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Is it possible to have a $3 \times 3$ matrix that is both orthogonal and skew-symmetric?

I know it has something to do with the odd order of the matrix and it is not possible to have such a matrix. But what is the reason?

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  1. Othogonal matrices have their eigenvalues on the unit circle.
  2. Skew-symmetric Matrices have their eigenvalues on the imaginary axis.
  3. Matrices with real entries have complex-conjugate pairs of eigenvalues.

The only points where the unit circle intersects with the imaginary axis are $i$ and $-i$, which make up one perfect complex-conjugate pair. But your matrix needs $3$ eigenvalues, so we are missing one. It cannot have a complex partner, so it must be on the real axis. The only point where the real axis intersects with the imaginary axis is $0$, which is not on the unit circle. You have three constraints, but these never meet at one point.

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    $\begingroup$ Really nice answer! $\endgroup$ – Vincent Jan 17 '17 at 9:23
  • $\begingroup$ Interesting summary of eigenvalues but taking them into account it would not be enough to notice that skew-symmetric dim. 3x3 has 0 eigenvalue, orthogonal has not? $\endgroup$ – Widawensen Feb 2 '17 at 19:24
  • $\begingroup$ But the argument of the zero eigenvalue comes from the pure imaginary eigenvalues that have to pair up. That's why one is left out. That step is needed to see, why $3 \times 3$ skew-symmetric is singular. $\endgroup$ – Laray Feb 2 '17 at 22:45
  • $\begingroup$ Yes, the zero complements the pair of pure imaginary eigenvalues values. So all ways (almost) lead to the singularity of sk.-sym. $\endgroup$ – Widawensen Feb 3 '17 at 7:39
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    $\begingroup$ You are right, that's why I put the argument in there. I could say "Skew-matrices of odd order are singular, duh!" but I personally like to remember that by the properties of Eigenvalues on the axis. That's why I went the extra mile. $\endgroup$ – Laray Feb 3 '17 at 7:46
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No, an orthogonal matrix has determinant $\pm 1$ whereas a skew symmetric matrix of order 3 has determinant $0$.

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No, it is not possible, for real matrices. Let $A$ be an $n \times n$ real matrix, with $n$ being odd. Suppose it is skew-symmetric, that is, $A^T = -A$. Then, $A^T A = -A^2$. If the matrix is also orthogonal, $I = -A^2$. Now take determinants to get $$\det(A)^2 + 1 = 0$$ The determinant is imaginary, giving a contradiction, since we assumed $A$ to be a real matrix.

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    $\begingroup$ $\det(-B) = (-1)^n \det(B)$, so this is only true for odd $n$. $\endgroup$ – Hetebrij Jan 17 '17 at 9:53

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