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Prove or give a counterexample

There is no $A \in \Bbb R^{3 \times 3}$ such that $A^2 = -\Bbb I_3$

Here's my attempt

I suspect that the statement is true. To prove it, I used contradiction

Assuming $A^2 = -\Bbb I_3$ holds, we take the determinant on both sides to get

$$ \det A^2= (\det A)^2= (-1)^3 \det(\Bbb I_3) \Rightarrow \det A = \sqrt{-1}$$

So the determinant of $A$ is ill-defined over the real numbers, which is a contradiction (as the determinant of square matrices is well-defined over the real numbers).

Do you all agree?

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    $\begingroup$ Correct ! A bit more elegant is to state $\det(A^2)=\det(A)^2\ge 0$ and together with $\det(-I)=-1$ , we can easily rule out $A^2=-I$ $\endgroup$
    – Peter
    Jul 8 at 9:38
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    $\begingroup$ Note that $3$ can be replaced by any odd number and you arrive at the exercise given in the title. $\endgroup$
    – Peter
    Jul 8 at 9:43
  • $\begingroup$ @Peter It's a very minor point, but I actually don't find your phrasing any clearer than what JD_PM wrote originally. I think it was entirely fine to begin with. $\endgroup$ Jul 8 at 11:25
  • $\begingroup$ @JoshuaP.Swanson The point is that in the title we have an arbitary odd number $n$. In this case, showing the claim just for $n=3$ is not enough. There could be another odd number $n$ for which a solution exists. But since the proof is completely analogue for an arbitary odd number , this can easily be repaired. $\endgroup$
    – Peter
    Jul 8 at 11:29
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Here's another related argument. The characteristic polynomial of $A$ is a degree 3 polynomial with real coefficients, so it's got a root. Hence $A$ has at least one real eigenvalue $\lambda$ with eigenvector $\vec{v}$. But then $$-\vec{v} = -\mathbb{I}_3\vec{v} = A^2 \vec{v} = \lambda^2 \vec{v}$$ so $\lambda^2 = -1$, which is a contradiction.

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  • $\begingroup$ Nice alternative proof (+1) $\endgroup$
    – Peter
    Jul 8 at 9:46
  • $\begingroup$ Again , $3$ can be replaced by any odd number. $\endgroup$
    – Peter
    Jul 8 at 9:50

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