Use the result of Exercise 11 to prove the following: If \(A\) is a \(2 \times 2\) matrix with complex entries, then \(A\) is similar over \(C\) to a matrix of one of the two types \[ \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \qquad \begin{bmatrix} a & 0 \\ 1 & a \end{bmatrix}. \]
Solution
The characteristic polynomial of \(A\) splits over \(\mathbb C\). If it has two distinct roots \(a,b\), eigenvectors for those roots form a basis and \(A\) is similar to \(\operatorname{diag}(a,b)\).
Otherwise the characteristic polynomial is \((x-a)^2\). Put \(N=A-aI\). Then the characteristic polynomial of \(N\) is \(x^2\), so Cayley–Hamilton gives \(N^2=0\). If \(N=0\), then \(A=aI\). If \(N\ne0\), Exercise 11 gives a basis in which \[ N=\begin{bmatrix}0&0\\1&0\end{bmatrix}, \] and hence \[ A=aI+N=\begin{bmatrix}a&0\\1&a\end{bmatrix}. \] These are exactly the two stated types.