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\begin{document}

\begin{center}
\textbf{Math 3013}\\[0pt]
SECOND EXAM \\[0pt]
November 3, 2020
\end{center}

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

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

\begin{itemize}
\item Writing these vectors as the rows of a matrix $\mathbf{A}$, we can
determine if the vectors are linearly independent by checking if the R.E.F.
of $\mathbf{A}$ has any non-zero row vectors. \ If R.E.F.$\left( \mathbf{A}%
\right) $ has any zero row vectors, then the original set of vectors are
linearly dependent.%
\begin{equation*}
\mathbf{A}=\left[ 
\begin{array}{ccccc}
1 & 2 & 1 & 1 & 0 \\ 
0 & 1 & 2 & 1 & 1 \\ 
-1 & -1 & 1 & 0 & 1 \\ 
-1 & 0 & 3 & 1 & 2%
\end{array}%
\right] \quad \underrightarrow{\text{ row reduction }}\quad \left[ 
\begin{array}{ccccc}
1 & 0 & -3 & -1 & -2 \\ 
0 & 1 & 2 & 1 & 1 \\ 
0 & 0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 0%
\end{array}%
\right] =R.R.E.F.\left( \mathbf{A}\right) 
\end{equation*}%
Since we have two zero row vectors, the original set of matrices are \textbf{%
not linearly independent}.
\end{itemize}

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

\begin{itemize}
\item The row reduction calculation in part (a)\ also produces a basis for
the row space of $\mathbf{A}$, which is also a basis for the subspace
generated by the original set of vectors. \ Since we have two basis vectors,
the dimension of the subspaces is $2$.
\end{itemize}

2. Write the definitions (as stated in class) of the following notions. (5
pts each)

(a) A \textbf{subspace} $W$ of $\mathbb{R}^{n}.$

\begin{itemize}
\item A \emph{subspace} $W$ of $\mathbb{R}^{n}$ is a subset of $\mathbb{R}%
^{n}$ such that

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

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

(b) A \textbf{basis }for a subspace $W$ of $\mathbb{R}^{n}$.

\begin{itemize}
\item A \emph{basis} for a subspace $W$ of $\mathbb{R}^{n}$ is a set of
vectors $\left\{ \mathbf{b}_{1},\ldots ,\mathbf{b}_{k}\right\} $ such that

\begin{itemize}
\item every vector $\mathbf{w}\in W$ can be expressed as%
\begin{equation}
\mathbf{w}=c_{1}\mathbf{b}_{1}+\cdots +c_{k}\mathbf{b}_{k}  \tag{*}
\end{equation}

\item for each $\mathbf{w}\in W$, the coefficients $c_{1},\ldots ,c_{k}$ in
(*) are unique.
\end{itemize}
\end{itemize}

(c) A set of \textbf{linearly independent vectors }

\begin{itemize}
\item A set of vectors $\left\{ \mathbf{v}_{1},\ldots ,\mathbf{v}%
_{k}\right\} $ are \emph{linearly independent} if the only solution of%
\begin{equation*}
x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots +x_{k}\mathbf{v}_{k}=\mathbf{0%
}
\end{equation*}%
is%
\begin{equation*}
x_{1}=0\ ,\ x_{2}=0\ ,\ \cdots \ ,\ x_{k}=0
\end{equation*}
\end{itemize}

(d) A \textbf{linear transformation}

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

\begin{itemize}
\item $\lambda \in \mathbb{R}\ ,\ \mathbf{v}\in V\quad \Rightarrow \quad
T\left( \lambda \mathbf{v}\right) =\lambda T\left( \mathbf{v}\right) $

\item $\mathbf{v}_{1},\mathbf{v}_{2}\in V\quad \Rightarrow \quad T\left( 
\mathbf{v}_{1}+\mathbf{v}_{2}\right) =T\left( \mathbf{v}_{1}\right) +T\left( 
\mathbf{v}_{2}\right) $\newpage 
\end{itemize}
\end{itemize}

3. Given that the matrix $\displaystyle\mathbf{A}=\left[ 
\begin{array}{cccc}
2 & 4 & 0 & 2 \\ 
-1 & -2 & 1 & -3 \\ 
0 & 0 & 1 & 2%
\end{array}%
\right] $ \quad row reduces to \quad $\left[ 
\begin{array}{cccc}
1 & 2 & 0 & 1 \\ 
0 & 0 & 1 & -2 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right] $

(a) (5 pts) Find a basis for the row space of $\mathbf{A}$.

\begin{itemize}
\item The non-zero rows of $R.E.F.\left( \mathbf{A}\right) $ provide a basis
for the row space of $\mathbf{A:}$%
\begin{equation*}
\text{basis for row space = }\left\{ \left[ 1,2,0,1\right] \ ,\ \left[
0,0,1,-2\right] \right\} 
\end{equation*}
\end{itemize}

(b) (5 pts) Find a basis for the column space of $\mathbf{A}$.

\begin{itemize}
\item The columns of $\mathbf{A}$ that correspond to the columns of its
R.E.F. that contain pivots will provide a basis for the column space of $%
\mathbf{A:}$%
\begin{equation*}
\text{basis for column space = }\left\{ \left[ 
\begin{array}{c}
2 \\ 
-1 \\ 
0%
\end{array}%
\right] \ ,\ \left[ 
\begin{array}{c}
0 \\ 
1 \\ 
1%
\end{array}%
\right] \right\} 
\end{equation*}
\end{itemize}

(c) (5 pts)\ Find a basis for the null space of $\mathbf{A}$.

\begin{itemize}
\item The null space of $\mathbf{A}$ is the solution set of the homogeneous
linear system $\mathbf{Ax}=\mathbf{0}$. \ We are given the R.R.E.F. of the
coefficient matrix $\mathbf{A}$, so we can immediately write down the
solutions of $\mathbf{Ax}=\mathbf{0:}$%
\begin{equation*}
\left. 
\begin{array}{r}
x_{1}+2x_{2}+x_{4}=0 \\ 
x_{3}-2x_{4}=0 \\ 
0=0%
\end{array}%
\right\} \quad \Rightarrow \quad \left\{ 
\begin{array}{l}
x_{1}=-2x_{2}-x_{4} \\ 
x_{2}\ \text{is a free variable} \\ 
x_{3}=2x_{4} \\ 
x_{4}\text{ is a free variable}%
\end{array}%
\right. 
\end{equation*}%
\begin{equation*}
\mathbf{x}=\left[ 
\begin{array}{c}
-2x_{2}-x_{4} \\ 
x_{2} \\ 
2x_{4} \\ 
x_{4}%
\end{array}%
\right] =x_{2}\left[ 
\begin{array}{c}
-2 \\ 
1 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{4}\left[ 
\begin{array}{c}
-1 \\ 
0 \\ 
2 \\ 
1%
\end{array}%
\right] 
\end{equation*}%
and 
\begin{equation*}
\text{basis for }NullSp\left( \mathbf{A}\right) =\left\{ \left[ -2,1,0,0%
\right] \ ,\ \left[ -1,0,2,1\right] \right\} 
\end{equation*}
\end{itemize}

(d) (5 pts)\ What is the rank of $\mathbf{A}$?

\begin{itemize}
\item The rank of $\mathbf{A}$ is the common dimension of its row and column
spaces. \ Since both the row and column spaces have two basis vectors%
\begin{equation*}
\text{rank}\left( \mathbf{A}\right) =2
\end{equation*}
\end{itemize}

4. Consider the following linear transformation:

\hspace{.5in} $T:\mathbb{R}^{3} \rightarrow \mathbb{R}^{3}:$ $T\left(
[x_{1}\; ,\; x_{2}\; ,\; x_{3} \right) =\left[ x_{1} +2 x_{2}+x_{3}\; , \;
x_{1} + x_{2} \; , \;x_{1}-x_{3}\right] $.

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

\begin{itemize}
\item We have%
\begin{equation*}
\mathbf{A}_{T}=\left[ 
\begin{array}{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}{ccc}
1 & 2 & 1 \\ 
1 & 1 & 0 \\ 
1 & 0 & -1%
\end{array}%
\right] 
\end{equation*}
\end{itemize}

(b) (5 pts) Find a basis for the range of $T$.

\begin{itemize}
\item We have $Range\left( T\right) =ColSp\left( \mathbf{A}_{T}\right) $. \
Thus, we find a basis for $Range\left( T\right) $ by finding a basis for the
column space of $\mathbf{A}_{T}$%
\begin{equation*}
\mathbf{A}_{T}=\left[ 
\begin{array}{ccc}
1 & 2 & 1 \\ 
1 & 1 & 0 \\ 
1 & 0 & -1%
\end{array}%
\right] \quad \underrightarrow{\ \text{row reduction }}\quad \left[ 
\begin{array}{ccc}
1 & 0 & -1 \\ 
0 & 1 & 1 \\ 
0 & 0 & 0%
\end{array}%
\right] =R.R.E.F.\left( \mathbf{A}_{T}\right) 
\end{equation*}%
Grabbing the columns of $\mathbf{A}_{T}$ corresponding to the columns of its
R.R.E.F. that have pivots we find 
\begin{equation*}
\text{basis for Range}\left( T\right) =\left\{ \left[ 1,1,1\right] \ ,\ %
\left[ 2,1,0\right] \right\} 
\end{equation*}
\end{itemize}

(c) (5 pts) Find a basis for the kernel of $T$.

\begin{itemize}
\item 
\begin{itemize}
\item $\ker \left( T\right) =NullSp\left( \mathbf{A}_{T}\right) =$ solution
set of $\mathbf{A}_{T}\mathbf{x}=\mathbf{0}$. \ Using the R.R.E.F. of $%
\mathbf{A}_{T}$ that we just calculated we find the solutions of $\mathbf{A}%
_{T}\mathbf{x}=\mathbf{0}$ must be vectors of the form%
\begin{equation*}
\mathbf{x}=\left[ 
\begin{array}{c}
x_{1} \\ 
x_{2} \\ 
x_{3}%
\end{array}%
\right] =\left[ 
\begin{array}{c}
x_{3} \\ 
-x_{3} \\ 
x_{3}%
\end{array}%
\right] =x_{3}\left[ 
\begin{array}{c}
1 \\ 
-1 \\ 
1%
\end{array}%
\right] 
\end{equation*}%
and so 
\begin{equation*}
\text{basis for }\ker \left( T\right) =\left\{ \left[ 1,-1,1\right] \right\} 
\end{equation*}
\end{itemize}
\end{itemize}

\newpage

5. (15 pts) Use cofactor expansions to compute the determinant of $\mathbf{A}%
=\left[ 
\begin{array}{cccc}
1 & 0 & 0 & 1 \\ 
1 & 0 & 1 & 1 \\ 
1 & 2 & 1 & 1 \\ 
1 & 0 & 0 & 2%
\end{array}%
\right] $

\begin{itemize}
\item The second column has only one non-zero entry. \ So we'll carry out a
cofactor expansion along the second column of $\mathbf{A}$%
\begin{eqnarray*}
\det \left( \mathbf{A}\right)  &=&0+0+\left( 2\right) \left( -1\right)
^{3+2}\det \left( 
\begin{array}{ccc}
1 & 0 & 1 \\ 
1 & 1 & 1 \\ 
1 & 0 & 2%
\end{array}%
\right)  \\
&=&\left( 2\right) \left( -1\right) \left( 0+\left( 1\right) \left(
-1\right) ^{2+2}\det \left( 
\begin{array}{cc}
1 & 1 \\ 
1 & 2%
\end{array}%
\right) +0\right)  \\
&=&\left( 2\right) \left( -1\right) \left( 1\right) \left( 1\right) \left(
2-1\right)  \\
&=&-2
\end{eqnarray*}
\end{itemize}

6. (15 pts) Use the row reduction method to determine the determinant of $%
\mathbf{A}=\left[ 
\begin{array}{cccc}
0 & 0 & 0 & 1 \\ 
0 & 2 & 2 & 2 \\ 
0 & 2 & 1 & -1 \\ 
1 & 1 & 1 & 1%
\end{array}%
\right] $

\begin{itemize}
\item The following sequence of elementary row operations converts $\mathbf{A%
}$ to row echelon form%
\begin{eqnarray*}
\mathbf{A} &\mathbf{=}&\left[ 
\begin{array}{cccc}
0 & 0 & 0 & 1 \\ 
0 & 2 & 2 & 2 \\ 
0 & 2 & 1 & -1 \\ 
1 & 1 & 1 & 1%
\end{array}%
\right] \ \underrightarrow{R_{1}\longleftrightarrow R_{4}}\ \left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 2 & 2 & 2 \\ 
0 & 2 & 1 & -1 \\ 
0 & 0 & 0 & 1%
\end{array}%
\right]  \\
&&\left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 2 & 2 & 2 \\ 
0 & 2 & 1 & -1 \\ 
0 & 0 & 0 & 1%
\end{array}%
\right] \ \underrightarrow{R_{3}\rightarrow R_{3}-R_{2}\ }\ \left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 2 & 2 & 2 \\ 
0 & 0 & -1 & -3 \\ 
0 & 0 & 0 & 1%
\end{array}%
\right] 
\end{eqnarray*}%
We have%
\begin{equation*}
\det \left( \mathbf{A}\right) =\left( -1\right) ^{r}\frac{1}{\lambda
_{1}\ldots \lambda _{k}}\det \left( R.E.F.\left( \mathbf{A}\right) \right) 
\end{equation*}%
Here $r$ is the number of row interchanges used in the row reduction, and $%
\lambda _{1},\ldots ,\lambda _{k}$ are the row rescalings used. \ In our
case, $r=1$ and we didn't use any row rescalings. \ Hence,%
\begin{eqnarray*}
\det \left( \mathbf{A}\right)  &=&\left( -1\right) \det \left( \left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 2 & 2 & 2 \\ 
0 & 0 & -1 & -3 \\ 
0 & 0 & 0 & 1%
\end{array}%
\right] \right) =\left( -1\right) \left( 1\right) \left( 2\right) \left(
-1\right) \left( 1\right)  \\
&=&2
\end{eqnarray*}
\end{itemize}

\end{document}
