Prove that $A_{5 \times 5}$ is a diagonalizable if $\operatorname{rank}(I+A)=\operatorname{rank}(2I-A)=3$ and $\det A=12$

Question

Prove that $A_{5 \times 5}$ is a diagonalizable if $\operatorname{rank}(I+A)=\operatorname{rank}(2I-A)=3$ and $\det A=12$.

Because $\operatorname{rank}(I+A)=\operatorname{rank}(2I-A)<5$ then: $$ \det (I+A)=\det (2I-A)=0 $$ therefore there're two eigenvalues: $t_1=-1, t_2=2$.

We know that geometric multiplicity of $I+A$ is $n-\operatorname{rank}(I+A)=2$ likewise for $2I-A$.

Because the algebraic multiplicity is greater than or equal to the geometric and $\deg p(t)=5$, the characteristic polynomial is: $$ p(t)=(t+1)^2(t-2)^2(t-x) $$

Because $\det A=12$ this means that the free variable of $p(t)$ is $-12$.

I understand everything up until this point: because the free variable is $-12$ then: $$ 1\cdot 2^2\cdot (-x)=-12 \Rightarrow x=3 \quad (\ast) $$ How do we get the equation $(\ast)$?

Answer 1

Hint : You already know the free variable of the polynomial $p(t)$. What is the free variable of $(t+1)^2(t-2)^2(t-x)$?

Answer 2

The characteristic polynomial is $p(t)=\det(tI-A)$. Therefore the constant term in $p$ is $p(0)=\det(-A)$. As the size of $A$ is odd, $\det(-A)=-\det A=-12$.

In general, if $A$ is $n\times n$, the constant term in the characteristic polynomial of $A$ is $(-1)^n\det(A)$.

Answer 3

From the ranks you've found the eigenvalues $\lambda_1=-1,\lambda_2=2$ have geometric multiplicity of 2,since the algebraic multiplicity is greater than or equal to the geometric multiplicity you know that $\lambda_1=-1,\lambda_2=2$ have at least algebraic multiplicity of $2$. But you've also got that $detA=\lambda_1\lambda_2\lambda_3\lambda_4\lambda_5$ from there you have

$12=\lambda_1\lambda_2\lambda_3\lambda_4\lambda_5$

Where $\lambda_1=\lambda_2=-1$ and $\lambda_3=\lambda_4=2$.

$12=(-1)^22^2\lambda_5 \rightarrow \lambda_5=\frac{12}{4}=3$