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\begin{document}
\examtitle{MATH 2270-2}{Sample Exam 2}{S2018}
\thispagestyle{empty}
\studentinfo
\finishfirstpage
\instructions{\begin{quote}\sf
No books or notes. No electronic devices, please.\\
These problems have credits 10 to 25, which is an estimate of the time
required to write the solution.
\\
This sample will be edited during the semester to reflect course content. Expect
to see problems that appeared on the 2017 final exam.
\\
The number of problems on Exam 2 is ten (10). This sample has more than 30 problems, in order to
illustrate possible problem types. See the 2016 and 2017 Exam2 (PDF) questions and answers for
a typical exam with 10 questions.
\vspace*{1pc}
\end{quote}
}


\setcounter{page}{1}
% Problems Start Here % ----------------------------------------------------- %
\anotherpage
\problem{5}{
Define matrix $A$ and vector $\vec{b}$ by the equations
$$A=\left(\begin{array}{rrrr}
 -2 &  3\\
  0 & -4 \\
 \end{array}\right), \quad \vec{b}=\colvector{-3 \\ 5}.$$
For the system $A\vec{x}=\vec{b}$, find $x_1$, $x_2$ by Cramer's Rule,
 showing {\bf all details} (details count 75\%).}{}{
$x_1=\Delta_1/\Delta$, $x_2=\Delta_2/\Delta$, $\Delta=\det\twobytwo{-2 & 3}{0 & -4}=8$, $\Delta_1=\det\twobytwo{-3 & 3}{5 & -4}=-3$, $\Delta_2=\det\twobytwo{-2 & -3}{0 & 5}=-10$, $x_1=\frac{-3}{8}$, $x_2=\frac{-10}{8}=\frac{-5}{4}$.
%
%A:=Matrix(
%  [[ -2 , 3  ],
%  [0 ,  4 ]]);
%  b:=Vector([-3,1]);
%  with(linalg):linsolve(A,b); det(A);
%  # answers =
%
}

\anotherpage
\problem{5}{
 Assume given $3\times 3$ matrices $A$, $B$. Suppose
 $E_3E_2E_1A=BA^2$ and $E_1$, $E_2$, $E_3$ are
 elementary matrices representing respectively a multiply by $3$, a swap and a combination. Assume $\det(B)=3$.
 Find all possible values of $\det(-2A)$.
}{}{
Start with the determinant product theorem $|FG|=|F||G|$. Apply it to obtain
$|E_3||E_2||E_1||A|=|B||A|^2$. Let $x=|A|$ in this equation and solve for $x$. You will need to know that
$|E_1|=3$, $|E_2|=-1$, $|E_3|=1$. Let $C=-2A$. Then $|C|=|(-2I)A|=|-2I||A|=(-2)^3x$. The answer is $|C|=-8x$, where
$x$ is the solution of $-3x=3x^2$. Then $|C|=0$ or $|C|=8$.
}

\anotherpage
\problem{5}{
Let $A=\twobytwo{2&1}{0&3}$. Show the details of two different methods for finding $A^{-1}$.
}{}{The two methods are (1) $A^{-1}=\frac{\adj(A)}{|A|}$ and (2)
For $C=<A|I>$, then $\rref(C)=<I|A^{-1}>$. Expected details not supplied here.}

\anotherpage

\problem{5}{
Find a factorization $A=LU$ into lower and upper triangular matrices for
the matrix $A=\threebythree{2&1&0}{1&2&1}{0&1&2}$.
}{}{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 $E_2E_1A = U=\threebythree{2&1&0}{0&\frac32 & 1}{0 & 0 & \frac43}$. Let $L=E_1^{-1}E_2^{-1}=\threebythree{1&0&0}{\frac12&1&0}{0&\frac23&1}$.}
\anotherpage

\problem{5}{
Explain how the {\bf span theorem} applies to show that the set $S$ of
all linear combinations of the functions $\cosh x, \sinh x$ is a
subspace of the vector space $V$ of all continuous functions on $-\infty
< x < \infty$.
}{}{The span theorem says $\SPAN(\vec{v}_1,\vec{v}_2)$ is
a subspace of $V$, for any two vectors in $V$. Choose the two vectors to
be $\cosh x, \sinh x$.}

\problem{5}{
Write a proof that the subset $S$ of all solutions $\vec{x}$ in $\Rr^n$
to a homogeneous matrix equation $A\vec{x}=\vec{0}$ is a subspace of
$\Rr^n$. This is called the {\bf kernel theorem}.
}{}{(1) Zero is in $S$ because $A\vec{0}=\vec{0}$; (2) If $A\vec{v}_1=\vec{0}$ and $A\vec{v}_2=\vec{0}$, then
$\vec{v}=\vec{v}_1+\vec{v}_2$ satisfies $A\vec{v}=A\vec{v}_1 + A\vec{v}_2 = \vec{0}+\vec{0}=\vec{0}$. So $\vec{v}$ is in $S$;
(3) Let $\vec{v}_1$ be in $S$, that is, $A\vec{v}_1=\vec{0}$. Let $c$ be a constant. Define $\vec{v}=c\vec{v}_1$. Then
$A\vec{v}=A(c\vec{v}_1)=cA\vec{v}_1=(c)\vec{0}=\vec{0}$. Then $\vec{v}$ is in $S$. This completes the proof.
}

\anotherpage

\problem{5}{
Using the subspace criterion, write two hypotheses that imply that a set
$S$ in a vector space $V$ is not a subspace of $V$. The full statement
of three such hypotheses is called the {\bf Not a Subspace Theorem}.
}{}{(1) If the zero vector is not in $S$, then $S$ is not a subspace.
(2) If two vectors in $S$ fail to have their sum in $S$, then $S$ is not
a subspace. (3) If a vector is in $S$ but its negative is not, then $S$ is not a subspace.
}

\problem{5}{
Report which columns of $A$ are pivot columns: $A=\threebythree{0&1&1}{0&1&2}{0&0&0}$.
}{}{Zero cannot be a pivot column (no leading one in $\rref(A)$). The
other two columns are not constant multiples of one another, therefore
they are independent and will become pivot columns in $\rref(A)$. Then: pivot columns =2,3.
}

\problem{5}{
Find the complete solution $\vec{x} = \vec{x}_h + \vec{x}_p$ for the nonhomogeneous system
$$\threebythree{0&1&1}{0&1&2}{0&0&0}\colvector{x_1 \\ x_2 \\ x_3} = \colvector{2 \\ 3 \\ 0}.$$
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 $\begin{array}{cccc} 0&1&0&1\\ 0&0&1&1\\ 0&0&0&0\end{array}$. Then $x_1=t_1,
x_2=1, x_3=1$ is the general solution in scalar form. The partial
derivative on $t_1$ gives the homogeneous solution basis vector
$\colvector{1\\ 0\\ 0}$. Then $\vec{x}_h=c_1\colvector{1\\ 0\\ 0}$ Set
$t_1=0$ in the scalar solution to find a particular solution
$\vec{x}_p=\colvector{0\\ 1 \\ 1}$.
}

\problem{5}{
Find the reduced row echelon form of the matrix $A=\threebythree{0&1&1}{0&0&0}{0&1&2}$.
}{}{It is the matrix $\threebythree{0&1&0}{0&0&1}{0&0&0}$.
}

\problem{5}{
A $10\times 13$ matrix $A$ is given and the homogeneous system $A\vec{x}=\vec{0}$ is transformed to
reduced row echelon form. There are $7$ lead variables. How many free variables?
}{}{Because $\vec{x}$ has 13 variables, then the rank plus the nullity is 13. There are 6 free variables.}

\problem{5}{
The rank of a $10\times 13$ matrix $A$ is $7$. Find the nullity of $A$.
}{}{There are 13 variables. The rank plus the nullity is 13. The nullity is 6. }

\anotherpage


\problem{5}{
%%%Not on the final exam S2012:
Given a basis $\vec{v}_1=\begin{pmatrix}3\\2\end{pmatrix},
\vec{v}_2=\begin{pmatrix}4\\4\end{pmatrix}$ of $\Rr^2$, and
$\vec{v}=\begin{pmatrix}10\\4\end{pmatrix}$, then
$\vec{v}=c_1\vec{v}_1+c_2\vec{v}_2$ for a unique set of coefficients
$c_1,c_2$, called the {\em coordinates of $\vec{v}$ relative to the
basis $\vec{v}_1, \vec{v}_2$}. Compute $c_1$ and $c_2$.
}{}{}

\anotherpage
% chapter 3
\problem{5}{
        Determine independence or dependence for the list of vectors
\[\begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}, \begin{pmatrix} 4\\ 0\\
4\end{pmatrix}, \begin{pmatrix} 3 \\ 2
  \\1\end{pmatrix}\]
}{}{Possible tests are the rank test, determinant test, pivot theorem.
Let $A$ denote the augmented matrix of the three column vectors.  The
determinant is 32, nonzero, so the vectors are independent. The pivot
theorem also applies. The $\rref(A)$ is the identity matrix, so all
columns are pivot columns, hence the three columns are independent. The
rank test applies because the rank is 3, equal to the number of columns,
hence independence. }

\anotherpage
\problem{5}{%
Check the independence tests which apply to prove that $1$, $x^2$, $x^3$ are independent in the vector space $V$ of
all functions on $-\infty < x < \infty$.
\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 Euler Solution Test & Any finite set of distinct Euler solution 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 apply to the given functions, while the others
apply only to fixed vectors.}

\anotherpage

\problem{5}{%
Define $S$ to be the set of all vectors
$\vec{x}$ in $\Rr^{3}$ such that $x_1+x_3=0$ and $x_3+x_2=x_1$.
Prove that
$S$ is a subspace of $\Rr^{3}$.
}{}
{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 $B\vec{x}=\vec{0}$. }

\anotherpage
\problem{5}{
The $5\times 6$ matrix $A$ below has some
independent columns. Report the independent columns of $A$, according to
the Pivot Theorem.
$$A=\left(\begin{array}{rrrrrr}
  0 & 0 &  0 & 0 &   0 & 0\\
 -3 & 0 &  0 & -2 &  1 & -1\\
 -1 & 0 &  0 & 0 &   1 &  0\\
  6 & 0 &  0 & 6 &   0 &  3\\
  2 & 0 &  0 & 2 &   0 &  1
 \end{array}\right)$$
}{}
%A:=matrix([
%  [ 0,  0, 0,  0,  0 , 0 ],
%  [-3,  0, 0, -2,  1 , -1],
%  [-1,  0, 0,  0,  1 , 0],
%  [ 6,  0, 0,  6,  0 , 3],
%  [ 2,  0, 0,  2,  0 , 1]
%    ]);
%  rref(A);
{Find {\small $\rref(A)= \left( \begin {array}{cccccc}
   1&0&0&0&-1&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\end {array} \right)
$}. The pivot columns are 1 and 4.}

\anotherpage
\problem{5}{
Let $A$ be an $m\times n$ matrix with independent columns. Prove that
$A^TA$ is invertible.
}{}{The matrix $B=A^TA$ has dimension $n\times n$. We prove that the nullspace of $B=A^TA$ is the zero vector.\\
Let $\vec{x}$ belong to $\Rr^n$.  Assume $B\vec{x}=\vec{0}$, then multiply this equation by $\vec{x}^T$ to obtain $\vec{x}^TA^TA\vec{x}=\vec{x}^T\vec{0}=0$. Therefore, $\|A\vec{x}\|^2=0$, or $A\vec{x}=\vec{0}$.
If $A$ has independent columns, then the nullspace of $A$ is the zero vector, so $\vec{x}=\vec{0}$.
We have proved that the nullspace of $B=A^TA$ is the zero vector.\\
An $n\times n$ matrix $B$ is invertible if and only if its nullspace is the zero vector. So $B=A^TA$ is invertible.
 }

\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.}

\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 conclusion follows.}



%\section*{Chapter 5: Eigenanalysis}

\problem{5}{Consider a $3\times 3$ real matrix $A$
%$$A=\left(\begin{array}{rrrr}
%  0 & 4 & -1 & 0 \\
% -4 & 0 & -2 & 1 \\
%  0 & 0 &  5 & 1 \\
%  0 & 0 & -3 & 1
% \end{array}\right)$$
%
%A:=matrix([[0 , 4 , -1 ],
%  [-4 , 0 , -2 ],
%  [0 , 0 ,  5  ]]);eigenvals(A);
%
%       # 5, 4 I, -4 I
with eigenpairs
$$\left(-1 ,  \colvector{5\\ 6\\ -4}\right),
\quad \left(2i, \colvector{i\\ 2\\ 0}\right),
\quad \left(-2i, \colvector{-i\\ 2\\ 0}\right).$$
\begin{quote}

 Display an invertible matrix $P$ and a
 diagonal matrix $D$ such that $AP=PD$.

% (2) [10\%] Display a matrix product formula for $A$. To save time, {\bf do not evaluate any matrix products}.
\end{quote}
}{}{The columns of $P$ are the eigenvectors and the diagonal entries of $D$ are the eigenvalues, taken in the same order.}

\problem{5}{Find the eigenvalues of the matrix
$A=\left(\begin{array}{rrrrr}
 0 & -12 & 3 & 0 \\
 0 &  1 & -1 & 0 \\
 0 &  1 & 3 & 0 \\
 0 &  5 & 1 & 3
 \end{array}\right)$.
 \newline
 To save time, {\bf do not} find
 eigenvectors!
%A4:=matrix([ # F2010
%    [0, -12,  3, 0],
%    [ 0,  1,  -1, 0],
%    [ 0 , 1,  3, 0],
%    [ 0,  5,  1, 3]]);eigenvals(A4);
% ans == 0, 2, 2, 3
}{}{The characteristic polynomial is $\det(A-rI)=(-r)(3-r)(r-2)^2$. 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. }

\problem{5}{The matrix
$A=\left(\begin{array}{rrrrr}
 0 & -12 & 3  \\
 0 &  1 & -1  \\
 0 &  1 & 3
 \end{array}\right)$
 has eigenvalues $0,2,2$ but it is not diagonalizable, because
 $\lambda=2$ has only one eigenpair. Find an eigenvector for $\lambda=2$.\newline
To save time, {\bf don't find the eigenvector for $\lambda=0$}.
%A5:=matrix([ # F2010
%    [0, -12,  3],
%    [ 0,  1,  -1],
%    [ 0 , 1,  3]]);eigenvects(A5);
% ans ==
}{}{Because $A-2I=\left(\begin{array}{rrr}
-2 & -12 & 3  \\
 0 & -1 & -1  \\
 0 &  1 &  1
 \end{array}\right)$ has last frame $B=\left(\begin{array}{rrr}
 1 &  0 & -15/2  \\
 0 &  1 &  1  \\
 0 &  0 &  0
 \end{array}\right)$, then there is only one eigenpair for $\lambda=2$, with
 eigenvector $\vec{v}=\colvector{15\\ -2\\ 2}$.}

\problem{5}{Find the two eigenvectors corresponding to complex eigenvalues $-1 \pm 2i$ for
 the $2\times 2$ matrix $A=\twobytwo{-1 & 2}{-2&-1}$.
}{}{$\left( -1+2i,\colvector{-i \\ 1}\right)$, $\left( -1-2i,\colvector{i \\ 1}\right)$}
% A6:=matrix([[-1,2],[-2,-1]]); eigenvects(A6);

\problem{5}{ Let $A=\twobytwo{-7 & 4}{-12 & 7}$. Circle possible eigenpairs of $A$.
$$
 \left(1,\colvector{1\\ 2}\right) ,\quad  \left(2,\colvector{2\\ 1}\right) ,\quad
 \left(-1,\colvector{2\\ 3}\right).
$$
% P:=transpose(matrix([[1,2],[2,3]]));D1:=diag(1,-1);A:=evalm(P&*D1&*inverse(P));
% # ans is A:=matrix([[-7,4], [-12,7]]);
}{}{The first and the last, because the test $A\vec{x}=\lambda\vec{x}$ passes in both cases.}

\problem{5}{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 $-1$, $1$, $5$. Find the eigenvalues of $A=2I+3B$.
% 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 $B=\diag(-1,1,5)$. In this
case, $A=2I+3B=\diag(2,2,2)+\diag(-3,3,15)=\diag(-1,5,17)$ and the eigenvalues of $A$ are $-1,5,17$.}


\problem{5}{Let $A$ be a $3\times 3$ matrix with
eigenpairs
$$
 (4,\vec{v}_1),\quad (3, \vec{v}_2), \quad
 (1, \vec{v}_3).
 $$
Let $P$ denote the augmented matrix of the eigenvectors $\vec{v}_2$, $\vec{v}_3$, $\vec{v}_1$, in exactly that order. Display the answer for $P^{-1}AP$. Justify the answer with a sentence.
}{}{Because $AP=PD$, then $D=P^{-1}AP$ is the diagonal matrix of eigenvalues, taken in the order determined by the eigenpairs $(3, \vec{v}_2), (1, \vec{v}_3), (4,\vec{v}_1)$. Then $D=\threebythree{3&0&0}{0&1&0}{0&0&4}$.}

\problem{5}{The matrix $A$ below has eigenvalues $3$, $3$ and $3$.
Test $A$ to see it is diagonalizable, and if it is, then display Fourier's model for $A$.
$$
 A=\left(\begin{array}{rrrr}
 4 & 1 & 1 \\
 -1 & 2 & 1 \\
 0 & 0 &  3
 \end{array}\right)$$
}{}{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 ,  3  ]]);eigenvals(A);
% \hrulefill
%

\problem{5}{ Assume $A$ is a given $4\times 4$ matrix
with eigenvalues $0$, $1$, $3\pm 2i$. Find the eigenvalues of
$4A-3I$, where $I$ is the identity matrix.
}{}{Such a matrix is diagonalizable, because of four distinct eigenvalues.
Then $4B-3I$ has the same eigenvalues for all matrices $B$ similar to $A$. In particular,
$4A-3I$ has the same eigenvalues as $4D-3I$ where $D$ is the diagonal matrix with entries
$0,1,3+2i,3-2i$. Compute $4D-3I=\fourbyfour{-3&0&0&0}{0&1&0&0}{0&0&9+8i&0}{0&0&0&9-8i}$. The answer is $0,1,9+8i,9-8i$.}

\problem{5}{Find the eigenvalues of the matrix
$A=\left(\begin{array}{rrrrr}
 0 & -2 & -5 & 0 & 0 \\
 3 & 0 & -12 & 3 & 0 \\
 0 & 0 &  1 & -1 & 0 \\
 0 & 0 &  1 & 3 & 0 \\
 0 & 0 &  5 & 1 & 3
 \end{array}\right)$.
 \newline
 To save time, {\bf do not} find
 eigenvectors!
%A4:=matrix([ # S2010
%    [ 0,  -2, -5,  0, 0],
%    [3,  0, -12,  3, 0],
%    [ 0,  0,  1,  -1, 0],
%    [ 0 , 0 , 1,  3, 0],
%    [ 0,  0,  5,  1, 3]]);eigenvals(A4);
% ans == sqrt(6)*I, -sqrt(6)*I, 2, 2, 3
}{}{The characteristic polynomial is $\det(A-rI)=(r^2+6)(3-r)(r-2)^2$. The eigenvalues are $2,2,3,\pm \sqrt6 i$. 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 $B=\left(\begin{array}{rrrrr}
 0 & -2 & -5 & 0  \\
 3 & 0 & -12 & 3  \\
 0 & 0 &  1 & -1  \\
 0 & 0 &  1 & 3
 \end{array}\right)$. 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 $3,\pm\sqrt6 i, 2,2$. It is also possible to directly find the eigenvalues of $B$ by cofactor expansion of $|B-rI|$.
}


\problem{5}{Consider a $3\times 3$ real matrix $A$
%$$A=\left(\begin{array}{rrrr}
%  0 & 4 & -1 & 0 \\
% -4 & 0 & -2 & 1 \\
%  0 & 0 &  5 & 1 \\
%  0 & 0 & -3 & 1
% \end{array}\right)$$
%
%A:=matrix([[0 , 4 , -1 ],
%  [-4 , 0 , -2 ],
%  [0 , 0 ,  5  ]]);eigenvals(A);
%
%       # 5, 4 I, -4 I
with eigenpairs
$$\left(3 ,  \colvector{13\\ 6\\ -41}\right),
\quad \left(2i, \colvector{i\\ 2\\ 0}\right),
\quad \left(-2i, \colvector{-i\\ 2\\ 0}\right).$$
\begin{quote}

 (1) [50\%] Display an invertible matrix $P$ and a
 diagonal matrix $D$ such that $AP=PD$.

 (2) [50\%] Display a matrix product formula for $A$, but do not evaluate the matrix products, in order to save time.
\end{quote}
}{}{(1) $P=\threebythree{13&i&-i}{6&2&2}{-41&0&0}$, $D=\threebythree{3&0&0}{0&2i&0}{0&0&-2i}$.
(2) $AP=PD$ implies $A=PDP^{-1}$.}

\problem{5}{Assume two $3\times 3$ matrices
$A$, $B$ have exactly the same characteristic equations. Let $A$
have eigenvalues $2$, $3$, $4$. Find the eigenvalues of $(1/3)B-2I$,
where $I$ is the identity matrix.
}{}{Because the answer is the same for all matrices similar to $A$ (that is, all $B=PAP^{-1}$)
then it suffices to answer the question for diagonal matrices. We know $A$ is diagonalizable, because it has distinct eigenvalues. So we choose $D$ equal to the diagonal matrix with entries $2,3,4$. Compute
$\frac13 D -2I = \threebythree{\frac23-2 & 0 & 0}{0 & \frac33-2 & 0}{0 & 0 & \frac43 - 2}$. Then the eigenvalues are $-\frac43, -1, -\frac23$.}

\problem{5}{Let $3\times 3$ matrices $A$ and $B$ be
related by $AP=PB$ for some invertible matrix $P$. Prove that the
roots of the characteristic equations of $A$ and $B$ are identical.
}{}{The proof depends on the identity $A-rI = PBP^{-1}-rI = P(B-rI)P^{-1}$ and the determinant product theorem $|CD|=|C||D|$. We get $|A-rI|=|P||B-rI||P^{-1}|=|PP^{-1}||B-rI|=|B-rI|$. Then $A$ and $B$ have exactly the same characteristic equation, hence exactly the same eigenvalues.}


\problem{5}{Find the eigenvalues of the matrix $B$:
$$B=\left(\begin{array}{rrrr}
 2 & 4 & -1 & 0 \\
 0 & 5 & -2 & 1 \\
 0 & 0 &  4 & 1 \\
 0 & 0 &  1 & 4
 \end{array}\right)$$
}{}{The characteristic polynomial is $\det(B-rI)=(2-r)(5-r)(5-r)(3-r)$. The eigenvalues are $2,3,5,5$.

It is possible to directly find the eigenvalues of $B$ by cofactor expansion of $|B-rI|$.

An alternate method is described below, which depends upon a determinant product theorem for special block matrices, such as encountered in this example.

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 is that $B$ has eigenvalues equal to the eigenvalues of $B_1$ and $B_3$. These are quickly found by Sarrus' Rule applied to the two $2\times 2$ determinants $|B_1-rI|=(2-r)(5-r)$ and $|B_3-rI|=r^2-8r+15=(5-r)(3-r)$.
}

% A:=matrix( [[2, 4 , -1 , 0 ],
%  [0 , 25 , -2 , 1 ],
%  [0 , 0 ,  4 , 1 ],
%  [0 , 0 ,  1 , 4]]);
% evs==2,5,5,3

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\begin{center} No new questions beyond this point. Please check back at the course web site until 2 April, for corrections and added sample exam problems.\end{center}

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