Is there a faster way to diagonalize this matrix?

Question

I'm asked to diagonalize this matrix for homework: $$\left[\begin{matrix}3&0&0&0\\ 0&2&0&0\\ 0&0&2&0\\1&0&0&3 \end{matrix}\right]$$ But since it's already almost diagonalized except for that $1$ entry, is it legit to say that we could row reduce and still have the same matrix except the 1 entry is gone? Because by simply taking the $4th$ row and subtracting $3$ times the first row we get the same exact matrix less the $1$ entry. But I'm not sure if that's a "legal" mathematical action. I wouldn't think so but I want to ask before I spend a bunch of time finding eigenvalues and determinants.

Answer 1

You're out of luck. This matrix is not diagonalizable.

Answer 2

Hint: After rearranging, it suffices to consider diagonalizing $\pmatrix{3&0\\1&3}$, which is not possible.

Answer 3

Since this matrix is triangular, (which represent an endomorphism $f$), then its eigenvalues are the entries on its diagonal i.e. $3,2,2,3$ and if we denote by $\mathcal{B}=(e_1,e_2,e_3,e_4)$ the standard basis then $e_2$, $e_3$ and $e_4$ are eigenvectors associated to the eigenvalues $2,2$ and $3$ respectively.

Now if this matrix is diagonalisable then there's an eigenvector $$v=a_1e_1+a_2e_2+a_3e_3+a_4e_4$$ such that $f(v)=3v$ and we can verify easly that this equality is impossible, hence this matrix isn't diagonalisable.