How can I tell if a matrix is diagonalizable knowing only the trace, one eigenvalue, and a result of the characteristic polynomial?

Question

Given $A$, a $3 \times 3$ matrix, and:

$\mathrm{tr}(A) = −2$,

$\mathrm{rk}(A−2I)< 3$,

$\chi_A(1) = −8$

($\mathrm{tr}$: trace, $\mathrm{rk}$: rank, and $\chi_A(x)$: characteristic polynomial)

How can I tell if the matrix is diagonalizable? What are the eigenvalues? I know that given $\mathrm{rk}(A−2I)< 3$, $2$ must an eigenvalue with multiplicity $1$, at least. But I don't know what information does "$\chi_A(1)=−8$" provide me.

Answer 1

You have two missing eigenvalues, other than the known eigenvalue $2$. The first equation tells you that their sum is $-4$. The last equation gives you another equation: since $\chi_A(t) = (t-2)(t-\lambda)(t-\mu)$, plugging in $t=1$ allows you to solve for $\lambda$ and $\mu$.

Answer 2

The characteristic polynomial is $(X-2)(X^2+bX+c)$. The trace is the coefficient of $X^2$ that is:$b-2=-2$.

The last equation enables you to find $c$.

Answer 3

We have that

• $\lambda_1+\lambda_2+\lambda_3=-2$

then we can exclude

• $\lambda_1=\lambda_2=\lambda_3=2$

suppose

• $\lambda_1=\lambda_2=2\implies \lambda_3=-6$

then

• $\chi_A(\lambda)=(\lambda-2)^2(\lambda+6)\implies \chi_A(1)=7 $

therefore $\lambda_1\neq \lambda_2 \neq \lambda_3$.