I'm having problems with this diagonalisation for it seems "paradoxical" to me.
The matrix reads
$$ \begin{pmatrix} 1 & b & 0 \\ 0 & b & 0 \\ 1 & 0 & 1 \end{pmatrix} $$
If I compute a simple Gauss reduction, I obtain the matrix
$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
Which is... already diagonal?
If $b = 1$ to say, I get the identity matrix, which is symmetric and diagonalizable.
If $b \neq 0$ in general, I can always divide the second row by $b$ and what I said above applies.
Yet if $b = 0$ well it is still a symmetric matrix hence it's diagonalizable.
Am I right here, or am I missing something?
I'm asking because my initial attack wasn't with a Gauss reduction. I got the characteristic polynomial as $p(\lambda) = (\lambda-1)^2(\lambda -b)$ whence when I compute the geometric multiplicity in the case of $b\neq 0$ (since I have a double $\lambda = 1$ eigenvalue, I get that the matrix, for $b \neq 0$ and $\lambda = 1$ reads
$$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} $$
which rank is two, hence the geometric multiplicity for $\lambda = 1$ and $b\neq 0$ is $1$, hence not diagonalizable.
I am surely missing something...
Thank you!
What has Gauss algorithm to do with diagonalization ?
By doing elementary row-operations, you are actually left multiplying your matrix with elementary matrices.
This means
$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & 1 \end{pmatrix} = E_1 E_2 ... E_n \begin{pmatrix} 1 & b & 0 \\ 0 & b & 0 \\ 1 & 0 & 1 \end{pmatrix} $$ where the $E_i$ are the elementary matrices corresponding to your row operations.
If you want to diagonalize, you need to find an invertible matrix $P$ such that $P^{-1} \begin{pmatrix} 1 & b & 0 \\ 0 & b & 0 \\ 1 & 0 & 1 \end{pmatrix} P$ is diagonal, which is different.
EDIT: From the row operations you have done you can conclude here is that $0$ is not an eigenvalue except if $b=0$,since the rank is preserved by row operations.