# Diagonalizability of a certain $4\times4$ matrix

Question $\bf 3.$ Determine if the following matrix is diagonalizable. (explain your answer) $$A=\pmatrix{ 1 & 4 & -2 & 3 \\ 3 & -3 & 0 & 4 \\ 1 & 1 & 1 & -1 \\ 0 & 5 & -5 & 2 \\ }$$

The problem is, it's a $4\times 4$ matrix and I don't want to find the characteristic polynomial. I've been trying to find a trick but can't see anything. I could probably check if det $A = 0$ then that would tell me $0$ is an eigenvalue, but that's still not enough.

Well, I gave up and did it online.

So $4$ distinct eigenvalues $\implies$ diagonalizable... but if there's another way let me know.

• As your link shows $\det A\ne0$. – user63181 Aug 5 '14 at 17:55
• You could perform row reduction then column reduction to see if the diagonal matrix you're left with has distinct eigenvalues. – Robert Wolfe Aug 5 '14 at 18:56

What you can do is: Compute the characteristic polynomial, here: $$\chi_A(t) = 126+77 \lambda -42 \lambda ^2-\lambda ^3+\lambda ^4$$ and its derivative $$\chi_A'(\lambda) = 77 - 84\lambda - 3\lambda^2 + 4\lambda^3$$ Now compute ${\rm gcd}(\chi_A, \chi_A')$, if it is $1$ (as here), the polynomial has distinct zeros (and hence $A$ distinct eigenvalues), and is therefore diagonizable (over $\mathbb C$).
$A$ is diagonalizable if and only if $A$ satisfy one of the below polynomials $$(t-a)=0$$ or $$(t-a)(t-b)=0$$ or $$(t-a)(t-b)(t-c)=0$$ or $$(t-a)(t-b)(t-c)(t-d)=0$$ where respectively sets $\{a\}$or $\{a,b\}$or $\{a,b,c\}$or $\{a,b,c,d\}$ are set of all eigenvalues of $A$ in every case.
• This is exactly how the asker reached the conclusion that $A$ is diagonalizable. The question, however, is whether there exists some other way of reaching this conclusion. – Omnomnomnom Aug 5 '14 at 18:10