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Let $\lambda$ be an eigenvalue of $A$. Then $\lambda^4$ is an eigenvalue of $A^4$.

Let $v$ be the corresponding eigenvector of $\lambda^4$. Then $A^4 v = \lambda^4 v \Rightarrow Iv=\lambda^4 v \Rightarrow \lambda^4=1 \Rightarrow \lambda= \pm 1,\pm i$.

Since $A$ is a real matrix, if $i$ (or $-i$) is it's eigenvalue, then $-i$ (respectively $i$) is also it's eigenvalue. Also $A \neq \pm I$.

Therefore possible characteristic polynomials of $A$ are $(x-1)(x+1)^2,(x+1)(x-1)^2,(x-1)(x^2+1),(x+1)(x^2+1)$.

Now if possible, $x^2+1$ will be a minimal polynomial of $A$ for the characteristic polynomials $(x-1)(x^2+1),(x+1)(x^2+1)$ only. But any such minimal polynomial must have factors $(x-1)$ and $(x+1)$ respectively.

Hence $A^2+ I=0$ is not possible.

Are there any mistakes in my proof? I am learning to handle proofs containing characteristic polynomials, minimal polynomials, eigenvalues etc. Thanks.

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Your reasoning is correct. I just want to point out a possible generalization to your problem, using the core part of your reasoning. If $p$ is a polynomial without real roots, and $A$ is a real matrix with odd dimension, then $p(A) = 0 $ never holds.

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if you want to figure it out by yourself with the proof that you presented

The characteristic polynomial of $A$ has odd degree, so has a real root $\lambda $ and a respective eighenvector $v$. Then $ 0 = p(A) v = p(\lambda ) v \Rightarrow p(\lambda ) = 0 $, which is impossible.

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  • $\begingroup$ you mean in this case $p(\lambda)=\lambda^2+1$ , of course ? Anyway you have presented interesting theorem, it could be added as the answer also to my unanswered question math.stackexchange.com/questions/2317086/…, I suppose.. $\endgroup$ – Widawensen Jun 21 '17 at 9:49
  • $\begingroup$ The last sentence I would present in a little extended form (which I present here mainly for myself). .. because $\lambda$ is the real eigenvalue for $A$ with the eigenvector $v$ so $p(\lambda)$ is the eigenvalue for $p(A)$ with the same eigenvector $v$ so we have $ p(A)v=p(\lambda)v$ . In this case $p(A)=A^2+I$, but if $ A^2+I=0 $ then consequently $p(\lambda)v=0$ and it should follow $\lambda^2+1=0$. However it is not possible. $\endgroup$ – Widawensen Jun 21 '17 at 10:48
  • $\begingroup$ Thanks for the feedback and generalizing the problem! $\endgroup$ – Error 404 Jun 21 '17 at 12:10
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In this case indeed it's impossible that $A^2=-I$ because $\det(A^2) \geq 0$ but $\det(-I_{3 \times 3})=-1$.

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  • $\begingroup$ Anyway the final OP conclusion seems to be O.K. $\endgroup$ – Widawensen Jun 21 '17 at 8:58
  • $\begingroup$ I didn't think this way. You have very much short and elegant proof than mine. But I want a feedback about my proof. This is because I want to learn to handle proofs having char polynomial, minimal polynomial, eigenvalues etc. Thanks! +1 $\endgroup$ – Error 404 Jun 21 '17 at 9:12
  • $\begingroup$ @Mathmore I see, probably someone more experienced in Linear Algebra can describe your procedure in more detailed way. There are different ways which lead to the same result, I wonder sometimes whether the linear algebra is the most noticeable part of mathematics where it is visible that results can come from very different directions.. $\endgroup$ – Widawensen Jun 21 '17 at 9:24
  • $\begingroup$ Truly. This is why I love Maths. Different ways to arrive at same answer. :) $\endgroup$ – Error 404 Jun 21 '17 at 12:07
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Your reasoning is correct.

If you need a counterexample, here is the simplest possible. Set $$ A=\left(\begin{array}{rrr}0&1&0 \\ -1&0&0\\ 0&0&1\end{array}\right). $$ Then $A^4=I$, but $$ A^2+I=\left(\begin{array}{rrr}0&0&0 \\ 0&0&0\\ 0&0&2\end{array}\right). $$

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  • $\begingroup$ Thanks for the feedback and the counterexample! $\endgroup$ – Error 404 Jun 23 '17 at 3:53

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