2
$\begingroup$

The Takagi decomposition provides a canonical form for a complex symmetric matrix $S$ under $U \mapsto USU^T$ where $UU^* = I$.

Question: Is there an anti-Takagi decomposition? I.e. Is there a canonical form for a Hermitian matrix $H$ under $P \mapsto PHP^*$ where $PP^T = I$?

Another way of putting it: Given a Hermitian form $H$ and non-degenerate symmetric bilinear form $B$ over a complex vector space $V$, is there a canonical $B$-orthonormal basis for $H$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Not necessarily --- at least if $H$ is degenerate.

Indeed, let $H=\left[\matrix{1& i\\ -i& 1}\right]$. Its diagonal form $D=PHP^*$ should contain a zero row, so that one row of $P$ should be in the (left) kernel of $H$, i.e. it should be proportional to $[1,i]$. But then the corresponding diagonal entry in $PP^\top$ vanishes.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I'm not saying the canonical form has to be diagonal... I know it can't always be. $\endgroup$
    – wlad
    Commented May 31, 2022 at 12:43
  • $\begingroup$ In my opinion (and I did pose the question) this doesn't really answer my question. $\endgroup$
    – wlad
    Commented May 31, 2022 at 12:43
  • $\begingroup$ Thanks anyway! I should've clarified that it need not be a diagonal canonical form. Really, this answer should be a comment though. But w/e. $\endgroup$
    – wlad
    Commented May 31, 2022 at 12:44
  • 1
    $\begingroup$ OK, I’ll delete my answer —- as soon as you clarify this in the question, otherwise you may get another such nor-an-answer.. But I am really interested in a canonical form you would expect from a matrix $[\xi^{I-j}]$ where $|\xi|=1$. $\endgroup$
    – Ilya Bogdanov
    Commented May 31, 2022 at 14:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .