For Problem DD-10:

(a) If $A = \left[\begin{array}{cc}a&b\\ b&d\end{array}\right]$, we can write $p _ A(\lambda) = \lambda ^ 2 - t \lambda + \Delta$, where $t = a + d$ and $\Delta = ad - b ^ 2$. The quadratic formula says the roots are $\lambda = {\frac 1 2}(t \pm \sqrt{t^2 - 4 \Delta})$. Simplifying the discriminant (the part inside the square root), we get $t^2 - 4 \Delta = (a+d)^2 - 4 (ad - b^2) = a^2 + 2 ad + d ^ 2 - 4 ad + 4 b^2 = (a-d) ^ 2 + (2b) ^ 2$.

(b) Since the discriminant is the sum of two squares and so is $\geq 0$, the square root is real. Therefore the eigenvalues are real and are ${\frac 1 2}(a + d \pm \sqrt{(a+d)^2 + (2b)^2})$.

(c) As suggested, the roots are equal when $(a-d) ^ 2 + (2b) ^ 2 = 0$. That happens when $a = d$ and $b = 0$, which is when $A = \left[\begin{array}{rr}a&0\\ 0&a\end{array}\right]$, a scalar matrix.

For Problem DD-11:

$\langle$   u$, A$v$\rangle =$   u$^ t A$   v and $\langle A$u$,$   v$\rangle = (A$u$) ^ t$   v$=$   u$^ t A ^ t$   v$=$   u$^ t A$   v (since $A$ is symmetric), $= \langle$   u$, A$v$\rangle$.