Can a matrix be not a multiple of identity, have repeated eigen values and still be diagonalizable?

Question

The question: Diagonalisability of 2×2 matrices with repeated eigenvalues suggests that if a matrix has all its eigen values distinct, it must be diagonalizable. However, any multiple of the identity matrix will have all of its eigen values the same and yet be diagonalizable. I conjecture that if a general $n \times n$ matrix has some non-zero off diagonal elements, and has any multiplicity of eigen values, it will not be diagonalizable. I haven't been able to find a counterexample to this. Can it be proven (or disproven with a counter example).

EDIT: Sorry, I was actually looking for a stochastic matrix (rows must sum to one) with these properties. I'll add another - matrix has to be full rank (so no zero eigen values). If no one answers with such an example in the next few hours, I'll accept the current one.

Answer 1

Try for example, $$ \pmatrix{1 & 1 & 1\cr 1 & 1 & 1\cr 1 & 1 & 1\cr}$$

In fact any real $n \times n$ symmetric matrix is diagonalizable, but these can have repeated eigenvalues.

EDIT: A full rank stochastic example is $$ \pmatrix{1/2 & 1/4 & 1/4\cr 1/4 & 1/2 & 1/4\cr 1/4 & 1/4 & 1/2\cr} $$

Answer 2

A simple counterexample: $$\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}.$$

Answer 3

Let $P$ be any full rank stochastic matrix (i.e. rows sum to one) which is not a multiple of the identity matrix. Then the following matrix will also not be a multiple of the identity matrix and have repeated eigenvalues $$ Q = \begin{pmatrix} P & 0 & 0\\ 0 & P & 0\\ 0 & 0 & P\\ \end{pmatrix} $$ Here, $Q$ has the matrix "$P$" replicated on its diagonal and has zeros outside the $P$-blocks.

Notice that each $P$-block describes a distinct subcomponent of the larger markov chain described by $Q$. And since the eigenvalues of each $P$-block are the same, $Q$ has distinct eigenvectors with the same eigenvalue. $Q$ is also stochastic.