-2I)=E_2$ which has basis
$\{(-1,0,1),(-2,1,1)\}$.
\end{solution}
\vfill
\end{itemize}
\vfill
\part[3]
Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that
$A=QDQ^{-1}$.
\begin{solution}
We use Gram-Schmidt to perform an orthogonal basis for each
eigenspace.
An orthonormal basis for the eigenspace corresponding to
$\lambda=-1$ is $\{(1/3,1/3,1/3)\}$.
An orthonormal basis for $\lambda=2$ is
$\{\frac{1}{\sqrt{2}}(-1,0,1),\sqrt{\frac{2}{3}}(-1/2,1,-1/2)\}$.
\end{solution}
\vfill
\vfill
\vfill
\part[3]
Find all $k\in \mathbb{R}$ such that $A-kI_3$ is not invertible.
\begin{solution}
$k=-1,2$.
\end{solution}
\vfill
\end{parts}
\newpage
\question
Let $v=(2,2,1)$ and $T:\mathbb{R}^3\to\mathbb{R}^3$ be defined by
$T(x)=\proj_v x$.
\begin{parts}
\part[4]
Find an orthogonal basis for $\mathbb{R}^3$ that contains $v$. (Hint:
first find a basis for $\mathbb{R}^3$ that contains $v$.)
\begin{solution}
$\{(2,2,1),(1,0,-2),(0,1,-2)\}$.
\end{solution}
\vfill
\vfill
\vfill
\vfill
\vfill
\vfill
\vfill
\part[4]
There exists a matrix $A$ such that $T(x)=Ax$. Find the eigenvalues of
$A$ and a basis for each eigenspace of $A$. (Hint: see part (a).)
\begin{solution}
The eigenspace corresponding to $1$ is spanned by $(2,2,1)$.
The eigenspace corresponding to $0$ is spanned by
$(1,0,-2),(0,1,-2)$.
\end{solution}
\vfill
\vfill
\vfill
\vfill
\part[4]
Let $e_1=(1,0,0)$. Evaluate the following:
\begin{itemize}
\item
$Ae_1$
\begin{solution}
This is $T(e_1)=\proj_{v} e_1 = (4/9,4/9,2/9)$.
\end{solution}
\vfill
\item
$A^2e_1$
\begin{solution}
Doing two projections is the same as one.
\end{solution}
\vfill
\item
$A^{100}e_1$
\begin{solution}
Doing one hundred projections is the same as one.
\end{solution}
\vfill
\end{itemize}
\vfill
\end{parts}
\end{questions}
\end{document}
