For Problem U-8: (a) $\left[\begin{array}{rr}a&b\ c&d\end{array}\right]\left[\begin{array}{r}1\ 1\end{array}\right] = \left[\begin{array}{r}(a+b)\ (c+d)\end{array}\right]$, the vector of row sums. So $A \left[\begin{array}{r}1\ 1\end{array}\right] = 1 \cdot \left[\begin{array}{r}1\ 1\end{array}\right]$ if and only if the row sums are both 1.

(b) Let $A$ be a stochastic $2 \times 2$ matrix, meaning that the column sums are 1. Part (a) was about row sums being 1, so $A ^ t$ has the eigenvalue 1, i.e., 1 is a root of the characteristic polynomial of $A$. By Observation 2, $A$ itself has the same characteristic polynomial and so also has the eigenvalue 1.

(c) By (b) we just look for an eigenvector for the eigenvalue 1. $A - 1 \cdot I = \left[\begin{array}{rr}-.8&.7\ .8&-.7\end{array}\right]$; $(A-I)$v$=$   0 says $\left[\begin{array}{rr}-.8&.7\ .8&-.7\end{array}\right]\left[\begin{array}{r}x\ y\end{array}\right] = \left[\begin{array}{r}0\ 0\end{array}\right]$; one solution is v$= \left[\begin{array}{r}x\ y\end{array}\right] = \left[\begin{array}{r}7\ 8\end{array}\right]$. But with stochastic matrices, usually it is best to have vectors whose entries sum to 1, so we can scale $\left[\begin{array}{r}7\ 8\end{array}\right]$ to get $\left[\begin{array}{r}\frac 7{15}\ \frac 8{15}\end{array}\right]$.

For Problem U-9: $A$   v$= 0$   v just says $A$   v$=$   0, so v is in the kernel of $\tau _ A$ = the null space of $A$. In this case, $A$ is singular, since for a nonsingular matrix $A$ we would have $A$v$=$   0$\Rightarrow$   v$= A ^ {-1}$0$=$   0.

Important: Notice that an eigenvalue can be 0 but an eigenvector can't be 0.