for which a, the matrix A is diagonalizable?

Question

A = $ \begin{pmatrix} 2a+3 & 0 & 0 \\ -a-3 & a & a+3 \\ a & a & a+3 \\ \end{pmatrix} $

Characteristic polynomial: $ \begin{vmatrix} t-(2a+3) & 0 & 0 \\ a+3 & t-a & -a-3 \\ -a & -a & t-(a+3) \\ \end{vmatrix} $

Can anyone solve this and show me for which a, the matrix A is diagonalizable?

I tried already for about 30minutes and no success

Answer 1

Call the characteristic polynomial $p(t)$. If you expand the determinant along the first row we find: $$p(t) = (t-(2a+3)) [ (t-a)(t-(a+3)) - (a+3)(a) ]$$

The polynomial inside the braces is quadratic, so you can factor it using at the quadratic equation if necessary. If for all $a$ your roots are distict, then this matrix is always diagonalizable. Then for the cases there they are not, you need to take a closer look.

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In the comments it was found that $p(t) = t(t-(2a+3))^2$. We now need to determine if there are two eigenvectors corresponding to $2a+3$, since we know there is one eigenvector corresponding to the zero eigenvalue. This ammounts to finding all vectors $(x,y,z)^T$ satisfying: $$\left( \begin{array}{ccc} 0& 0& 0\\ a+3& a+3& -(a+3)\\ -a& -a& a\end{array}\right) \left( \begin{array}{c}x\\y\\c\end{array}\right) = \left( \begin{array}{c}0\\0\\0\end{array}\right)$$

(The matrix was obtained by setting $t=(2a+3)$)

By inspection we see that $(1,0,1)^T$ and $(0,1,1)^T$, since if you add the first and third rows you get the zero vector and adding the second and third row gives the same.

Answer 2

$$p(t)=(t-(2a+3))\Big((t-a)(t-(a+3))-a(a+3)\Big)=t(t-(2a+3))(t-(2a+3))$$ For $a\neq-\frac{3}{2}$ eigenvalues are $0$ and $2a+3$ with algebraic multiplicity respectively $1$ and $2$. You need to check the geometric multiplicity of $2a+3$ : $$\dim\ker \begin{pmatrix} 0 & 0 & 0 \\ a+3 & a+3 & -a-3 \\ -a & -a & a \\ \end{pmatrix}$$ which is $2$ (and thus coincide with the alg. mult.) for any $a\neq-\frac{3}{2}$: so for all these values of $a$ $A$ is diagonalizable.

For $a=-\frac{3}{2}$ you have as unique eigenvalue 0 with alg. mult. $3$ and geom. mult. $2$: thus $A$ is not diagonalizable in this case.

Answer 3

So I'll give you the answer and let you work through it. It's easy to see (with a computer) that the eigenvalues and vectors are

$$ \lambda_1 = 0 , v_1 = ( 0, -3/a -1 , 1) $$ $$ \lambda_2 = 2a+3 , v_2 = ( 1,0,1 )$$ $$ \lambda_3 = 2a+3 , v_3 = (-1,1 ,0) $$

So we're only going to have issues if $a = -3/2$. Why? Since this means we have 1 eigenvalue of $0$ with multiplicity 3. So we check the eigenbasis, is it complete?

$$ \dim \left \{ v_1,v_2,v_3 \right \} = 3 ????? $$

Well, clearly $v_2 + v_3 = v_1$ so we only have a dimension of 2, thus it is not diagonalizalbe for that value of $a$.