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\section*{Strang: Chapter 6}
\begin{description}

\item[{\bf Section 6.1}.] Exercises 1, 2, 3, 4, 5, 6, 9, 16, 25, 32

\item[{\bf Section 6.2}.] Exercises 4, 11, 12, 15, 16, 18, 20, 26

\item[{\bf Section 6.3}.] Exercises  4, 10, 18, 19

\item[{\bf Section 6.4}.] Exercises  5, 7, 11, 14, 21, 23

\item[{\bf Section 6.5}.] Exercises  3, 8, 10, 23, 24, 35

\item[{\bf Problem week13-1}.] Find the eigenvalues of the Markov matrix
$A =\twobytwo{.90& .15}{.10 &.85}$.
The sum of the eigenvalues is the trace of $A$. What is the steady state eigenvector for the eigenvalue $\lambda_1=1$?
See Exercise 8.3-1.

\item[{\bf Problem week13-2}.] Prove that the square $M^2$ of a Markov matrix $M$ is also a Markov matrix. See Exercise 8.3-9.


\item[{\bf Problem week13-3}.] If $A$ is a Markov matrix, then does $I + A + A^2+ \cdots$ add up to the {\em resolvent} $(A-I)^{-1}$? See Exercise 8.3-17.

\item[{\bf Section 6.6}.] Exercises 3, 17, 20

\item[{\bf Section 6.7}.] Exercises 1, 4, 5, 6

\end{description}


\subsection*{Some Answers}

\bigskip

{\bf 6.1}. Exercises 1, 3, 6, 16, 32 have textbook answers.

{\bf 6.1-2}. A has $\lambda_1 = -1$ and $\lambda_2 = 5$ with eigenvectors $x_1=(-2,1)$ and $x_2 =(1,1)$. The
matrix $A + I$ has the same eigenvectors, with eigenvalues increased by $1$ to $0$ and $6$.
That zero eigenvalue correctly indicates that $A + I$ is singular.

{\bf 6.1-4}.  A has $\lambda_1 =-3$ and $\lambda_2 = 2$ (check trace = $-1$ and determinant = $-6$) with $x_1 = (3,-2)$ and $x_2 = (1,1)$. $A^2$ has the same eigenvectors as $A$, with eigenvalues $\lambda_1^2 =9$
and $\lambda_2^2=4$.

{\bf 6.1-5}. A and B have eigenvalues $1$ and $3$. $A+B$ has $\lambda_1 = 3$, $\lambda_2 = 5$. Eigenvalues of $A+B$
are not equal to eigenvalues of $A$ plus eigenvalues of $B$.

{\bf 6.1-9}. (a) Multiply by $A$: $A(Ax)=A(\lambda x) = \lambda Ax$ gives $A^2x=\lambda^2 x$ (b) Multiply by
$A^{-1}$: $x= A^{-1}Ax = A^{-1}(\lambda x)=\lambda A^{-1}x$ gives $A^{-1}x = \frac{1}{\lambda} x$ (c) Add $Ix = x$:
$(A + I)x=(\lambda+ 1)x$.

{\bf 6.1-25}. With the same $n$ eigenpairs $(\lambda_i,x_i)$, then $x=c_1 x_1+\cdots+c_n x_n$ implies $Ax= c1\lambda_1 x_1+\cdots+c_n\lambda_n x_n$ and
 $Bx= c1\lambda_1 x_1+\cdots+c_n\lambda_n x_n$, therefore
$Ax=Bx$ for all vectors $x$, which implies $A=B$.




\bigskip

{\bf 6.2}. Exercises 4, 12, 15, 26 have textbook answers.


{\bf 6.2-11}. (a) True (no zero eigenvalues) (b) False (repeated $\lambda= 2$ may have only one line of
eigenvectors) (c) False (repeated $\lambda$ may have a full set of eigenvectors).

{\bf 6.2-16}. $\Lambda=\twobytwo{1&0}{0&0.2}$, $S=\twobytwo{1&1}{1&-1}$, $\Lambda^k \to \twobytwo{1&0}{0&0}$, $S\Lambda^kS^{-1} \to \frac12\twobytwo{1&1}{1&1}$, which is the steady state.

{\bf 6.2-18}. $A^k = \frac12\twobytwo{1+3^k&1-3^k}{1-3^k&1+3^k}$

{\bf 6.2-20}. This proof works when A is
diagonalizable, $A=S\Lambda S^{-1}$:
$$\det(A)= \det( S)\det(\Lambda)\det(S^{-1})= \det(\Lambda)=\lambda_1\cdots\lambda_n$$






\bigskip

{\bf 6.3}. Exercise 4  has a textbook answer.

{\bf 6.3-10}. $A=\twobytwo{0&1}{4&5}$. $\lambda^2-5\lambda - 4=0$ is the characteristic equation of $A$ with roots
$\frac52\pm\frac12\sqrt{41}$. Check the characteristic equation by substitution of $y=e^{\lambda x}$ into the differential equation $y''-5y'-4y=0$.

{\bf 6.3-18}.  Differentiate the matrix series for $e^{At}$ as though $A$ was a scalar to get the calculus answer
$A+A^2t+A^3t^2/2 + \cdots$ which is exactly $A$ times the infinite series  for $e^{At}$.

{\bf 6.3-19}. $e^{Bt}=I+Bt$ (because $B^2$, $B^3$, \ldots are all the zero matrix). Then $e^{Bt}=\twobytwo{1&-4t}{0&1}$. Check $\frac{d}{dt}e^{Bt}=\twobytwo{0&-4}{0&0}$ and $Be^{Bt}=B(I+Bt)=B+B^2t=B+\mbox{zero matrix}=\twobytwo{0&-4}{0&0}$.


\bigskip

{\bf 6.4}. Exercises   5, 11, 14, 21, 23 have textbook answers.

{\bf 6.4-7}. (a) $\twobytwo{1& 2 }{2& 1}$
has eigenvalues $-1$ and $3$. (b) Each pivot has the same signs as the $\lambda$s (c) trace
$=\lambda_1  + \lambda_2 = 2$, so $A$ cannot have two negative eigenvalues.

\bigskip

{\bf 6.5}. Exercises 3, 8, 10, 24 have textbook answers.

{\bf 6.5-23}. $x^2/a^2 + y^2/b^2$ is $x^TAx$ when $A = \diag(1/a^2, 1/b^2)$. Then $\lambda_1 = 1/a^2$ and $\lambda_2 = 1/b^2$ so $a= 1/\sqrt{\lambda_1}$ and $b=1/\sqrt{\lambda_2}$. The ellipse $9x^2 + 16y^2 = 1$ has axes with
half-lengths $a = 1/3$ and $b = 1/4$. The points $(1/3,0)$ and $(0,1/4)$ are at the ends of the axes.

{\bf 6.5-35}. Put parentheses in $x^TA^TCAx$ to get  $(Ax)^TC(Ax)$. Since $C$ is assumed positive definite,
this energy can drop to zero only when $Ax = 0$. Since $A$ is assumed to have independent
columns, then $Ax = 0$ only happens when $x = 0$. Thus $A^TCA$ has positive energy and it is
positive definite.

{\bf Strang}: My textbooks {\em Computational Science and Engineering} and {\em Introduction to Applied
Mathematics} start with many examples of $A^TCA$ in a wide range of applications.
I believe this is a unifying concept from linear algebra.


\bigskip

{\bf Problem week13-1}.  Eigenvalues $\lambda = 1, 0.75$; $(A - I)x = 0$ gives the steady state $x=(.6, .4)$ with
$Ax = x$.

{\bf Problem week13-2}. $M^2$ is still nonnegative; multiply $M$ on the left by $y=[1,\ldots,1]$ (all ones) to obtain $yM=y$. Then multiply $yM=y$ on the right by $M$ to
find $yM^2 = y$, which implies that the columns of $M^2$ add to $1$.

{\bf Problem week13-3}. No, $A$ has an eigenvalue $\lambda=1$ and $(I-A)^{-1}$ does not exist.


\bigskip

{\bf 6.6}. Exercise 17 has a textbook answer.

{\bf 6.6-3}. $B =\twobytwo
{1& 0}
{0& 0}
=\twobytwo
{1& 0}
{1& 1}^{-1}
\twobytwo
{1& 0}
{1& 0}
\twobytwo
{1& 0}
{1& 1}
=M^{-1}AM$;

$B=\twobytwo
{1& -1}
{-1& 1}
=\twobytwo
{1& 0}
{0& -1}^{-1}
\twobytwo
{1& 1}
{1& 1}
\twobytwo
{1& 0}
{0& -1}
$
;

$B =\twobytwo
{4& 3}
{2& 1}
=\twobytwo
{0& 1}
{1& 0}^{-1}
\twobytwo
{1& 2}
{3& 4}
\twobytwo
{0& 1}
{1& 0}$
.

{\bf 6.6-20}. (a) $A = M^{-1}BM$ implies $A^2= AA=M^{-1}B^2M$. So $A^2$ is similar
to $B^2$. (b) $A^2$ equals $(-A)^2$ but $A$ may not be similar to $-B$ (it could be!).
(c) $\twobytwo
{3& 1}
{0& 4}$
is diagonalizable to
$\twobytwo
{3& 0}
{0& 4}$
because $\lambda_1 \ne \lambda_2$, so these matrices are similar.
(d) $\twobytwo
{3& 1}
{0& 3}$
has only one eigenvector, so it is not diagonalizable (e) $PAP^T$ is similar to A.

\bigskip

{\bf 6.7}. Exercises 1, 4, 5 have textbook answers.

{\bf 6.7-6}.  $AA^T =\twobytwo
{2& 1}
{1& 2}$
has $\sigma_1^2= 3$ with $u_1 =
\frac{1}{\sqrt2} \colvector{1 \\ 1}$ and
$\sigma_2^2= 1$ with $u_2 =
\frac{1}{\sqrt2} \colvector{1 \\ -1}$.

$A^TA =\threebythree
{1& 1& 0}
{1& 2& 1}
{0& 1& 1}$
has
$\sigma_1^2= 3$ with $v_1 =
\frac{1}{\sqrt6}\colvector{1\\ 2\\ 1}$,
$\sigma_2^2=1$
with $v_2 =
\frac{1}{\sqrt2}\colvector{1\\ 0\\ -1}$
and
$v_3 =
\frac{1}{\sqrt3}\colvector{1\\ -1\\ 1}$.
Then
$$\left(\begin{array}{rrr} 1 &1& 0 \\ 0& 1& 1 \end{array}\right)=
\AUG(u_1 , u_2) \left(\begin{array}{rrr} \sqrt3 &0& 0 \\ 0& 1& 0 \end{array}\right) \AUG(v_1,v_2,v_3)^T$$


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