Let $A$ be a $3\times 3$ matrix with characteristic polynomial $x^3-3x+a$, for what values of $a$ given matrix must be diagonalizable.

Question

Let $A$ be a $3\times 3$ matrix with characteristic polynomial $x^3-3x+a$. For what values of $a$ given matrix must be diagonalizable.

I am talking about diagonalizability over reals. Efforts:

If a matrix has distinct eigenvalues, then matrix is diagonalizable. But how can I use this information.

I also know that if char. polynomial is $x^3+a_2x^2+a_1x+a_0$, then $-a_2=\mbox{trace}(A) \mbox{ and } -a_0=\det(A).$

I am not able to proceed.

Can someone please give a hint?

Thanks.

Answer 1

Hint: compute the discriminant of the characteristic polynomial.

Answer 2

Let $p(x)=x^3-3x+a$. If $p(x)$ has a multiple root, then that root will also be a root of $p'(x)$. But the roots of $p'(x)$ are $\pm1$. So,\begin{align}p(x)\text{ has a multiple root}&\iff\pm1\text{ is a root of }p(x)\\&\iff a=\pm2.\end{align}So, check only these two cases.

Answer 3

Hint: A cubic has three distinct real roots iff its discriminant is positive. The discriminant of $x^3-3x+a$ is $\Delta=-27 (a^2 - 4)$.