I thought this one was pretty neat, but I haven't been able to solve it, here it goes:
Let $A$ be a linear operator on a four dimensional complex vector space that satisfies $P(A) = A^4 + 2A^3 - 2A - I = 0.$ Given that $\dim(\text{Im}\,(A+I)) = 2,$ and $|\text{Tr}\,(A)| = 2,$ find the Jordan form of $A.$
Here's what I've been thinking. First, $P(A)$ must be the characteristic polynomial of $A$ because it is a 4-th order monic polynomial with $A$ as a root. If that's the case, we see that $Tr(A) = -2$ because of the $A^3$ coefficient in $P(A).$ Furthermore, we see that $\det(A) = 1,$ due to the coefficient of the $I$ term in $P(A).$ Thus, I conclude that $A$ is invertible. Since $A$ is invertible, it may be diagonalized to the form \begin{equation} P^{-1}AP = \begin{pmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 & \lambda_2 & 0 & 0 \\ 0 & 0 & \lambda_3 & 0 \\ 0 & 0 & 0 & \lambda_4 \end{pmatrix} \end{equation} Given that $rk(A+I) = 2,$ I conclude that 2 eigenvalues (say $\lambda_1$ and $\lambda_2$) are $-1.$ I check that $P(-1) = 0$ and it works out.
Then I have written that $Tr(A) = \sum \lambda_i = -2 + \lambda_3 + \lambda_4 = -2,$ and $det(A) = \Pi \lambda_i = \lambda_3 \lambda_4 = 1,$ from the trace and determinant properties. This leads to the conclusion that $\lambda_3 = i = -\lambda_4.$ However, $P(\pm i) \neq 0,$ so this must be wrong (I think).
If I were correct, and $A$ is diagonalizable, wouldn't the Jordan form just be a diagonal matrix? Can anyone find where I made a mistake in my logic, or if someone sees a more direct way?
Since $p(t)=(t-1)(t+1)^3$, you know that the eigenvalues are $1$ and $-1$. Since the algebraic multiplicity of $1$ is one, that would correspond to a $1\times1$ block in the Jordan form. The question is what happens with $-1$.
The fact that $\dim(\text{Im}\,(A+I))=2$ tells you that the geometric multiplicity of $-1$ is two, since $\dim\ker(A+I))=2$ (via the rank-nullity theorem). So $-1$ has two Jordan blocks. As these blocks correspond to three rows and columns in $A$ has has size one and the other size two. So the Jordan form of $A$ is $$ \begin{bmatrix} 1&0&0&0\\0&-1&0&0\\ 0&0&-1&1\\ 0&0&0&-1 \end{bmatrix} $$