4

Let $A$ be a real antisymmetric matrix. Is it true that $A$ must be orthogonally similar to its transpose (i.e to $-A$)?

Note: It's known that every matrix is similar to its transpose. It's also known, that in general, not every real matrix is orthogonally similar to its transpose.

However, in the example given in the reference above, $A$ is not antisymmetric.

Community
  • 1
Asaf Shachar
  • 25,111
  • In general, an antisymmetric $A$ will be unitarily similar to its transpose. Not sure about orthogonal similarity, though. – Ben Grossmann Jan 04 '16 at 19:25
  • It seems it is true for $2\times 2$ matrices (if I have not misunderstood the question). In general, we want $A$ to be real and such that $A^T=-A$. We want to see whether there exist an orthogonal matrix $O$ ($O O^T=O^T O=\mathbf{1}$) such that $$ A = -OAO^T\ . $$ (continue) – Pierpaolo Vivo Jan 04 '16 at 20:43
  • Now take $$ A=\begin{pmatrix} 0 & \alpha\ -\alpha & 0\ \end{pmatrix} $$ ($\alpha\in\mathbb{R}$ and $\alpha\neq 0$) and parametrize the $2\times 2$ orthogonal matrix as $$ O=\begin{pmatrix} \cos\theta & \sin\theta\ \sin\theta & -\cos\theta\ \end{pmatrix}\ . $$ Now, the matrix $A+OAO^T$ is indeed equal to the null matrix $\forall\theta$. So any orthogonal matrix $2\times 2$ with $\det O =-1$ does the job. Not sure about general size, though... – Pierpaolo Vivo Jan 04 '16 at 20:43

1 Answers1

2

The answer is yes. All real skew-symmetric matrices are orthogonally similar to their real Jordan forms (see, e.g. Corollary 2.5.14 of Horn and Johnson's Matrix Analysis). So, the conclusion follows from either one of the following two observations:

  1. The transpose of a skew-symmetric matrix $A$ is skew-symmetric. Hence both $A$ and $A^T$ are orthogonally similar to the same real Jordan form.
  2. The $2\times2$ skew-symmetric block $\pmatrix{0&-y\\ y&0}$ (that appears in the real Jordan form of a skew-symmetric matrix) is permutation-similar to its transpose.
user1551
  • 139,064