Tell if matrix is diagonalizable

Question

Consider the matrix $$\begin{pmatrix} 1&1&1&1 \\ 1&2&3&4 \\ 1&-1&2&-2 \\ 0&0&1&-2 \end{pmatrix}$$ Is it diagonalizable over $\mathbb R$?

I tried to compute the eigenvalues, but I couldn't factor the characteristic polynomial. So I put it in WolframAlpha and found out it wasn't the product of linear factors. So this matrix is not even triangularizable (over $\mathbb R$). But is it possible to tell straight away that this matrix is not diagonalizable without using software or making too much calculations?

Answer 1

It is diagonalizable over ${\Bbb C}$ but not over ${\Bbb R}$. There are two real evals and two complex conjugated. But roots are not very nice. I think the easiest way to go is to compute the characteristic polynomial (but a computer helps): $p(s) = s^4 -3 s^2 +s^2 +10s -13$ and study this. For example, $p(s)$ and $p'(s)$ have no non-trivial common factor which implies that roots are distinct (so the matrix is diagonalizable over ${\Bbb C}$).

Answer 2

First find the characteristic polynomial. Then apply Sturm's theorem, which gives the number of real roots. If this is lower than the degree of the polynomial, the matrix is not diagonalizable.

The method involves some polynomial division, but you can get there without the use of software.