These days, I met a problem on linear algebra:
Suppose $A,B$ are real matrices. If there's a complex unitary matrix $U$ such that $U^*AU=B$, where $U^*=\overline U^\top$, namely, the conjugate transpose of $U$, then there's a real orthogonal matrix $O$ such that $O^\top AO=B$.
I tried but failed. It seems that I lack some advanced tools. By some search work, I've found an answer. However, I want to determine whether my preceding thoughts would work, so I post a question here to look for some requests.
Note that if $P$ is unitary, then $P^{-1}=P^*$ and if $P$ is orthogonal, then $P^{-1}=P^\top$, therefore either conditions is equivalent to $AP=PB$, where $P=U$ or $O$. Consider the matrix space $M_n(\mathbb R)$ and its complexification $M_n(\mathbb C)$. The matrices $P$ such that $AP=PB$ form a subspace of $M_n$. Denote $O_n\subseteq M_n(\mathbb R)$ the group of orthogonal matrices, and $U_n\subseteq M_n(\mathbb C)$ the group of unitary matrices. We consider $U_n$ the complexification of $O_n$. The original problem (somewhat generalized) could be described as follows:
Suppose $V\subseteq M_n(\mathbb R)$ is a subspace, and $^\mathbb CV$ is its complexication in $M_n(\mathbb C)$. If the complexification of $V\cap O_n$, i.e. $^\mathbb CV\cap U_n$ is non-empty, then $V\cap O_n\neq\emptyset$ per se.
I have tried some further generalizations, but eventually I found it's wrong. In order to explicate my idea clearly, I write it here:
If the intersection of complexification of quadratic hypersurfaces is non-empty, then the intersection of themselves is non-empty too, where quadratic hypersurface is the zero set of a quadratic form $r(x)=\langle x,x\rangle$, where $\langle\cdot,\cdot\rangle$ is a bilinear form (not necessarily symmetric), and the complexification of such a hypersurface is the zero set of $^\mathbb Cr(x+iy)=\langle x,x\rangle+\langle y,y\rangle+i\langle x,y\rangle-i\langle y,x\rangle\triangleq{}^\mathbb C\langle x+iy,x+iy\rangle$
Note that the constraints of $O_n$ and $U_n$ are just of this kind.
It's wrong, since $x^2=1,y^2=1,xy+yx=0$ has no solution, but $\bar xx=1,\bar yy=1, \bar xy+\bar yx=0$ has solution $(x,y)=(1,i)$.
Intuitively, the failure comes from $\bar xy+\bar yx=0$, whose zero set is somewhat singular. By contrast, at least, $SO_n$ and $SU_n$ are well-known analytic manifolds.
I want to learn more on the topic of the following schema:
Suppose there's a complexification functor betwen $\mathbb R^N$ and $\mathbb C^N$. Some structured $W$ is non-empty iff its complexification $^\mathbb CW$ is non-empty.
Any idea? Thanks!
