Under what conditions is the following matrix diagonalisable Let 
$$A=\begin{bmatrix} 0&1/2&1/2 \\ 0&0&1/2 \\ 1&1/2&0 \end{bmatrix}.$$
Could anyone advise me on how to show $M=(1-m)A+mS,$ where $S_{ij}=\frac{1}{3},$ is not diagonalisable for $0\leq m < 1 \ ?$ Do we compute the characteristic polynomial of $M$, show it has $2$ distinct eigenvalues and the dimension of eigenspace corresponding to each distinct eigenvalue is $1 \ ?$
 A: The matrix $M$ is
$$M=\frac{1}{6}\left(
\begin{array}{ccc}
 2 m & 3-m & 3-m \\
 2 m & 2 m & 3-m \\
 6-4 m & 3-m & 2 m \\
\end{array}
\right).$$
Now compute the characteristic polynomial
$$f(x)=\frac14(m-1-2x)^2 (x-1).$$
This shows that $M$ has eigenvalues $1$ and $\frac{m-1}{2}$. Now check that $\dim\ker\left(M-\left(\frac{m-1}{2}\right)I\right)=1$, for example by row reducing the matrix $M-\left(\frac{m-1}{2}\right)I$ (here $I$ is the $3\times 3$ identity matrix). This shows that $\frac{m-1}{2}$ corresponds to a $2\times 2$ Jordan block.
I did this with a computer, so this sort of obscured the part where you use $m\neq 1$. If you look at the Jordan normal form for $M$, you see that it is similar to
$$\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & \frac{m-1}{2} & 1 \\
 0 & 0 & \frac{m-1}{2} \\
\end{array}
\right)$$
via conjugation by
$$\left(
\begin{array}{ccc}
 \frac{m-3}{2 (m-2)} & 0 & -\frac{2}{m-1} \\
 \frac{1}{2-m} & -1 & \frac{2}{m-1} \\
 1 & 1 & 0 \\
\end{array}
\right).$$
Of course, this matrix assumes $m\neq 1$. If $m=1$, then in fact we get the Jordan normal form
$$\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right),$$
and so $M$ is diagonalizable if and only if $m=1$.
