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 \ne 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 $$a \ne d $$.