# Proving there is no squared $n\times n$ (n odd) real matrix yielding minus the identity matrix

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?

• 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 $A^2=-I$ Jul 8, 2021 at 9:38
• Note that $3$ can be replaced by any odd number and you arrive at the exercise given in the title. Jul 8, 2021 at 9:43
• @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. Jul 8, 2021 at 11:25
• @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. Jul 8, 2021 at 11:29

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.
• Again , $3$ can be replaced by any odd number. Jul 8, 2021 at 9:50