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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. Give the definitions of the following linear algebraic concepts:

(a) (5 pts)\ a \textbf{subspace} of a vector space $V$.

\begin{itemize}
\item A subspace of vector space $V$ is a subset $W$ of $V$ such that

\begin{itemize}
\item[(i)] whenever $\lambda\in\mathbb{R}$, $\mathbf{w}\in W\ ,\quad
\quad\left(  \lambda\mathbf{w}\right)  \in W$

\item[(ii)] whenever $\mathbf{w}_{1},\mathbf{w}_{2}\in W\ ,\ \quad\left(
\mathbf{w}_{1}+\mathbf{w}_{2}\right)  \in W$
\end{itemize}
\end{itemize}

(b) (5 pts) a \textbf{basis} for a subspace $W$ of a vector space $V$

\begin{itemize}
\item A set of vectors $\left\{  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right\}
$ is a basis for a subspace $W$ if

\begin{itemize}
\item $W=span\left(  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right)  $

\item If $\mathbf{w}=c_{1}\mathbf{v}_{1}+\cdots+c_{k}\mathbf{v}_{k}$, then
coefficients $c_{1},\ldots,c_{k}$ are unique.
\end{itemize}
\end{itemize}

(c) (5 pts) a set of \textbf{linearly independent} vectors

\begin{itemize}
\item A set of vectors $\left\{  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right\}
$ is a linearly independent set if the only solution of%
\[
x_{1}\mathbf{v}_{1}+\cdots+x_{k}\mathbf{v}_{k}=\mathbf{0}%
\]
is%
\[
x_{1}=0,\ldots,x_{k}=0
\]

\end{itemize}

(d) (5 pts) a \textbf{linear transformation} between two vector spaces $V$ and
$W$.

\begin{itemize}
\item A linear transformation between two vector space $V$ and $W$ is a
function $T:V\rightarrow W$ such that

\begin{itemize}
\item[(i)] $T\left(  \lambda\mathbf{v}\right)  =\lambda T\left(
\mathbf{v}\right)  $ for all $\lambda\in\mathbb{R}$ and all $\mathbf{v}\in V$

\item[(ii)] $T\left(  \mathbf{v}_{1}+\mathbf{v}_{2}\right)  =T\left(
\mathbf{v}_{1}\right)  +T\left(  \mathbf{v}_{2}\right)  $ for all
$\mathbf{v}_{1},\mathbf{v}_{2}\in V$
\end{itemize}
\end{itemize}

2. For each of the following augmented matrices, describe the solution space
of the corresponding linear system. (Determine if there are solutions; and, if
there are solutions, how many free parameters are needed to describe the
general solution.)

(a) (5 pts) $\left[
\begin{array}
[c]{cccc}%
1 & 0 & 2 & 1\\
0 & 1 & 0 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
2\\
1\\
0
\end{array}
\right.  \right]  \quad$

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

(b) (5 pts)$\left[
\begin{array}
[c]{cccc}%
1 & 0 & 4 & 2\\
0 & 2 & 0 & 3\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
2\\
0\\
0
\end{array}
\right.  \right]  $

\begin{itemize}
\item 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}

(c) (5 pts) $\left[
\begin{array}
[c]{ccc}%
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
0\\
3\\
0
\end{array}
\right.  \right]  $

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

\newpage

3. (10 pts)\ Compute the inverse of $\mathbf{A}=$ $\left[
\begin{array}
[c]{ccc}%
0 & 1 & 0\\
1 & -2 & 0\\
0 & 2 & 1
\end{array}
\right]  $

\begin{itemize}
\item
\begin{align*}
\left[  \mathbf{A}\mid\mathbf{I}\right]    & =\left[
\begin{array}
[c]{ccc|ccc}%
0 & 1 & 0 & 1 & 0 & 0\\
1 & -2 & 0 & 0 & 1 & 0\\
0 & 2 & 1 & 0 & 0 & 1
\end{array}
\right]  \quad\underrightarrow{R_{1}\longleftrightarrow R_{2}}\quad\left[
\begin{array}
[c]{ccc|ccc}%
1 & -2 & 0 & 0 & 1 & 0\\
0 & 1 & 0 & 1 & 0 & 0\\
0 & 2 & 1 & 0 & 0 & 1
\end{array}
\right]  \\
& \quad\underrightarrow{R_{3}\rightarrow R_{3}-2R_{2}}\quad\left[
\begin{array}
[c]{ccc|ccc}%
1 & -2 & 0 & 0 & 1 & 0\\
0 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & -2 & 1
\end{array}
\right]  \\
\quad\underrightarrow{R_{1}\rightarrow R_{1}+2R_{2}}\quad\left[
\begin{array}
[c]{ccc|ccc}%
1 & 0 & 0 & 2 & 1 & 0\\
0 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & -2 & 1
\end{array}
\right]    & =\left[  \mathbf{I}\mid\mathbf{A}^{-1}\right]
\end{align*}
and so%
\[
\mathbf{A}^{-1}=\left[
\begin{array}
[c]{|ccc}%
2 & 1 & 0\\
1 & 0 & 0\\
0 & -2 & 1
\end{array}
\right]
\]

\end{itemize}

4. (10 pts) Let $W=\left\{  \left[  x,y\right]  \in\mathbb{R}^{2}\mid
x+y=3\in\mathbb{R}\right\}  $. Prove or disprove that $W$ is a subspace of
$\mathbb{R}^{2}$.

\begin{itemize}
\item Every vector in $W$ is of the form $\left[  x,-x\right]  $,
$x\in\mathbb{R}$. \ If $\mathbf{w}=\left[  -x,x\right]  \in W$, we have%
\[
\lambda\in\mathbb{R}\text{, }\mathbf{w}=\left[  x,-x\right]  \in
W\quad\Rightarrow\quad\left(  \lambda\mathbf{w}\right)  =\left[  \lambda
x,-\lambda x\right]  \in W
\]
so $W$ is closed under scalar multiplication. 

If $\mathbf{w}_{1}=\left[  x,-x\right]  $, $\mathbf{w}_{2}=\left[
y,-y\right]  \in W$, then%
\[
\left(  \mathbf{w}_{1}+\mathbf{w}_{2}\right)  =\left[  x+y,-\left(
x+y\right)  \right]  \in W
\]
and so $W$ is also closed under vector addition.

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.
\begin{align*}
\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\underrightarrow{%
\begin{array}
[c]{c}%
R_{2}\rightarrow R_{2}-2R_{1}\\
R_{3}\rightarrow R_{3}-R_{1}\\
R_{4}\rightarrow R_{4}-3R_{1}%
\end{array}
}\quad\left[
\begin{array}
[c]{cccc}%
1 & -1 & -1 & 1\\
0 & 1 & 0 & 2\\
0 & 1 & 0 & 2\\
0 & 2 & 0 & 4
\end{array}
\right]  \\
& \quad\underrightarrow{%
\begin{array}
[c]{c}%
R_{3}\rightarrow R_{3}-R_{2}\\
R_{4}\rightarrow R_{4}-2R_{2}%
\end{array}
}\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]
\end{align*}
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}

\newpage

6. Consider the following linear transformation: $T:\mathbb{R}^{3}%
\rightarrow\mathbb{R}^{2}:$ $T\left(  \left[  x_{1},x_{2},x_{3}\right]
\right)  =\left[  x_{2}-x_{1}\,,\,x_{1}-x_{2}\right]  $.

(a) (10 pts) Find a matrix that represents $T$.

\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
\begin{align*}
Ker\left(  T\right)    & =NullSp\left(  \mathbf{A}_{T}\right)  =NullSp\left(
\left[
\begin{array}
[c]{ccc}%
-1 & 1 & 0\\
1 & -1 & 0
\end{array}
\right]  \right)  \\
& =NullSp\left(  RREF\left(  \mathbf{A}_{T}\right)  \right)  =NullSp\left(
\left[
\begin{array}
[c]{ccc}%
1 & -1 & 0\\
0 & 0 & 0
\end{array}
\right]  \right)  \\
& =\left\{  \mathbf{x}=\left[  x_{1},x_{2},x_{3}\right]  \in\mathbb{R}^{3}\mid
x_{1}-x_{2}=0\right\}  \\
& =\left\{  \left[
\begin{array}
[c]{c}%
x_{2}\\
x_{2}\\
x_{3}%
\end{array}
\right]  \mid x_{2},x_{3}\in\mathbb{R}\right\}  \\
& =span\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{align*}
and so%
\[
\text{basis for }Ker\left(  T\right)  =\left\{  \left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  \ ,\ \left[
\begin{array}
[c]{c}%
0\\
0\\
1
\end{array}
\right]  \right\}
\]

\end{itemize}

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]  $

\begin{itemize}
\item We have%
\[
p_{\mathbf{A}}\left(  \lambda\right)  =\det\left(  \mathbf{A}-\lambda
\mathbf{I}\right)  =\det\left(  \left[
\begin{array}
[c]{ccc}%
2-\lambda & 1 & 0\\
0 & 0-\lambda & 1\\
0 & 0 & 2-\lambda
\end{array}
\right]  \right)  =\left(  2-\lambda\right)  ^{2}\left(  -\lambda\right)
\]


So the eigenvalues of $\mathbf{A}$ are $2$ and $0$. \ Since we have two
factors of $\left(  2-\lambda\right)  $ in $p_{\mathbf{A}}\left(
\lambda\right)  $ the eigenvalue $\lambda=2$ has algebraic multiplicity $2$.
Since we have only one factor of $\left(  0-\lambda\right)  $ in
$p_{\mathbf{A}}\left(  \lambda\right)  $, the eigenvalue $\lambda=0$ has
algebraic multiplicity $1.$

$2$-eigenspace%
\begin{align*}
E_{2}  & =NullSp\left(  \mathbf{A}-\left(  2\right)  \mathbf{I}\right)
=NullSp\left(  \left[
\begin{array}
[c]{ccc}%
0 & 1 & 0\\
0 & -2 & 1\\
0 & 0 & 0
\end{array}
\right]  \right)  =\left[
\begin{array}
[c]{ccc}%
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}
\right]  \\
& =span\left(  \left[
\begin{array}
[c]{c}%
1\\
0\\
0
\end{array}
\right]  \right)
\end{align*}
Since the $2$-eigenspace of $\mathbf{A}$ has only one basis vector, the
geometric multiplicity of the eigenvalue $2$ is $1$.

$0$-eigenspace%
\begin{align*}
E_{0}  & =NullSp\left(  \mathbf{A}-\left(  0\right)  \mathbf{I}\right)
=NullSp\left(  \left[
\begin{array}
[c]{ccc}%
2 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 2
\end{array}
\right]  \right)  =NullSp\left(  \left[
\begin{array}
[c]{ccc}%
1 & \frac{1}{2} & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}
\right]  \right)  \\
& =span\left(  \left[
\begin{array}
[c]{c}%
-\frac{1}{2}\\
1\\
0
\end{array}
\right]  \right)
\end{align*}
Since the $0$-eigenspace of $\mathbf{A}$ has only one basis vector, the
geometric multiplicity of the eigenvalue $0$ is $1$.
\end{itemize}

(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}

\newpage

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 First, we need to find the eigenvalues and eigenvectors of $\mathbf{A}%
$.
\[
p_{\mathbf{A}}\left(  \lambda\right)  =\det\left(  \left[
\begin{array}
[c]{cc}%
3-\lambda & 1\\
1 & 3-\lambda
\end{array}
\right]  \right)  =\left(  3-\lambda\right)  ^{2}-1=\lambda^{2}-6\lambda
+8=\left(  \lambda-2\right)  \left(  \lambda-4\right)
\]
So the eigenvalues of $\mathbf{A}$ are $2$ and $4$. \ Next, we look for the
corresponding eigenvectors:

$2$-eigenspace%
\begin{align*}
E_{2}  & =NullSp\left(  \left[
\begin{array}
[c]{cc}%
3-2 & 1\\
1 & 3-2
\end{array}
\right]  \right)  =NullSp\left(  \left[
\begin{array}
[c]{cc}%
1 & 1\\
0 & 0
\end{array}
\right]  \right)  \\
& =span\left(  \left[
\begin{array}
[c]{c}%
-1\\
1
\end{array}
\right]  \right)  \quad\Rightarrow\quad\mathbf{v}_{\lambda=2}=\left[
\begin{array}
[c]{c}%
-1\\
1
\end{array}
\right]
\end{align*}
$4$-eigenspace%
\begin{align*}
E_{4}  & =NullSp\left(  \left[
\begin{array}
[c]{cc}%
3-4 & 1\\
1 & 3-4
\end{array}
\right]  \right)  =NullSp\left(  \left[
\begin{array}
[c]{cc}%
1 & -1\\
0 & 0
\end{array}
\right]  \right)  \\
& =span\left(  \left[
\begin{array}
[c]{c}%
1\\
1
\end{array}
\right]  \right)  \quad\Rightarrow\quad\mathbf{v}_{\lambda=4}=\left[
\begin{array}
[c]{c}%
1\\
1
\end{array}
\right]
\end{align*}
We use the eigenvalues of $\mathbf{A}$ to construct the diagonal matrix
$\mathbf{D}$ and then use the corresponding eigenvectors as the columns of the
matrix $\mathbf{C}$. Thus,%
\[
\mathbf{D}=\left[
\begin{array}
[c]{cc}%
2 & 0\\
0 & 4
\end{array}
\right]  \qquad,\qquad\mathbf{C}=\left[
\begin{array}
[c]{cc}%
-1 & 1\\
1 & 1
\end{array}
\right]
\]

\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 First, we'll construct an orthogonal basis $\left\{  \mathbf{o}%
_{1},\mathbf{o}_{2}\right\}  $ for $W$. \ We set%
\[
\mathbf{o}_{1}=\mathbf{v}_{1}=\left[  1,-1,0\right]
\]
The second orthogonal basis vector is formed by removing from $\mathbf{v}_{2}%
$, the component that runs parallel to $\mathbf{o}_{1}=\mathbf{v}_{1}$:%
\[
\mathbf{o}_{2}=\mathbf{v}_{2}-\frac{\mathbf{o}_{1}\cdot\mathbf{v}_{2}%
}{\mathbf{o}_{1}\cdot\mathbf{o}_{1}}\mathbf{o}_{1}=\left[  1,0,1\right]
-\frac{\left(  1+0+0\right)  }{\left(  1+1+0\right)  }\left[  1,-1,0\right]
=\left[
\begin{array}
[c]{ccc}%
\frac{1}{2} & \frac{1}{2} & 1
\end{array}
\right]
\]
To get an orthonormal basis $\left\{  \mathbf{n}_{1},\mathbf{n}_{2}\right\}  $
from $\left\{  \mathbf{o}_{1},\mathbf{o}_{2}\right\}  $ we just have to divide
$\mathbf{o}_{1}$ and $\mathbf{o}_{2}$ by their lengths, so that we get
orthogonal basis vectors of unit length:%
\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{\sqrt{2}}{3}\ ,\ \frac{\sqrt{2}}{3}\ ,\ \frac
{2\sqrt{2}}{3}\right]
\end{align*}

\end{itemize}

\newpage

10. (15 pts)Let $\mathbf{v}=\left[  2,1,0\right]  $ and let $W=span\left(
\left[  1\ ,1,0\right]  ,\ \left[  1,0,1\right]  \right)  $. Find the
orthogonal decomposition $\mathbf{v}=\mathbf{v}_{W}+\mathbf{v}_{\perp}$ of
$\mathbf{v}$ with respect to the subspace $W$.

\begin{itemize}
\item Since $\left\{  \left[  1\ ,1,0\right]  ,\ \left[  1,0,1\right]
\right\}  $ are linearly independent, they'll provide a basis for $W$. 

\item Next, we need a basis for $W^{\perp}=\left\{  \mathbf{x}\in
\mathbb{R}^{3}\mid\mathbf{x}\cdot\mathbf{w}=0\text{ for all }\mathbf{w}\in
W\right\}  $
\begin{align*}
W^{\perp}  & =NullSp\left(  \left[
\begin{array}
[c]{ccc}%
1 & 1 & 0\\
1 & 0 & 1
\end{array}
\right]  \right)  =NullSp\left(  \left[
\begin{array}
[c]{ccc}%
1 & 0 & 1\\
0 & 1 & -1
\end{array}
\right]  \right)  \\
& =span\left(  \left[  -1,1,1\right]  \right)
\end{align*}


\item Combining the basis for $W^{\perp}$ with the basis for $W$, we obtain a
basis $B=\left\{  \left[  1\ ,1,0\right]  ,\ \left[  1,0,1\right]  ,\left[
-1,1,1\right]  \right\}  $ for $\mathbb{R}^{3}$

\item Next, we find the coordinate vector $\mathbf{v}_{B}$ of $\mathbf{v}$
with respect to the basis $B$. \ This means solving the linear system%
\[
x_{1}\left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  +x_{2}\left[
\begin{array}
[c]{c}%
1\\
0\\
1
\end{array}
\right]  +x_{3}\left[
\begin{array}
[c]{c}%
-1\\
1\\
1
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
2\\
1\\
0
\end{array}
\right]
\]
This linear system has the following augmented matrix%
\[
\quad\left[  \mathbf{A}\mid\mathbf{b}\right]  =\quad\left[
\begin{array}
[c]{cccc}%
1 & 1 & -1 & 2\\
1 & 0 & 1 & 1\\
0 & 1 & 1 & 0
\end{array}
\right]
\]
which row reduces to%
\[
\left[  \mathbf{I}\mid\mathbf{v}_{B}\right]  =\left[
\begin{array}
[c]{cccc}%
1 & 0 & 0 & \frac{4}{3}\\
0 & 1 & 0 & \frac{1}{3}\\
0 & 0 & 1 & -\frac{1}{3}%
\end{array}
\right]
\]
And so%
\[
x_{1}=\frac{4}{3}\quad,\quad x_{2}=\frac{1}{3}\quad,\quad x_{3}=-\frac{1}{3}%
\]
Thus,%
\[
\mathbf{v}=\left(  \frac{4}{3}\left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  +\frac{1}{3}\left[
\begin{array}
[c]{c}%
1\\
0\\
1
\end{array}
\right]  \right)  -\frac{1}{3}\left[
\begin{array}
[c]{c}%
-1\\
1\\
1
\end{array}
\right]
\]
The first two terms live in $W$ and their sum is $\mathbf{v}_{W}$, the last
term lives in $W^{\perp}$ and it corresponds to $\mathbf{v}_{\perp}$. \ Thus,%
\[
\mathbf{v}=\mathbf{v}_{W}+\mathbf{v}_{\perp}%
\]
with%
\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}
