Let $A$ be a $n\times n$ real matrix $A$ such that $A^2=-I$. Such an $A$ cannot be,

  1. Orthogonal.
  2. Invertible.
  3. Skew-symmetric.
  4. Symmetric.
  5. Diagonalizable.

I tried to figure out the answer by looking at the [possible] determinant of $A$, using the fact $\det(AB)=\det(A)\det(B)$. So $\det(A^2)=(\det(A))^2=(-1)^n\det(I)$. If $n$ is even, then $\det(A)=\pm1$ and if odd then $\det(A)=\pm i$. There I stuck, if $n$ is even, I can try some guessing, but when $n$ is odd, the determinant becomes complex and I have no logic to go forward. Is there any other approach that might be more correct? How can I do this? Help me out.

  • 1
    $\begingroup$ Such $A$ certainly is invertible, with inverse $A^3$. Notice also that $\det A$ must be real since $A$ is real. $\endgroup$
    – user296602
    Apr 25, 2016 at 3:23
  • $\begingroup$ @T.Bongers I did not get the inverse thing. Can you elaborate a bit please? $\endgroup$ Apr 25, 2016 at 5:05
  • 2
    $\begingroup$ $A^4=I$ then $A^3 = A^{-1}$ (but I don't know whether this helps...) $\endgroup$ Apr 25, 2016 at 7:42

2 Answers 2


We can use the following characterization of diagonalizable:

A matrix or linear map is diagonalizable over the field F if and only if its minimal polynomial is a product of distinct linear factors over F.

$A^2+I=0$ implies that $A$ satisfies the polynomial $x^2+1$ which is the minimal polynomial since it is irreducibe over the reals. This minimal polynomial is not a product of distinct linear factors over $\mathbb{R}$, thus $A$ is not diagonalizable.

Invertible, skew-symmetric, orthogonal can be all ruled out with the example $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

Symmetric is false, by stity's comment below, since symmetric matrices are diagonalizable.

Alternatively, you can observe that it is false for $n=2$ by the computation: $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&c\\b&d\end{pmatrix}=\begin{pmatrix}a^2+b^2&\dots\\\dots&\dots\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ is not possible over the reals.

  • 1
    $\begingroup$ You could add that since every real symmetric matrix is diagonalizable, A is not symmetric $\endgroup$
    – stity
    Apr 25, 2016 at 3:39
  • $\begingroup$ Good point. Thanks $\endgroup$
    – yoyostein
    Apr 25, 2016 at 3:40

I experimented with rotation matrices and powers and found the following solution for $\small n=4$ where $\small w = \sqrt{1/3}$ :

     0   w  -w  w
A=  -w   0   w  w
     w  -w   0  w
    -w  -w  -w  0


      -3*w^2       .       .       .
           .  -3*w^2       .       .
  A^2=     .       .  -3*w^2       .  = - I
           .       .       .  -3*w^2

And also,

  • $\small A$ is invertible
  • $\small A^{-1} = A^T$ (or $\small A \cdot A^T=I $ ) which is a condition for orthogonal/orthonormal matrices.
  • Also $\small A^T = - A$ which means, $\small A$ is skew-symmetric.
  • Using complex numbers, a diagonalization is possible and gives the diagonal $\small D = [1,1,-1,-1] \cdot \sqrt{-3} $

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