For which $a$ is a matrix $A$ diagonalizable?

Question

Say I have a matrix $A_a$ with

$$A_a:= \left(\begin{array}{c} 2 & a+1 & 0 \\ -a & -3a & -a \\ a & 3a+2 & a+2 \end{array}\right)$$

I was wondering if there was an easy way to determine for which $a$ the matrix would be diagonalizable.

I tried to determine its eigenvalues first.
$det(A_a - \lambda I_3) = 0$ gave me a rather complex formula, from where on I don't know how to continue.

$0 = 7a \lambda - 6a - 4 \lambda + 4 \lambda^2 - 2a\lambda^2 - \lambda^3 - a^2\lambda + 2a^2$

Answer 1

Given:

$$A_a:= \left(\begin{array}{c} 2 & a+1 & 0 \\ -a & -3a & -a \\ a & 3a+2 & a+2 \end{array}\right)$$

We find the eigenvalues using $|A_a - \lambda I| = 0$, yielding:

$$-a^2 \lambda +2 a^2-2 a \lambda ^2+7 a \lambda -6 a-\lambda ^3+4 \lambda ^2-4 \lambda = 0 \implies (\lambda-2)(\lambda - (1-a-\sqrt{a+1}))(\lambda - (1-a+\sqrt{a+1}))=0$$

This gives us the three eigenvalues:

$$\lambda_{1,2,3} = 2,-a-\sqrt{a+1}+1,-a+\sqrt{a+1}+1$$

Hint: try the cases $a = 0$ and $a = -1$ (one is good and the other is not)! Do you see why those two values were chosen? Think algebraic multiplicity of the eigenvalues.

Generally, the eigenvectors are found by $[A-\lambda_i I]v_i = 0$, so for $\lambda_1 = 2$, the RREF is:

$$\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)v_1=0$$

This gives $v_1 = (-1,0,1)$.

For $\lambda_2 = -a-\sqrt{a+1}+1$, the RREF is:

$$\left( \begin{array}{ccc} 1 & 0 & \frac{-a+\sqrt{a+1}-1}{a} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{array} \right)v_2=0$$

This gives $v_2 = \left(-\dfrac{-a+\sqrt{a+1}-1}{a},-1,1\right)$.

We can immediately write (due to conjugates) $v_3 = \left(-\dfrac{-a-\sqrt{a+1}-1}{a},-1,1\right)$.

Answer 2

If you factor the characteristic polynomial, you find it is $$ (2a^2-6a)-(a^2-7a+4)\lambda+(4-2a)\lambda^2-\lambda^3= (2-\lambda)(\lambda^2+(2a-2)\lambda+a^2-3a) $$ If you know this polynomial has three distinct roots, then the matrix is diagonalizable. A root can be repeated only if either

• the discriminant of $f(\lambda)=\lambda^2+(2a-2)\lambda+a^2-3a$ is zero, or
• $2$ is also a root of $f(\lambda)$, that is $f(2)=0$.

The first case gives $$ 4(a-1)^2-4(a^2-3a)=0 $$ that is $$ a^2-2a+1-a^2+3a=0 $$ or $$ a=-1 $$ The second case gives $$ 4+2(2a-2)+a^2-3a=0 $$ or $$ 4+4a-4+a^2-3a=0 $$ that is, $$ a^2+a=0 $$ which has the roots $0$ and $-1$.

So, when $a\ne0$ and $a\ne-1$, the matrix is diagonalizable.

For $a=-1$ the characteristic polynomial is $(2-\lambda)^3$ and discussing the eigenspace is easy.

For $a=0$ the eigenvalues are $2$ (double) and $0$ single, so you have to discuss the eigenspace relative to $2$.