For Problem DD-1: Assume (3). For (1): All matrices have real entries. Since $D$ has real eigenvalues (the diagonal entries), since $A \sim D$, and since similar matrices have the same eigenvalues, $A$ has real eigenvalues.

For (2): Notice that it's OK if eigenvalues are repeated (i.e., have multiplicity greater than 1). For the diagonal matrix $D$ and a given $\lambda$, the eigenspace of $\lambda$ is the span of all $e _ i$ with $D _ {ii} = \lambda$. Since the standard basis vectors are perpendicular (orthogonal) to each other, the eigenspaces are perpendicular. What is the relationship between the eigenvectors of $A$ and those of $D$? $Dv = \lambda v$ says $(P^{-1} A P) v = \lambda v$, or $A Pv = \lambda Pv$. Therefore the eigenspace of $A$ for $\lambda$ is the eigenspace of $D$ for $\lambda$ rotated by $\tau _ R$. Since $R$ is a rotation matrix, $\tau _ R$ is rigid and preserves angles, so the eigenspaces of $A$ are also perpendicular to one another.

Note. Maybe that was the hard way to show (2), since it's easy to prove (2) directly, as on p. DD 5.

Also notice that for an eigenspace $E _ \lambda$, $\lambda$ doesn't have to be an an eigenvalue, but if it's not, then $E _ \lambda = \{$0$\}$.

For Problem DD-2:

$p _ A(\lambda) = \lambda ^ 2 - 15 \lambda + 50 = (\lambda - 10)(\lambda - 5)$, so eigenvalues are $10$ and $5$. For $\lambda = 10$ we have $A - 10I = \left[\begin{array}{rr}-4&2\\ 2&-1\end{array}\right]$ and an eigenvector is $\left[\begin{array}{r}1\\ 2\end{array}\right]$, which scales to $\left[\begin{array}{r}\frac 1{\sqrt 5}\\ \frac 2{\sqrt 5}\end{array}\right]$ of length 1. For $\lambda = 5$ we have $A - 5 I = \left[\begin{array}{rr}1&2\\ 2&4\end{array}\right]$ so an eigenvector is $\left[\begin{array}{r}2\\ -1\end{array}\right]$, which scales to $\left[\begin{array}{r}\frac 2{\sqrt 5}\\ - \frac 1{\sqrt 5}\end{array}\right]$. However, the matrix $\left[\begin{array}{rr}\frac 1{\sqrt 5}& \frac 2{\sqrt 5}\\ \frac 2{\sqrt 5}& -\frac 1{\sqrt 5}\end{array}\right]$ has determinant $-1$, so let's negate the second column to get the answer $P = \left[\begin{array}{rr}\frac 1{\sqrt 5}& -\frac 2{\sqrt 5}\\ \frac 2{\sqrt 5}& \frac 1{\sqrt 5}\end{array}\right]$. Now $P^{-1} A P = D$ for $D = \left[\begin{array}{rr}10&0\\ 0&5\end{array}\right]$.