Let $A \in \mathbb R^{5 \times 5}$. Prove that $A^{4} + I \neq O$

Question

Let $A \in \mathbb R^{5 \times 5}$. Prove that $$A^{4} + I \neq O$$

I understand that I have to start from $A^{4}+I= 0$ and come up with a contradiction. Is it the Cayley Hamilton I have to use?

Answer 1

No, you need not to use Cayley–Hamilton theorem. Note that $-I_5=A^4$ implies that $$(-1)^5=\det(-I_5)=\det(A^4) = \det(A)^4.$$ Now use the fact that $A$ has real coefficients. Can you take it from here?

Answer 2

Since $5$ is odd, $A$ has some real eigenvalue $\lambda$. Therefore $\lambda^4+1$ is an eigenvalue of $A^4+\operatorname{Id}$. Since $\lambda^4+1\neq0$, $A^4+\operatorname{Id}\neq0$.

So, no, you don't need Cayley-Hamilton.

Answer 3

Since it is real it characteristic polynomial has at least one real egienvalue since it dgree is odd. But if $t$ is it egienvalue, then $$Av = tv\implies A^4v = -v =t^4v$$

So we have $t^4+1=0$ which is not true for real values.

Answer 4

Since eigenvalues are zeros of a real polynomial the complex ones occur in conjugate pairs. Hence in a $5 \times 5$ real matrix there must be at least one real eigenvalue. If $\lambda$ is real, $x \neq 0$ and $Ax =\lambda x$ then $A^{4}x =(\lambda )^{4}x$. So $A^{4}=-I$ would lead to a contradiction, right?.