True or false? If the triangular matrix \(A\) is similar to a diagonal matrix, then \(A\) is already diagonal.
Solution
The assertion is \(\boxed{\text{false}}\).
1. A concrete counterexample is\[ A=\begin{pmatrix}1&1\\0&2\end{pmatrix}. \]
The matrix is upper triangular but is not diagonal because its \((1,2)\) entry is \(1\). Its two eigenvalues are the distinct diagonal entries \(1\) and \(2\). Corresponding eigenvectors are \[ v_1=\begin{pmatrix}1\\0\end{pmatrix}, \qquad v_2=\begin{pmatrix}1\\1\end{pmatrix}, \] because \(Av_1=v_1\) and \(Av_2=2v_2\).
Let \[ P=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \] Its columns are \(v_1,v_2\), so \(P\) is invertible, and the two eigenvector identities give \[ AP=P\begin{pmatrix}1&0\\0&2\end{pmatrix}. \] Multiplying by \(P^{-1}\) on the left yields \[ P^{-1}AP=\begin{pmatrix}1&0\\0&2\end{pmatrix}. \] Thus \(A\) is similar to a diagonal matrix although \(A\) itself is not diagonal, which disproves the assertion [@HK71].