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..

  • $\begingroup$ Check I had edited. Is now am I correct? and it is not diagonalizable over R or over C? and what about its invertiblity? $\endgroup$ – Akash Patalwanshi Oct 1 '17 at 15:33

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

  • $\begingroup$ As both the eigenvalues are complex so is invertible over $\mathbb{R}$ ? $\endgroup$ – Akash Patalwanshi Oct 1 '17 at 16:07
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    $\begingroup$ I have no idea where you're coming from with that comment. We can say that none of the eigenvalues are zero, so the matrix is invertible over any field containing the rationals. More to the point is the simple observation that the characteristic polynomial is defined as $\det(A - xI)$. $\endgroup$ – Omnomnomnom Oct 1 '17 at 16:11
  • $\begingroup$ Sir, I mean to say, as none of the eigenvalues are real then how can be matrix is invertible over real? $\endgroup$ – Akash Patalwanshi Oct 1 '17 at 16:13
  • $\begingroup$ I don't see the connection between those two ideas. Why should one thing be connected to the other? $\endgroup$ – Omnomnomnom Oct 1 '17 at 16:14
  • $\begingroup$ I had a typo in my earlier comment, see the latest version. $\endgroup$ – Omnomnomnom Oct 1 '17 at 16:15

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?

  • $\begingroup$ Sir, So can we say $A$ is diagonalizable Over $\mathbb{C}$? and from your reply how can we say $A$ is invertible or not? $det(A) ≠0$ but can we say it is invertible over $\mathbb{R}$? $\endgroup$ – Akash Patalwanshi Oct 1 '17 at 15:37

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