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\begin{document}
\examtitle{MATH 2270-2}{Final Exam}{Spring 2012}

\studentinfo
\instructions{No books, notes or electronic devices, please.\\
The questions have credits which reflect the time required to write the solution.\\
If you must write a solution out of order or on the back side, then supply a road map.\\
Solutions are expected to include readable and convincing details. A correct answer without details earns 25\%.
}
\finishfirstpage
% Problems Start Here % ----------------------------------------------------- %


%\section*{Chapters 1 and 2}

% 2012, save for another version in 2013
%\problem{5}{
%Let $A=\twobytwo{1&2}{1&3}$. Show the details of two different methods for finding $A^{-1}=\twobytwo{3&-2}{-1&1}$.
%}{}{The two methods are (1) $A^{-1}=\frac{\adj(A)}{|A|}$ and (2)
%For $C=<A|I>$, then $\rref(C)=<I|A^{-1}>$. }
%

\subsection*{Essay Questions}
\instructions{{\bf Complete essay problems 1 and 2}. Expected completion time is 10 minutes.}


\problem{10}{
Define an Elementary Matrix. Display the fundamental matrix multiply equation which summarizes a sequence of swap, combination and multiply operations, transforming a matrix $A$ into a matrix $B$.
}{}{An elementary matrix is a matrix $E$ obtained from the identity matrix $I$ by applying one combination,
swap or multiply operation. The equation is
$$E_k\cdots E_2E_1A = B$$
where $E_1,E_2,\ldots,E_k$ are elementary matrices representing swap, multiply and combination operations that take $A$ into $B$.}

%\vspace{3.25in}

\problem{20}{
State the Fundamental Theorem of Linear Algebra. Include {\bf Part 1}:
The dimensions of the four subspaces, and {\bf Part 2}: The
orthogonality equations for the four subspaces.
}{}{Let $A$ denote an $m\times n$ matrix of rank $r$. Part 1. The dimensions of the nullspace($A$), colspace($A$),
rowspace($A$), nullspace($A^T$) are respectively $n-r$, $r$, $r$, $m-r$.\newline
Part 2. rowspace($A$) $\perp$ nullspace($A$), colspace($A$) $\perp$ nullspace($A^T$). Both can be summarized by rowspace $\perp$ nullspace, applied to both $A$ and $A^T$. }
\anotherpage

\subsection*{Problems 3 to 17}
\instructions{{\bf Complete as many of the remaining problems as time allows}. Credits are marked 10 to 40, which represents the amount of time needed to write the solution. Estimated time to complete the remaining 15 problems is 100 minutes. There are applications of theorems, but no proofs to write.}

\problem{30}{
Find a factorization $A=LU$ into lower and upper triangular matrices for
the matrix $A=\threebythree{2&1&0}{1&2&1}{0&1&1}$.
}{}{Let $E_1$ be the result of combo(1,2,-1/2) on $I$, and $E_2$ the result of combo(2,3,-2/3) on $I$.
Then $E_2E_1A = U=\threebythree{2&1&0}{0&\frac32 & 1}{0 & 0 & \frac13}$. Let $L=E_1^{-1}E_2^{-1}=\threebythree{1&0&0}{\frac12&1&0}{0&\frac23&1}$.}
%% A:=Matrix([[2,1,0],[1,2,1],[0,1,1]]);with(LinearAlgebra):LUDecomposition(A);
%% convert([%][2],list,nested=true);convert([%][3],list,nested=true);
%% L = [[1, 0, 0], [1/2, 1, 0], [0, 2/3, 1]]
%% U = [[2, 1, 0], [0, 3/2, 1], [0, 0, 1/3]]


\anotherpage
\problem{30}{
Determine which values of $k$ correspond to
%a {\large\sf unique solution}
%(1) a {\large\sf unique solution},
{\large\sf infinitely many solutions}
% and (3) {\large\sf no solution},
for the system $A\vec{x}=\vec{b}$ given by
$$A=\left(\begin{array}{ccc}
  1 & 4 &  k  \\
  0 & k-3 & k-3 \\
  1 & 4 &  1
 \end{array}\right), \quad \vec{b}=\colvectorC{1\\ -1 \\ k }.$$
% Check it!
}{}
{There is a unique solution for $\det(A)\ne 0$, which implies $k\ne 1$ and $k \ne 3$. 
Elimination methods with swap, combo, multiply give
{\small $\left( \begin {array}{cccc} 1&4&k&1\\ 0&k-3&k-3&-1\\ 0&0&1-k&k-1\end {array} \right)$}. Then
(1) Unique solution for three lead variables, equivalent to the determinant nonzero for the frame above, or $(k-3)(1-k)\ne 0$;
(2) No solution for $k=3$ [signal equation];
(3) Infinitely many solutions for $k=1$ [one free variable]. }

%\section*{Chapters 3, 4}

\anotherpage
\problem{30}{
Find the complete solution $\vec{x} = \vec{x}_h + \vec{x}_p$ for the nonhomogeneous system
$$\left(\begin{array}{rrrrr} 0&1&1&0\\ 0&1&2&0\\ 0&0&0&0\end{array}\right)\colvector{x_1 \\ x_2 \\ x_3\\ x_4} = \colvector{2 \\ 3 \\ 0}.$$
The homogeneous solution $\vec{x}_h$ is a linear combination of Strang's special solutions.
Symbol $\vec{x}_p$ denotes a particular solution.
}{}{The augmented matrix has reduced row echelon form (last frame) equal to
the matrix $\left(\begin{array}{ccccc} 0&1&0&1&0\\ 0&0&1&1&0\\ 0&0&0&0&0\end{array}\right)$. Then $x_1=t_1,
x_2=1, x_3=1, x_4=t_2$ is the general solution in scalar form. The partial
derivative on $t_1$ gives the homogeneous solution basis vector
$\colvector{1\\ 0\\ 0\\ 0}$. The partial
derivative on $t_2$ gives the homogeneous solution basis vector
$\colvector{0\\ 0\\ 0\\ 1}$. Then $\vec{x}_h=c_1\colvector{1\\ 0\\ 0\\ 0}+c_2\colvector{0\\ 0\\ 0\\ 1}$. Set
$t_1=t_2=0$ in the scalar solution to find a particular solution
$\vec{x}_p=\colvector{0\\ 1 \\ 1\\ 0}$.
}
%% A:=Matrix([[0,1,1,0],[0,1,2,0],[0,0,0,0]]); v:=<2,3,0>; C:=<A | v>;

%\anotherpage
%\problem{5}{
%Check the independence tests which apply to prove that three vectors are independent in the vector space $V=\Rr^4$ of
%all fixed vectors $\colvector{x_1\\ x_2 \\ x_3\\ x_4}$.
%\begin{center}\begin{tabular}{llp{0.6\textwidth}}
% \checkbox~ & \bf Wronskian test & Wronskian of $f_1,f_2,f_3$ nonzero at $x=x_0$ implies independence of $f_1,f_2,f_3$. \\
% \checkbox~ & \bf Rank test & Vectors $\vec{v}_1,\vec{v}_2,\vec{v}_3$ are independent if their augmented matrix has rank 3. \\
% \checkbox~ & \bf Determinant test & Vectors $\vec{v}_1,\vec{v}_2,\vec{v}_3$ are independent if their square augmented matrix has nonzero determinant. \\
% \checkbox~ & \bf Atom test & Any finite set of distinct atoms is independent.\\[1ex]
% \checkbox~ & \bf Pivot test & Vectors $\vec{v}_1,\vec{v}_2,\vec{v}_3$ are independent if their augmented matrix $A$ has 3 pivot columns.
%\end{tabular}\end{center}
%}{}{The first and fourth fail to apply in $V$, because they only apply to functions. The determinant test does not directly apply, because the augmented matrix of the three vectors is not square. All others
%apply to fixed vectors, hence they apply to three vectors in $V$. }

\anotherpage
\problem{20}{
Define $S$ to be the set of all vectors
$\vec{x}$ in $\Rr^{3}$ such that $x_1+x_3=0$ and $x_3+x_2=x_1$.
Apply a theorem which concludes that
$S$ is a subspace of $\Rr^{3}$. This implies stating the hypotheses and checking that they apply.
}{}
{Let $A=\left(\begin{array}{rrr}
  1 & 0 &  1  \\
 -1 & 1 &  1  \\
  0 & 0 &  0
 \end{array}\right)$. Then the restriction equations can be written as $A\vec{x}=\vec{0}$. Apply the nullspace theorem (also called
the kernel theorem), which says that the nullspace of a matrix is a subspace.

 Another solution: The given
restriction equations are linear homogeneous algebraic equations.
Therefore, $S$ is the nullspace of some matrix $B$, hence a subspace of $\Rr^3$.
This solution uses the fact that linear homogeneous algebraic equations can be written as a matrix equation $B\vec{x}=\vec{0}$, without actually finding the matrix. }

\anotherpage
\problem{20}{
The $5\times 7$ matrix $A$ below has some
independent columns. Report the independent columns of $A$, according to
the Pivot Theorem.
$$A=\left(\begin{array}{rrrrrrr}
  0 & 0 &  0 & 0 & 0 &   0 &  0\\
 -3 & 0 &  0 & 0 &-2 &   1 & -1\\
 -1 & 0 &  0 & 0 & 0 &   1 &  0\\
  6 & 0 &  0 & 0 & 6 &   0 &  3\\
  2 & 0 &  0 & 0 & 2 &   0 &  1
 \end{array}\right)$$
}{}
%A:=Matrix([
%  [ 0,  0, 0, 0, 0,  0 ,  0 ],
%  [-3,  0, 0, 0,-2,  1 , -1],
%  [-1,  0, 0, 0, 0,  1 ,  0],
%  [ 6,  0, 0, 0, 6,  0 ,  3],
%  [ 2,  0, 0, 0, 2,  0 ,  1]
%    ]);
%  rref(A);
{Compute {\small $\rref(A)= \left( \begin {array}{ccccccc}
   1&0&0&0&0&-1&0\\
   0&0&0&0&1&1&1/2\\
   0&0&0&0&0&0&0\\
   0&0&0&0&0&0&0
\\ 0&0&0&0&0&0&0\end {array} \right)
$}. The pivot columns are 1 and 5.}

\anotherpage
\problem{40}{
Let $S$ be the subspace of $\mathbb{R}^4$ spanned by the vectors
$\vec{v}_1=\begin{pmatrix}1\\1\\1\\0\end{pmatrix}$ and
$\vec{v}_2=\begin{pmatrix}1\\0\\0\\1\end{pmatrix}$. Find the Gram-Schmidt orthonormal basis
of $S$.
}{}{Let $\vec{y}_1=\vec{v}_1$ and
$\vec{u}_1=\frac{1}{\|\vec{y}_1\|}\vec{y}_1$. Then
$\vec{u}_1=\frac{1}{\sqrt3}\colvector{1\\ 1\\ 1\\ 0}$.
Let $\vec{y}_2 = \vec{v}_2$ minus the shadow projection of $\vec{v}_2$ onto the span of $\vec{v}_1$.
Then $$\vec{y}_2 = \vec{v}_2 -
\frac{\vec{v}_2\cdot\vec{v}_1}{\vec{v}_1\cdot\vec{v}_1}\vec{v_1}=\frac13\colvector{2 \\ -1 \\ -1 \\ 3}.$$
Finally, $\vec{u}_2=\frac{1}{\|\vec{y}_2\|}\vec{y}_2$.
We report the Gram-Schmidt basis:
$$\vec{u}_1=\frac{1}{\sqrt3}\colvector{1\\ 1\\ 1\\ 0 },\quad
\vec{u}_2 =\frac{1}{\sqrt{15}}\colvector{2 \\ -1 \\ -1 \\ 3}.$$
}

% 2012, save for another version 2013
%\anotherpage
%\problem{5}{
%Let $A$ be an $m\times n$ matrix with $A^TA$ invertible. Prove that the
%columns of $A$ are independent.
%}{}{The columns of $A$ are independent if and only if the nullspace of $A$ is the zero vector. If you don't know this result, then find it in Strang's book, or prove it yourself. \\
%Assume $\vec{x}$ is in the nullspace of $A$, $A\vec{x}=\vec{0}$, then multiply by $A^T$ to get $A^TA\vec{x}=\vec{0}$. Because $A^TA$ is invertible,
%then $\vec{x}=\vec{0}$, which proves the nullspace of $A$ is the zero vector. We conclude that the columns of $A$ are independent.}
%

% 2012, save for another version 2013
%\anotherpage
%\problem{5}{
%Let $A$ be an $m\times n$ matrix and $\vec{v}$ a vector orthogonal to
%the nullspace of $A$. Prove that $\vec{v}$ must be in the row space of $A$.
%}{}{The fundamental theorem of linear algebra is summarized by {\tt rowspace $\perp$ nullspace}.
%This relation implies {\tt nullspace $\perp$ rowspace}, because for subspaces $S$ we have $(S^\perp)^\perp = S$. The nullspace and the rowspace, being orthogonal, meet only in the common vector $\vec{0}$, and
%every vector $\vec{v}$ can be uniquely written as a sum $\vec{v}=\vec{v}_1|\vec{v}_2$, where $\vec{v}_1$ is in rowspace and $\vec{v}_2$ is in the nullspace. We use $0=\vec{v}\cdot\vec{v}_2=\vec{v}_1\cdot\vec{v}_2+\vec{v}_2\cdot\vec{v}_2$ together with orthogonality $\vec{v}_1\cdot\vec{v}_2=0$ to get $0=0+\vec{v}_2\cdot\vec{v}_2$, proving $\vec{v}_2=\vec{0}$. Then $\vec{v}=\vec{v}_1+\vec{0}$ implies $\vec{v}=\vec{v}_1=$ a vector in the rowspace. }

%\section*{Chapter 5}

\anotherpage
\problem{30}{
Define matrix $A$ and vector $\vec{b}$ by the equations
$$A=\left(\begin{array}{rrrr}
 -2 &  3 & 0  \\
  0 & -2 &  4  \\
  1 &  0 &  -2
 \end{array}\right), \quad \vec{b}=\colvector{1 \\ 2 \\ 3 }.$$
Find the value of $x_2$ by Cramer's Rule
in the system $A\vec{x}=\vec{b}$.
}{}
{$x_2=\Delta_2/\Delta$, $\Delta_2=\det\threebythree{-2 & 1 & 0}{0 & 2 & 4}{1 & 3 & -2}=36$, $\Delta=\det(A)=4$, $x_2=9$. }

\anotherpage
\problem{20}{
Display the entry in row 4, column 3 of the adjugate matrix of
 $A=\left(\begin{array}{rrrr}
 0 & 2 & -1 & 0\\
 0 & 0 &  4  & 1\\
 0 & 1 & 1 & 0 \\
  1 & 3 & -2 & 0
 \end{array}\right)
 $. \newline {\small\em The adjugate matrix is also called the adjoint matrix. It appears in the formula $C^{-1}=\dd\frac{\adj(C)}{\det(C)}$}.
% A:=Matrix([[0, 2, -1, 0], [0, 0, 4, 1], [0, 1, 1, 0], [1, 3, -2, 0]]);A^(-1);
% convert(A,list,nested=true);LinearAlgebra[Minor](A,3,4);
}{}
{The answer is the cofactor of $A$ in row 3, column 4 = $(-1)^{7}$ times the minor of $A$
in row 3, column 4. The minor equals $8$. The answer is $-8$. }


%\section*{Chapter 6}

\anotherpage
%$$A=\left(\begin{array}{rrrr}
%  0 & 4 & -1 & 0 \\
% -4 & 0 & -2 & 1 \\
%  0 & 0 &  5 & 1 \\
%  0 & 0 & -3 & 1
% \end{array}\right)$$
%
%A:=matrix([[0 , 4 , -1 ],
%  [-4 , 0 , -2 ],
%  [0 , 0 ,  5  ]]);eigenvals(A);
%
%       # 5, 4 I, -4 I
% (2) [10\%] Display a matrix product formula for $A$.
% To save time, {\bf do not evaluate any matrix products}.
\problem{10}{Consider a $3\times 3$ real matrix $A$ with eigenpairs
$$\left(-1 ,  \colvector{5\\ 6\\ -4}\right),
\quad \left(1, \colvector{1\\ 2\\ 0}\right),
\quad \left(2, \colvector{-1\\ 2\\ 0}\right).$$
 Display an invertible matrix $P$ and a
 diagonal matrix $D$ such that $AP=PD$.
}{}{The columns of $P$ are the eigenvectors and the diagonal entries of $D$ are the eigenvalues, taken in the same order.}

\anotherpage
\problem{30}{Find the eigenvalues of the matrix
$A=\left(\begin{array}{rrrrr}
 0 & -12 & 3 & 0 \\
 0 &  1 & 3 & 0 \\
 0 &  1 & -1 & 0 \\
 0 &  5 & 1 & 3
 \end{array}\right)$.
 \newline
 To save time, {\bf do not} find
 eigenvectors!
%A4:=Matrix([ # S2012
%    [0, -12,  3, 0],
%    [ 0 , 1,  3, 0],
%    [ 0,  1,  -1, 0],
%    [ 0,  5,  1, 3]]);linalg[eigenvals](A4);
% ans == 0, 2, 3, -2
}{}{The characteristic polynomial is $\det(A-rI)=(-r)(3-r)(r-2)(r+2)$. The eigenvalues are $0,2,-2,3$.
 Determinant expansion of $\det(A-\lambda I)$ is by the cofactor method along column 1.
This reduces it to a $3\times 3$ determinant, which can be expanded by the cofactor method along column 3. }


\anotherpage
\problem{20}{Let $I$ denote the $3\times 3$ identity matrix. Assume given two $3\times 3$ matrices
$B$, $C$, which satisfy $CP=PB$ for some invertible matrix $P$. Let $C$
have eigenvalues $-1$, $1$, $0$. Let $A=2I+3B$. Find the eigenvalues of $A^3$.
% ans:
}{}{Both $B$ and $C$ have the same eigenvalues, because $\det(B-\lambda I)=\det(P(B-\lambda I)P^{-1})=\det(PCP^{-1}-\lambda PP^{-1})=\det(C-\lambda I)$. Further, both $B$ and $C$ are diagonalizable. The answer is the same for all such matrices, so the computation can be done for a diagonal matrix $B=\diag(-1,1,0)$. In this
case, $A=2I+3B=\diag(2,2,2)+\diag(-3,3,0)=\diag(-1,5,2)$ and the eigenvalues of $A$ are $-1,5,2$.
To finish, observe that the eigenvalues of $A^3$ and $A$ are related, because $A\vec{v}=\lambda\vec{v}$
implies $A^3\vec{v}=\lambda^3\vec{v}$. Therefore, the eigenvalues of $A^3$ are $-1,125,8$. }

% 2012, save for 2013
%\anotherpage
%\problem{5}{The matrix $A$ below has eigenvalues $3$, $3$ and $4$.
%Test $A$ to see it is diagonalizable, and if it is, then display $P$ and $D$ in the
%diagonalization equation $AP=PD$.
%$$
% A=\left(\begin{array}{rrrr}
% 4 & 1 & 1 \\
% -1 & 2 & 1 \\
% 0 & 0 &  4
% \end{array}\right)$$
%}{}{Eigenvalue $4$ will have exactly one eigenpair. Failure to be diagonalizable can occur only for eigenvalue $3$. Compute $\rref(A-3I)=\threebythree{1&1&0}{0&0&1}{0&0&0}$. This has rank 2, nullity 1. There is just one eigenvector $\colvector{1\\ -1 \\ 0}$. No Fourier's model, not diagonalizable.
%}
%
%A:=Matrix([[4 , 1 , -1 ],
%  [-1 , 2 , -2 ],[0 , 0 ,  4  ]
%  ]);linalg[eigenvals](A);linalg[eigenvects](A);
% linalg[rref](A-3*Matrix(3,3,shape=identity));
% rref = [[1, 1, 0], [0, 0, 1], [0, 0, 0]]

% 2012, save for another version 2013
%\problem{5}{Find the eigenvalues of the matrix
%$A=\left(\begin{array}{rrrrr}
% 0 & -2 & -5 & 0 & 0 \\
% 1 & 0 & -12 & 3 & 0 \\
% 0 & 0 &  1 & -1 & 0 \\
% 0 & 0 &  1 & 3 & 0 \\
% 0 & 0 &  5 & 1 & 3
% \end{array}\right)$.
% \newline
% To save time, {\bf do not} find
% eigenvectors!
%%A4:=Matrix([ # S2010
%%    [ 0,  -2, -5,  0, 0],
%%    [1,  0, -12,  3, 0],
%%    [ 0,  0,  1,  -1, 0],
%%    [ 0 , 0 , 1,  3, 0],
%%    [ 0,  0,  5,  1, 3]]);linalg[eigenvals](A4);
%% ans == sqrt(2)*I, -sqrt(2)*I, 2, 2, 3
%}{}{The characteristic polynomial is $\det(A-rI)=(r^2+2)(3-r)(r-2)^2$. The eigenvalues are $2,2,3,\pm \sqrt2 i$. Determinant expansion is by the cofactor method along column 5. This reduces it to a $4\times 4$ determinant, which can be expanded as a product of two quadratics. In detail, we first get $|A-rI|=(3-r)|B-rI|$,
%where $B=\left(\begin{array}{rrrrr}
% 0 & -2 & -5 & 0  \\
% 1 & 0 & -12 & 3  \\
% 0 & 0 &  1 & -1  \\
% 0 & 0 &  1 & 3
% \end{array}\right)$. So we have one eigenvalue $3$, and we find the eigenvalues of $B$. Matrix $B$ is a block matrix $B=\left(\begin{array}{r|r} B_1 & B_2 \\ \hline 0 & B_3 \end{array}\right)$, where $B_1,B_2,B_3$ are all $2\times 2$ matrices. Then $B-rI=\left(\begin{array}{r|r} B_1-rI & B_2 \\ \hline 0 & B_3-rI \end{array}\right)$. Using the determinant product theorem for such special block matrices (zero in the left lower block) gives $|B-rI|=|B_1-rI|B_3-rI|$.
% So the answer for the eigenvalues of $A$ is $3$ and the eigenvalues of $B_1$ and $B_3$. We report $3,\pm\sqrt2 i, 2,2$.
%}

\anotherpage
\problem{10}{
The Cayley-Hamilton theorem suggests that there are real $2\times 2$ matrices $A$
  such that $A^2=A$.  Give an example of one such matrix $A$.
}{}{Choose any matrix whose characteristic equation is $\lambda^2-\lambda=0$.
Then $A^2-A=0$ by the Cayley-Hamilton theorem.}

\anotherpage
\problem{40}{
The spectral theorem says that a symmetric matrix $A$ can be factored into $A=QDQ^T$ where $Q$ is orthogonal and $D$ is diagonal. Find $Q$ and $D$ for the symmetric matrix
$A=\begin{pmatrix}2 & -1\\ -1 &2\end{pmatrix}$.
}{}{
Start with the equation $r^2-4r+3=0$ having roots $r=1,3$. Compute the eigenpairs  $(1,\vec{v}_1)$, $(3,\vec{v}_2)$ where
$\vec{v}_1=\colvector{1\\ 1}$ and $\vec{v}_2=\colvector{-1\\ 1}$. The two vectors are orthogonal but not of unit length. Unitize them to
get $\vec{u}_1=\frac{1}{\sqrt2}\vec{v}_1$,  $\vec{u}_2=\frac{1}{\sqrt2}\vec{v}_2$. Then $Q=<\vec{u}_1 | \vec{u}_2>=\frac{1}{\sqrt2}\twobytwo{1&-1}{1&1}$,
$D=\diag(1,3)$.
}

% 2012, save for 2013
%\problem{5}{
%Prove that a Jordan block of size $3\times 3$ has exactly one eigenvector.
%}{}{
%Write the matrix $A=\threebythree{c&1&0}{0&c&1}{0&0&c}$ where $c$ is the eigenvalue of $A$. This is the general form of
%a Jordan block of dimension $3$. We analyze $(A-cI)\vec{x}=\vec{0}$ in order to find all eigenvectors of $A$. This system
%in scalar form is $x_2=0, x_3=0, 0=0$. There is one free variable, therefore exactly one eigenvector.
%}

%\section*{Chapter 7}

% 2012, save for 2013
%\anotherpage
%\problem{5}{
%Give an example of a linear transformation that makes a clockwise rotation in the plane. Expected is a $2\times 2$ matrix $A$
%which represents the linear transformation.
%}{}{
%$\twobytwo{\cos \theta&\sin\theta}{-\sin\theta&\cos\theta}$ for
%$\theta>0$ rotates clockwise and for $\theta<0$ rotates counter clockwise.
%}

\anotherpage
\problem{30}{
Let the linear transformation $T$ from $\Rr^2$ to $\Rr^2$
be defined by its action on two independent vectors:
$$
T\left(\begin{pmatrix}3\\ 0\end{pmatrix}\right)=
\begin{pmatrix}4 \\ 4\end{pmatrix},
T\left(\begin{pmatrix}2\\ 1\end{pmatrix}\right)=
\begin{pmatrix}4 \\  2\end{pmatrix}.$$
Find the unique $2\times 2$ matrix $A$ such that $T$ is defined
by the matrix multiply equation $T(\vec{x})=A\vec{x}$.
}{}{
Let $A$ be the matrix for $T$, such that $T(\vec{v})=A\vec{v}$ (matrix multiply).
Then $A\twobytwo{3&2}{0&1}=\twobytwo{4&4}{4&2}$. Solving for $A$ gives
$A=\twobytwo{4&4}{4&2}\twobytwo{3&2}{0&1}^{-1}= \twobytwo{4/3& 4/3}{4/3& -2/3}$.
}


\anotherpage
\problem{10}{
Assume singular value decomposition
$A=U\Sigma V^T$ given by
$$\twobytwo{2&2}{-1&1}=
\twobytwo{1&0}{0&1}
\twobytwo{\sqrt8&0}{0&\sqrt2}
\left(\frac{1}{\sqrt2}\twobytwo{1&-1}{1&1}\right)^T.$$
Find a formula for the pseudo-inverse, but don't bother to multiply out matrices.
}{}{
The definition is $A^+ = V\Sigma^+U^T$. Identifying the matrices from the formula gives
$A^+ =
\left(\frac{1}{\sqrt2}\twobytwo{1&-1}{1&1}\right)
\twobytwo{\frac{1}{\sqrt8}&0}{0&\frac{1}{\sqrt2}}
\twobytwo{1&0}{0&1}^T$.
}



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% Problems End Here % ------------------------------------------------------- %

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