4.3.10

Consider the matrices $A = \begin{bmatrix} a & 0 \\ 0 & d \\ \end{bmatrix}$ and $B = \begin{bmatrix} a & b \\ 0 & d \\ \end{bmatrix}$, where $b\neq 0$. We will show that A is similar to B if and only if $a \ne d$.

Suppose that $a \ne d$. Then, we have two linearly independent eigenvectors $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$ and $\begin{bmatrix} \frac{b}{d-a} \\ 1 \\ \end{bmatrix}$ of B, with distinct eigenvalues a and d respectively.Letting S = $\begin{bmatrix} 1 & \frac{b}{d-a} \\ 0 & 1 \\ \end{bmatrix}$ be the matrix of eigenvectors, S is invertible, and we can compute directly that $SBS^{-1} = A$, thus A and B are similar.

Now suppose that $a = d$. Then $A = aI$, where I is the 2x2 identity matrix. Suppose for contradiction that A was similar to B, then there exists an invertible S such that $SAS^{-1} = B$, but since $A = aI$, we have that $SAS^{-1} = a*(SIS^{-1}) = aI = B$, so that b = 0, a contradiction. Thus if a = d, A and B are not similar, and hence A and B are similar if and only if $a \ne d$.