For which values is the matrix diagonalizable?

Question

I have the following $2\times 2$ matrix $$\left( \begin{array}{rrr} 0 & 1 \\ -1 & a \\ \end{array}\right)$$ with $a ∈ \mathbb{R}$, and then same matrix but with $a ∈ \mathbb{C}$. For which $a$ is the matrix diagonalizable?

Is there a condition i'm missing or a rule to use, because computing the eigenvalues and eingenspaces is annoying with the variable $a$.

Thanks for any type of help! :)

Answer 1

I do not fully agree with the other two answers (in the details), here is why :

A matrix is diagonalisable if its minimal polynomial is split with single roots.

The characteristic polynomial of $M=\begin{pmatrix}0&1\\-1&a\end{pmatrix}$ is

$\chi(t)=\det(M-tI)=\begin{vmatrix}0-t&1\\-1&a-t\end{vmatrix}=(-t)(a-t)-(-1)(1)=t^2-at+1$

The minimal polynomial $\mu(t)$ is a divisor of $\chi(t)$ but since we are in dimension $2$, $\mu(t)$ can only be of degree $0,1$ or $2$.

The degree is neither $0$ since $M\neq 0$ nor $1$ since $M-tI\neq 0$ for any $t$, consequently $\mu(t)=\chi(t)$.

The discriminant of $\mu(t)$ is $\Delta=a^2-4$.

• In the complex case, i.e. working in $M_2(\mathbb C)$, a polynomial is always split, so we only have to check the multiplicity of the roots.

We have a double root only in the case $a=\pm 2$ in which case $M$ is not diagonalisable.

• In the real case, i.e. working in $M_2(\mathbb R)$, a polynomial is split only when $\Delta\ge 0$, so compared to the complex case we have to exclude the interval $a\in(-2,2)$ in addition to the two values $\{-2,2\}$.

Answer 2

The only cases where it might not be diagonalizable are where there is a multiple eigenvalue, and for that to occur the discriminant of the characteristic polynomial must be $0$.

Answer 3

Let $A$ denote the desired matrix, and consider the characteristic polynomial of $A$ given by $f(t)=t^2-at+1$.

Notice that as Robert point, the matrix would be diagonalizable when the quadratic polynomial $f$ has no roots of multiplicity $2$. It is when $$\Delta=a^2-4\ne0\iff a\ne\pm2.$$

Hence the matrix is diagonalizable for all $a\in \mathbb{C}\setminus \lbrace 2,-2\rbrace $.

And we are done.

Remark.

Let $A\in M_{n\times n}(\mathbb{F})$ be an element of the ring of matrices with entries in the field $\mathbb{F}$, $I$ the identity of the ring of matrices and $t$ an indeterminate. The polynomial $$f(t)=\operatorname{det}(A-tI)$$ is called the characteristic polynomial of the matrix $A$.