A question regarding matrices and minimal polynomials.

Question

I am trying to answer the following question:

Let $A$ be a $3\times 3$ real matrix and $A^4=I$ and $A\neq \pm I$,then does it imply $A^2+I=O$?

Attempt: Since $A$ is $3\times 3$ real matrix,it has at least one eigenvalue $c\in \mathbb R$.Then $(x-c)|m(x)$,where $m$ is the minimal polynomial of $A$.Now,$A^2+I=O$ implies $x^2+1$ annihilates $A$ and hence $m(x)| x^2+1$ but then $(x-c)$ must be a factor of $(x^2+1)$ and hence $c$ must be a root of $x^2+1$ which is not possible as $c\in \mathbb R$.

Is this solution alright?

Answer 1

Our OP Kishalay Sarkar's demonstration looks fine to me; here's a somewhat similar and perhaps shorter argument:

Note that

$A^2 + I = 0 \tag 1$

implies that any eigevalue $\mu$ of $A$ satisfies

$\mu^2 + 1 = 0, \tag 2$

since

$Av = \mu v, \tag 3$

where

$v \ne 0 \tag 4$

is some $3$-vector, yields

$(\mu^2 + 1)v = \mu(\mu v) + v = \mu Av + v = A(\mu v) + v$ $= A(Av) + v = A^2v + Iv = (A^2 + I)v = 0, \tag 5$

which in light of (4) implies

$\mu^2 + 1 = 0; \tag 6$

but any $3 \times 3$ real matrix has at least one real eigenvalue, for its characteristic polynomial is a real cubic; thus (2) is impossible for some $\mu$ obeying (3); this contradiction in turn implies that (1) cannot bind.

Nota Bene: Observe the condition

$A^4 + I = 0 \tag 7$

is not needed in the above proof, nor is it used by Kishalay Sarkar. Similar remarks apply to the hypothesis

$A \ne \pm I. \tag 8$

Indeed, the essential assertion here is perhaps the more comprehensive:

Let $A$ be a real $n \times n$ matrix, where $n \in \Bbb N$ is odd; then

$A^2 + I \ne 0. \tag 9$

The proof may easily be generalized from what has been done above. End of Note.

Answer 2

All eigenvalues of $A$ are roots of $x^4-1$, so possible eigenvalues are $1,-1, i, -i$ and each has multiplicity 1 in the minimal polynomial. Since the minimal polynomial divides $x^4-1$, has real coefficients, the minimal polynomial either has degree 2 and roots $\pm 1$, or degree 2 and roots $\pm 1$ or degree $3$ and end equal to the characteristic polynomial. In the second and third cases the characteristic polynomial would be equal $(x\pm 1)(x^2+1)$. So it has 3 different roots. Hence the matrix $A$ is similar to the diagonal matrix $diag(1,i,-i)$ or $diag(-1,i,-i)$. In neither case it satisfies $A^2+I=0$. Concrete example $$A=\left(\begin{array}{lll}1&0&0\\0 & 0 & 1\\ 0 &1 &0\end{array}\right).$$