Are there diagonalisable endomorphisms which are not unitarily diagonalisable? I know that normal endomorphisms are unitarily diagonalisable. Now I'm wondering, are there any diagonalisable endomorphisms which are not unitarily diagonalisable? 
If so, could you provide an example? 
 A: $$\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$$ will do it. This matrix isn't normal (so by the Spectral theorem, it can't be unitarily diagonalisable) but it is diagonalisable with eigenvalues $1 \pm \sqrt{2}$.
A: Another way to look at it, though no really different in essence, is to consider the operator norm on ${\rm M}_{n}(\mathbb{C})$ induced by the Euclidean norm on $\mathbb{C}^{n}$ (thought of as column vectors). Hence $\|A \| = {\rm max}_{ v : \|v \| = 1} \|Av \|.$ Since the unitary transformations are precisely the isometries of $\mathbb{C}^{n},$ we see that conjugation by a unitary matrix does not change the norm of a matrix. If $A$ can be diagonalized by a unitary matrix, then it is clear from this discussion that $\| A \| = {\rm max}_{\lambda} |\lambda |,$ as $\lambda$ runs over the eigenvalues of $A$. Hence as soon as we find a diagonalizable matrix $B$ with $\| B \| \neq {\rm max}_{\lambda} |\lambda|,$ we know that $B$ is not diagonalizable by a unitary matrix. For example, the matrix $$B = \left( \begin{array}{clcr} 3 & 5\\0 & 2 \end{array} \right)$$ has largest eigenvalue $3,$ but $\|B \| > 5$  because
$B \left( \begin{array}{cc} 0 \\1 \end{array} \right) = \left( \begin{array}{cc} 5 \\2 \end{array} \right).$ Also, $B$ is diagonalizable, but we now know that can't be achieved via a unitary matrix.
