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Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,


I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?

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I believe so, but if I'm correct, some of the entries will have to be complex. – Ataraxia May 11 '13 at 17:03
@Zetta: the first condition is that $A$ has real entries. – Qiaochu Yuan May 12 '13 at 8:25
@QiaochuYuan I know. That condition wasn't there when I first wrote the comment, the user edited it after I wrote it. – Ataraxia May 12 '13 at 10:01
@doraemonpaul I can see why you added "matrix-equations". But why did you remove the "matrices" tag? Removing "adjoint" would have been more appropriate. – 1015 May 17 '13 at 1:04
@doraemonpaul Please, have a look at meta, where I posted a question about these retagging issues. We should try to avoid editing wars, so maybe the issue should be discussed at meta first, before any mass-retagging is done. – Martin Sleziak May 17 '13 at 8:23

When $A$ is an $n\times n$ real matrix, it is possible if and only if $n$ is even.

  1. If $n$ is odd, then $\det(A^2)=(\det A)^2\ge 0$ but $\det(-I)=-1$, so $A^2=-I$ is impossible.
  2. If $n$ is even, you can construct $A$ from the case $n=2$ easily.
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Take the determinant of both sides. You find that $(\det A)^2=-1$ so that $\det A=\pm i$ which is impossible for real $A$. (But trivial for complex $A$.)

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No. For $3\times 3$ matrices, the determinant of $-\mathbf{I}$ is $-1$, and $\mathrm{det}(\mathbf{AA}) = \mathrm{det}(\mathbf{A})^2$ is always nonnegative.

Of course, if you allow complex numbers then the answer is yes, e.g. if $\mathbf{A} = i\mathbf{I}$.

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The answer is no in $\mathbb{R}^{3\times 3}$. Recall that $\det AB=\det A\det B$ and so if $\det AA=\det -I$ then $(\det A)^2=-1$ but $\det A$ is a real number for all $A\in\mathbb{R}^{3\times 3}$ and so no such $A$ can exist.

This extends easily to all odd $n$ for $\mathbb{R}^{n\times n}$ and also suggest that solutions exist in $\mathbb{C}^{n\times n}$. Most notably, the matrix $A=\text{diag}(i,\ldots, i)$ has the property that $A^2=-I$.

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Here is an argument without using determinant or eigenvalues (yes, this is a deliberately complicated answer). We prove by contradiction. Suppose $A^2=-I_3$. Take $x\neq0$ and let $y=Ax$. If $y=kx$ for some $k\in\mathbb{R}$, then $$k^2x=ky=kAx=A(kx)=Ay=A^2x=-x\ \Rightarrow\ k^2=-1,$$ which is impossible. Therefore $x$ and $y$ are linearly independent. Extend them to a basis $\{x,y,z\}$ of $\mathbb{R}^3$. Let $w=Az$. By similar reasoning, $z$ and $w$ are linearly independent and hence $w\neq0$. Therefore $$w=ax+by+cz\tag{1}$$ for some $(a,b,c)^T\in\mathbb{R}^3\setminus0$. Then $-z=Aw=A(ax+by+cz)=ay-bx+cw$, i.e. $$cw=bx-ay-z.\tag{2}$$ As $x,y,z$ are linearly independent, $c\neq0$. Therefore $(2)$ implies that $$w=\frac{b}{c}x-\frac{a}{c}y-\frac{1}{c}z.\tag{3}$$ Comparing the coefficients of $z$ in $(1)$ and $(3)$, we get $c=-\frac1c$, which is impossible. Hence there does not exist a real matrix $A$ such that $A^2=-I_3$.

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