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..
Best Answer
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 $$ \det(A) = \det(A - 0I) = (0^2 + 1)^2 = 1 \neq 0 $$