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\begin{center}
Math 3013.25578\\ 
Solutions to Final Exam \\ 
May 2, 2022
\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. Suppose each of the following augmented matrices is a Row Echelon Form of
the augmented matrix of a linear system $\mathbf{Ax} = \mathbf{b}$. Describe
the original system (i.e., how many equations in how many unknowns) and
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) $\left[
\begin{array}
[c]{cccccc}%
0 & -1 & 0 & 1 & 2 & 1\\
0 & 0 & 2 & 0 & 1 & 0\\
0 & 0 & 0 & 2 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}
\right]  \quad$ : \ 4 equations in 5 unknowns. Solutions with 2 free
parameters ($x_{1}$ and $x_{5}$). 

(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\\
1\\
0
\end{array}
\right.  \right]  $ : 4 equations in 4 unknowns. There is no solution since
the 3rd row implies a contradictory equation.

(c) (5 pts) $\left[
\begin{array}
[c]{ccc}%
1 & 0 & 2\\
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
2\\
1\\
0
\end{array}
\right.  \right]  $: 4 equations in 3 unknowns. There is a unique solution
(since each column of the coefficient part of the augmented matrix in RREF has
a pivot).

\newpage

3. (10 pts) Solve the following linear system, expressing the solution set as
a hyperplane.
\begin{align*}
x_{1}-x_{3}+x_{4}  &  =1\\
x_{1}+x_{3}+x_{4}  &  =1\\
x_{1}+x_{4}  &  =1
\end{align*}


\begin{itemize}
\item We begin by writing down the augmented matrix for this linear system and
row reducing it to Reduced Row Echelon Form.%
\begin{align*}
\left[  \mathbf{A}|\mathbf{b}\right]   &  =\left[
\begin{array}
[c]{cccc|c}%
1 & 0 & -1 & 1 & 1\\
1 & 0 & 1 & 1 & 1\\
1 & 0 & 0 & 1 & 1
\end{array}
\right]  \quad\underrightarrow{%
\begin{array}
[c]{c}%
\quad R_{2}\rightarrow R_{2}-R_{1}\\
R_{3}\rightarrow R_{3}-R_{1}%
\end{array}
\quad}\quad\left[
\begin{array}
[c]{cccc|c}%
1 & 0 & -1 & 1 & 1\\
0 & 0 & 2 & 0 & 0\\
0 & 0 & 1 & 0 & 0
\end{array}
\right] \\
&  \underrightarrow{%
\begin{array}
[c]{c}%
\quad R_{2}\rightarrow\frac{1}{2}R_{2}\\
R_{3}\rightarrow R_{3}-R_{2}\\
R_{1}\rightarrow R_{1}+R_{2}%
\end{array}
\quad}\quad\left[
\begin{array}
[c]{cccc|c}%
1 & 0 & 0 & 1 & 1\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{array}
\right]
\end{align*}
Converting back to equations (and noting that $x_{2}$ and $x_{4}$ will be the
free parameters in the solution)%
\[
\left.
\begin{array}
[c]{r}%
x_{1}+x_{4}=1\\
x_{3}=0\\
0=0
\end{array}
\right\}  \quad\Rightarrow\quad\left\{
\begin{array}
[c]{l}%
x_{1}=1-x_{4}\\
x_{3}=0
\end{array}
\right.
\]
and so a solution vector must be of the form%
\[
\mathbf{x}=\left[
\begin{array}
[c]{c}%
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}%
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
1-x_{4}\\
x_{2}\\
0\\
x_{4}%
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
1\\
0\\
0\\
0
\end{array}
\right]  +x_{2}\left[
\begin{array}
[c]{c}%
0\\
1\\
0\\
0
\end{array}
\right]  +x_{4}\left[
\begin{array}
[c]{c}%
-1\\
0\\
0\\
1
\end{array}
\right]
\]

\end{itemize}

\newpage

4. (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 & -2 & 0 & 1
\end{array}
\right]   &  & \\
& \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 & -2 & 0 & 1
\end{array}
\right]     =\left[  \mathbf{I}\mid\mathbf{A}^{-1}\right]    \quad
\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}

5. (10 pts) Let $W=\left\{  \left[  x,y\right]  \in\mathbb{R}^{2}\mid
x+2y=3\right\}  $. Prove or disprove that $W$ is a subspace of $\mathbb{R}%
^{2}$.

\begin{itemize}
\item Consider the vector $\mathbf{w}=\left[  1,1\right]  $. $\mathbf{w}$
belongs to $W$ since $\left(  1\right)  +2\left(  1\right)  =3$. If we scalar
multiply $\mathbf{w}$ by the number $0$, we get%
\[
\lambda\mathbf{w}=\left[  0,0\right]
\]
which does not lie in $W$ since $\left(  0\right)  +2\left(  0\right)
=0\neq3$. This means $W$ is not closed under scalar multiplication and so $W$
\textbf{is not a subspace.}
\end{itemize}

\newpage

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

(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\\
1 & 2 & -1 & 1\\
2 & 3 & 0 & 3\\
3 & 4 & 1 & 3
\end{array}
\right]  \quad\underrightarrow{%
\begin{array}
[c]{c}%
R_{2}\rightarrow R_{2}-R_{1}\\
R_{3}\rightarrow R_{3}-2R_{1}\\
R_{4}\rightarrow R_{4}-3R_{1}%
\end{array}
}\quad\left[
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & -2 & 0\\
0 & 1 & -2 & 1\\
0 & 1 & -2 & 0
\end{array}
\right] \\
&  \quad\underrightarrow{%
\begin{array}
[c]{c}%
R_{3}\rightarrow R_{3}-R_{2}\\
R_{4}\rightarrow R_{4}-R_{2}%
\end{array}
}\quad\left[
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & -2 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}
\right]  \quad\underrightarrow{R_{1}\rightarrow R_{1}-R_{4}}\quad\left[
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 0\\
0 & 1 & -2 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}
\right] \\
&  \underrightarrow{R_{1}\rightarrow R_{1}-R_{2}\quad}\quad\left[
\begin{array}
[c]{cccc}%
1 & 0 & 3 & 1\\
0 & 1 & -2 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}
\right]
\end{align*}
Since the REF of $\mathbf{A}$ a zero row, 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
$RowSp\left(  \mathbf{A}\right)  =span\left(
[1,1,1,1],[1,2,-1,1],[2,3,0,2],\left[  3,4,1,3\right]  \right)  .$

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

7. 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) \quad \Longrightarrow \quad
\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{align*}
\end{itemize}

\newpage

8. (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

9. (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}

\newpage

10. Let $W=span\left(  \left[  1\ ,1,1\right]  ,\ \left[  0,1,1\right]
\right)  $.

(a) (7 pts) Find the orthogonal complement $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{itemize}
\item Note that the generators of $W$ are clearly linearly independent, and so
basis vectors for $W$. To lie in $W_{\perp}$, a vector $\mathbf{x}$ must be
perpendicular to each basis vector of $W$. Thus, setting $\mathbf{b}%
_{1}=\left[  1,1,1\right]  $, $\mathbf{b}_{2}=\left[  0,1,1\right]  $%
\[
\mathbf{x}\in W_{\perp}\quad\Rightarrow\quad\left\{
\begin{array}
[c]{c}%
\mathbf{b}_{1}\cdot\mathbf{x}=0\\
\mathbf{b}_{2}\cdot\mathbf{x}=0
\end{array}
\right.  \quad\Rightarrow\quad\left[
\begin{array}
[c]{ccc}%
\leftarrow & \mathbf{b}_{1} & \rightarrow\\
\leftarrow & \mathbf{b}_{2} & \rightarrow
\end{array}
\right]  \mathbf{x}=\mathbf{0}%
\]
Thus,%
\begin{align*}
W_{\perp}  &  =NullSp\left(
\begin{array}
[c]{ccc}%
\leftarrow & \mathbf{b}_{1} & \rightarrow\\
\leftarrow & \mathbf{b}_{2} & \rightarrow
\end{array}
\right)  =NullSp\left(  \left[
\begin{array}
[c]{ccc}%
1 & 1 & 1\\
0 & 1 & 1
\end{array}
\right]  \right)  =NullSp\left(  \left[
\begin{array}
[c]{ccc}%
1 & 0 & 0\\
0 & 1 & 1
\end{array}
\right]  \right) \\
&  =\left\{  \mathbf{x}\in\mathbb{R}^{3}\mid x_{1}=0\ ,\ x_{2}=-x_{3}\right\}
=span\left(  \left[
\begin{array}
[c]{c}%
0\\
-1\\
1
\end{array}
\right]  \right)
\end{align*}

\end{itemize}

(b) (8 pts) Let $\mathbf{v}=\left[  1,-2,0\right]  $. Find the orthogonal
decomposition $\mathbf{v}=\mathbf{v}_{W}+\mathbf{v}_{\perp}$; $\mathbf{v}%
_{W}\in W$, $\mathbf{v}_{\perp}\in W_{\perp}$, of $\mathbf{v}$ with respect to
$W$.

\begin{itemize}
\item $B=\left\{  \left[  1,1,1\right]  ,\left[  0,1,1\right]  \right\}
\cup\left\{  \left[  0,-1,1\right]  \right\}  $ will be a basis for the entire
vector space $\mathbb{R}^{3}$. \ Let us find the coordinates of $\mathbf{v}$
with respect to the basis $B$%
\[
c_{1}\left[
\begin{array}
[c]{c}%
1\\
1\\
1
\end{array}
\right]  +c_{2}\left[
\begin{array}
[c]{c}%
0\\
1\\
1
\end{array}
\right]  +c_{3}\left[
\begin{array}
[c]{c}%
0\\
-1\\
1
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
1\\
-2\\
0
\end{array}
\right]  \quad
\]
This linear system has the augmented matrix
\[
\left[  \mathbf{A}|\mathbf{b}\right]  =\left[
\begin{array}
[c]{ccc|c}%
1 & 0 & 0 & 1\\
1 & 1 & -1 & -2\\
1 & 1 & 1 & 0
\end{array}
\right]
\]
which in turn row reduces to%
\[
\left[
\begin{array}
[c]{ccc|c}%
1 & 0 & 0 & 1\\
0 & 1 & 0 & -2\\
0 & 0 & 1 & 1
\end{array}
\right]  =R.R.E.F.\left(  \left[  \mathbf{A}|\mathbf{b}\right]  \right)
\quad\Rightarrow\quad\left\{
\begin{array}
[c]{l}%
c_{1}=1\\
c_{2}=-2\\
c_{3}=1
\end{array}
\right.
\]
and so%
\begin{align*}
\mathbf{v}  &  =\left[
\begin{array}
[c]{c}%
1\\
1\\
1
\end{array}
\right]  -2\left[
\begin{array}
[c]{c}%
0\\
1\\
1
\end{array}
\right]  +\left[
\begin{array}
[c]{c}%
0\\
-1\\
1
\end{array}
\right] \\
&  =\left(  \left[
\begin{array}
[c]{c}%
1\\
1\\
1
\end{array}
\right]  -2\left[
\begin{array}
[c]{c}%
0\\
1\\
1
\end{array}
\right]  \right)  +\left[
\begin{array}
[c]{c}%
0\\
-1\\
1
\end{array}
\right] \\
&  =\mathbf{v}_{W}+\mathbf{v}_{\perp}%
\end{align*}
where%
\begin{align*}
\mathbf{v}_{W}  &  =\left(  \left[
\begin{array}
[c]{c}%
1\\
1\\
1
\end{array}
\right]  -2\left[
\begin{array}
[c]{c}%
0\\
1\\
1
\end{array}
\right]  \right)  =\left[
\begin{array}
[c]{c}%
1\\
-1\\
-1
\end{array}
\right]  \in W\\
\mathbf{v}_{\perp}  &  =\left[
\begin{array}
[c]{c}%
0\\
-1\\
1
\end{array}
\right]  \in W_{\perp}%
\end{align*}

\end{itemize}


\end{document}
