For Problem R-1:

A $90 ^ \circ$ rotation, as a transformation on $\mathbb{R}^2\rightarrow \mathbb{R}^2$, has no eigenvectors. (Later we'll see that it does have eigenvectors if complex numbers are allowed.)

A reflection in $\mathbb{R}^2$ has two eigenspaces (line of eigenvectors), one along the mirror line and one perpendicular to the mirror line.

A shear has just one eigenspace, and it doesn't help to allow complex numbers.

For Problem R-2:

(a) $P(D) = \left[\begin{array}{cc}p(1)&0\\ 0&p(2)\end{array}\right]$, so we want $p(1) = 0$ and $p(2) = 0$. Therefore we should use $p(x) = (x-1)(x-2) = x^2 - 3x + 2$.

(b) For the same reason, let $p(x) = (x-1)(x-2)(x-3) = x^3 - 6 x^2 + 11x - 6$.

For Problem R-3:

To review: Writing $\epsilon$ for an elementary matrix, when we do successive elementary row operations starting from $A$ we are making $\epsilon _ 1 A$, $\epsilon _ 2 \epsilon _ 1 A$, etc. So if $B$ is the end result after $k$ elementary row operations, then $B = MA$, where $M = \epsilon _ k \dots \epsilon _ 1$. We know $M$ is invertible since $M$ can be undone by the inverses of the elementary row operations: $M^{-1} = \epsilon _ 1 ^ {-1} \dots \epsilon _ k ^ {-1}$.

(a) If $A$ is row-reduced to $B$ then $MA = B$ as just explained. The isomorphism is simply $\tau _ M$, since $\tau _ M$ is invertible (so is an isomorphism) and takes each column of $A$ to the corresponding column of $B$.

One of you suggested that this point of view is another good way to see why linear relations among columns are preserved by row reduction even though the column space is changed: The column space is changed isomorphically!

(b) Since the isomorphism in (a) takes columns of $A$ to corresponding columns of $B$, a selection of columns that makes a basis for the column space of $A$ corresponds to the same thing for $B$.