# How can I prove that this matrix is diagonalizable?

How can I prove that for all $a \in \mathbb{R}$ the following matrix is diagonalizable? I computed the characteristic polynomial, but I couldn't decompose it into linear factors. This is the polynomial: $t^2-3t+2-a^2$ What should I do next?

Here is the matrix: \begin{bmatrix} 2 & a \\ a & 1 \end{bmatrix}

• Well, what do you usually do in that case? Sep 22, 2016 at 19:57
• I cry. I try to compute the eigenspace but that is not helping me much atm
– user355196
Sep 22, 2016 at 20:01
• You have a symmetric matrix. Does that mean anything? Sep 22, 2016 at 20:03
• What happens for $a=\sqrt 2$?
– N74
Sep 22, 2016 at 20:18

Find the zeors of the polynomial. If there are two of them, you are done else you have to take a closer look at the eigenspace and check if it is two-dimensional.

• But how do I found the zeros of this polynomial: t^2-3t+2-a^2=0. i found for a=0 that t=1 and 2. but i need to show that for all a this is okay. And now i showed it for a=0
– user355196
Sep 22, 2016 at 19:59
• @esma2708, you surely know a formula to find the roots of a degree two polynomial! Sep 22, 2016 at 20:00
• Holy moly... the abc-formula haahha thaaank you
– user355196
Sep 22, 2016 at 20:02

The discriminant of the polynomial is $$3^2-4(2-a^2)=1+4a^2$$ For all real values of $a$, the discriminant is positive. What does this tell you about the roots of the polynomial?

What is a sufficient criterion for diagonalizability? (In other words, you don't need to compute the eigenspaces, in this case.)

• I found 2 values of t. So it is diagonalizable, because we have two eigenvalues?
– user355196
Sep 22, 2016 at 20:22
• @esma2708 Yes, two distinct real eigenvalues make a $2\times 2$ matrix diagonalizable. Sep 22, 2016 at 21:02