Prove that $\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$ and $\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$ are similar iff $a \neq d$

Question

Prove that $\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$ and $\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$ are similar iff $a \neq d$.

The problem with this is that most properties that I've tried i.e. rank, determinant are the same if $a=d$. Showing this algebraically gets really ugly so this is not a good idea.

Answer 1

Let $t=\frac{b}{d-a}$. It's easy to check that $$ \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} \begin{bmatrix} 1 & -t \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}. $$ The converse follows from the fact that a scalar matrix commutes with any other matrix.