\def\bmatrix#1{\left[\matrix{#1}\right]}
\def\dmatrix#1{\left|\matrix{#1}\right|}
\def\fr#1#2{{#1\over #2}}

    {\bf Section 5.1}

\noindent{\bf 4.}

$P$ has two eigenvalues $c_1=0$ and $c_2=1,$ giving us corresponding
eigenvectors $v_1=(1,-1)$ and $v_2=(1,1).$
This gives a general solution
$$c_1v_1 + c_2v_2e^t,$$
and our initial condition $u(0)$ can be written as the combination
$v_1+4v_2,$
giving our particular solution
$$v_1 + 4v_2e^t.$$

\noindent{\bf 14.}

The matrix of ones has rank 1 because any pair of rows makes a linearly
dependent set (by including a duplicate).
$$\dmatrix{1-\lambda & 1 & 1 & 1\cr
           1 & 1-\lambda & 1 & 1\cr
           1 & 1 & 1-\lambda & 1\cr
           1 & 1 & 1 & 1-\lambda\cr} = 0\Longrightarrow
  \dmatrix{1-\lambda & 1 & 1 & 1\cr
           \lambda & -\lambda & 0 & 0\cr
           \lambda & 0 & -\lambda & 0\cr
           \lambda & 0 & 0 & -\lambda\cr} = 0\Longrightarrow
  \dmatrix{4-\lambda & 0 & 0 & 0\cr
           1 & -1 & 0 & 0\cr
           1 & 0 & -1 & 0\cr
           1 & 0 & 0 & -1\cr} = 0,
$$
by row operations (including multiplications), which do not affect a
zero determinant where $\lambda\neq0.$ Because this is triangular, the
determinant is the product of the diagonal, so $\lambda = 4.$
This eigenvalue corresponds to $(1,1,1,1)^T.$

The checkerboard matrix has rank 2 because any triplet of rows makes a
linearly dependent set (again by including a duplicate).

$$\dmatrix{-\lambda&1&0&1\cr
           1&-\lambda&1&0\cr
           0&1&-\lambda&1\cr
           1&0&1&-\lambda} = 0\Longrightarrow
  \dmatrix{-\lambda&0&\lambda&0\cr
           0&-\lambda&0&\lambda\cr
           0&1&-\lambda&1\cr
           1&0&1&-\lambda} = 0\Longrightarrow$$$$
  \dmatrix{-1&0&1&0\cr
           0&-1&0&1\cr
           0&0&-\lambda&2\cr
           0&0&2&-\lambda} = 0\Longrightarrow
  \dmatrix{-1&0&1&0\cr
           0&-1&0&1\cr
           0&0&-\lambda&2\cr
           0&0&0&-\lambda+4/\lambda} = 0
$$
This gives $\lambda^2 - 4 = 0 \to \lambda = \pm 2.$
These eigenvalues correspond to $(1,1,1,1)$ and $(1,-1,1,-1).$

\noindent{\bf 15.}

The rank of an $n\times n$ matrix of ones is 1 (giving $\lambda=0$ with
multiplicity $n-1$), and its remaining non-zero eigenvalue of $n$
(because $A-nI$ has the sum of every row, a linear combination, equal to
zero, therefore it is an eigenvalue).

The rank of an $n\times n$ checkerboard matrix is 2 because it has only
two unique columns (giving $\lambda=0$ with multiplicity $n-2$) and its
non-zero eigenvalues being $\pm\sqrt n,$ determined from the cofactor
expansion of $A-\lambda I$ where $A$ is the checkerboard matrix.

    {\bf Section 5.2}

\noindent{\bf 8.}

{\it (a)}
$Au = uv^Tu = u(v\cdot u) = (v\cdot u)u,$
giving that $u$ is an eigenvector with corresponding eigenvalue $v^Tu.$

{\it (b)}
The other eigenvalues are zero because there is one non-zero eigenvalue
already shown and the dimension of the null space of $A = A - 0I$ is
$n-r = n-1,$ completing the eigenvalue set.

{\it (c)}
The trace is equal to the sum of eigenvalues, which is $v^Tu + 0 =
v^Tu,$ and it's equal to the sum of the diagonal, of which each entry is
an element-wise product of $v$ and $u.$

\noindent{\bf 12.}

{\it (a)}
False.
$$\bmatrix{1&1&0\cr0&1&1\cr0&0&1}$$
is invertible but has only the eigenvector set mentioned.

{\it (b)}
True. Every eigenvalue corresponds to at least one eigenvector (counting
complex eigenvectors), so there must be only one (multiple) eigenvalue
of $A.$

{\it (c)}

A is not diagonalizable because its eigenvector set does not span $R^3.$

\noindent{\bf 32.}
$$A = \bmatrix{2&1\cr1&2}.$$
This matrix has eigenvalues $1$ and $3,$ and they correspond to
eigenvectors $(1,-1)$ and $(1,1)$ respectively, giving a diagonalization
with
$$S = \bmatrix{1&1\cr-1&1} \Longrightarrow S^{-1} = \fr12\bmatrix{1&-1\cr1&1}$$
and
$$\Lambda = \bmatrix{1&0\cr0&3}$$
$$A^k = S\Lambda^kS^{-1} =
\bmatrix{1&1\cr-1&1}\bmatrix{1&0\cr0&3^k}\fr12\bmatrix{1&-1\cr1&1} =
\fr12\bmatrix{1&1\cr-1&1}\bmatrix{1&-1\cr3^k&3^k} =
\fr12\bmatrix{1+3^k&-1+3^k\cr-1+3^k&1+3^k}.
$$

    {\bf Section 5.3}

\noindent{\bf 4.}

$$A = \bmatrix{1/2 & 1/2\cr 1 & 0}.$$

{\it (a)}
$A$ has characteristic polynomial $\lambda^2 - \lambda/2 - 1/2 =
(\lambda-1)(\lambda+1/2) \to \lambda = 1, -1/2,$ with eigenvectors
respectively $(1,1)$ and $(1/2, -1).$

{\it (b)}

$$A = -\fr23\bmatrix{1&1/2\cr 1&-1}\bmatrix{1&0\cr0&-1/2}
\fr13\bmatrix{2&1\cr2&-2}.$$
As $k\to\infty,$
$$A^k = \fr13\bmatrix{1&1/2\cr 1&-1}\bmatrix{1&0\cr0&-1/2}^k
\bmatrix{2&1\cr2&-2} =
\bmatrix{1&1/2\cr 1&-1}\bmatrix{2&1\cr0&0} =
\bmatrix{2&1\cr2&1}.$$

{\it (c)}

$$A\bmatrix{1\cr0} = -\fr13\bmatrix{2&1\cr2&1}\bmatrix{1\cr0} =
\bmatrix{2/3\cr2/3},$$
giving us the steady state limit of 2/3 from this initial condition.

\noindent{\bf 18.}

These matrices are stable/neutrally stable if both eigenvalues are less
than or equal to 1. Therefore, the maximal value will be when the larger
eigenvalue equals 1.

For the first matrix with characteristic equation
$\lambda^2 - (.2+a)\lambda + .2a + .64 = 0 \to
(\lambda-(.1+a/2))^2 = .01 + .1a + a^2/4 -.2a + .64,$
$\lambda = .1+a/2 \pm (.65-.1a+a^2/4) =
.75 - .4a + .25a^2 = 1 \to a = 2.081$

The eigenvalues of this matrix are $.2$ and $b,$ so the largest
neutrally stable value of $b$ is 1.

For the third matrix with characteristic equation $\lambda^2 -
2c\lambda + c^2 - .16 = 0,$ $\lambda = c \pm .4,$ giving maximum $c$ of
$0.6.$

\bye