Is it true to say that every matrix $A\in M_n(\mathbb{R})$ is similar (conjugate) to a matrix $B=(b_{ij})$ with $b_{ij}=-b_{ji}$ for all $i\neq j$?(With some abuse of terminology,a matrix $B$ with this property is called "Semi antisymmetric").
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4$\begingroup$ Just to be clear: there is no restriction on the diagonal of B, so this is a somewhat unusual usage of the term "antisymmetric matrix", right? $\endgroup$– Nathaniel JohnstonCommented Oct 5, 2017 at 0:23
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$\begingroup$ It seems that for an identity matrix all the conjugates are also identity matrices, so the statement can be true only if there is no restriction on the diagonal, e.g. the identity matrix is defined to be antisymmetric. $\endgroup$– Sergey DovgalCommented Oct 5, 2017 at 2:39
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$\begingroup$ I also assume that only "real" basis changes are desired, because otherwise it can be proven without much effort that the statement is true. $\endgroup$– Sergey DovgalCommented Oct 5, 2017 at 2:50
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$\begingroup$ @NathanielJohnston yes you are right. I called such matrix semi anti symmetric, sonce there is no restriction on the doagonal. $\endgroup$– Ali TaghaviCommented Oct 5, 2017 at 6:42
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$\begingroup$ @SergeyDovgal In my question there is no restriction on diagonal. The congugacy is assumed via real matrices not complex matrices. With such assumption, do you think the answer to my question is affirmative? $\endgroup$– Ali TaghaviCommented Oct 5, 2017 at 6:46
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1 Answer
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Yes. Every matrix can be written as the sum of a symmetric plus an antisymmetric one: $A = \frac{A+A^T}{2}+\frac{A-A^T}{2}$. Now change basis such that the symmetric part is diagonal.
answered Oct 5, 2017 at 7:01
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6$\begingroup$ This shows in addition that the conjugating matrix can be taken to be orthogonal. $\endgroup$ Commented Oct 5, 2017 at 7:06
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$\begingroup$ @Federico Thank you very much for this very excellent answer. $\endgroup$ Commented Oct 5, 2017 at 15:05
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3$\begingroup$ @FrançoisBrunault Thanks for this comment. As you wrote,it is essential to take the orthogonal conjugacy otherwise we may loose the antisymmetricity of the second matrix. $\endgroup$ Commented Oct 5, 2017 at 15:08