Is the matrix with this characteristic polynomial is diagonalizable?

Question

Question: given that $A$ is $4×4$ matrix with characteristic polynomial $(x^2 + 1)^2$ then is $A$ is diagonalizable and invertible?

My attempt: since we can write $p(x) = (x^2+1)^2 = x^4 +2x^2 +1$ then writing companion matrix of above polynomial, we have

$C(p)=\begin{bmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&-2\\0&0&1&0\end{bmatrix}$ then as we know minimal and characteristic polynomial of matrix $C(p)$ are same and is $(x^2+ 1)^2$.

$→$ minimal polynomials of $C(p)$ does not splits into distinct linear factors over $\mathbb{C}$. Hence $C(p)$ is not diagonalizable over $\mathbb{C}$.But $C(p)$ is one of the matrix corresponding to given characteristic polynomial.

Hence $A$ need not be diagonalizable over $\mathbb{C}$ and over $\mathbb{R}$ too.

Further as $0$ is not an eigenvalue of $A$ hence it is invertible.

Is am I correct? Further is $A$ is invertible over $\mathbb{C}$ or over $\mathbb{R}$ or over both? Please help me..

Answer 1

Indeed, because the minimal polynomial can't be factored into distinct linear factors, we can conclude that the matrix is not diagonalizable.

It's clear that $A$ is invertible. From the definition of the characteristic polynomial, we have $$ \det(A) = \det(A - 0I) = (0^2 + 1)^2 = 1 \neq 0 $$

Answer 2

By Cayley-Hamilton, $(A^2+I)^2=O$, that is $A^4+2A^2+I=O$, or $-A^4-2A^2=I$. The question remains, is $A$ invertible?