In joint work with Andreas Fischle (TU Dresden, Germany) and Patrizio Neff (U Essen, Germany) we needed to use the following statement: If $S$ is a real $n\times n$ matrix with $S^2$ symmetric, then there exists an orthogonal matrix $P$ such that $PSP^{-1}$ is block diagonal with blocks of size at most two. While we have a proof of this statement, we are wondering whether it already appears in the literature. Have you seen this result or its generalizations published anywhere?
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4$\begingroup$ I think this is the kind of thing that is proven over-and-over inside lemmas and so on, insofar as I can see that it would follow from not-quite-intro-level linear algebra... thus, probably not graduating to the status even of "lemma" in any noticeable contemporary literature. So, I'd bet there's no reasonable cite-able source in contemporary mathematics. $\endgroup$– paul garrettFeb 26, 2016 at 0:26
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$\begingroup$ I don't know, it could be an exercise in some textbook, although it could be just a touch too tricky for that. $\endgroup$– Lev BorisovFeb 26, 2016 at 1:11
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$\begingroup$ Just as a comment: in the case of $S$ being an involution, the statement is pretty much equivalent to the principal angles between subspaces story. $\endgroup$– Lev BorisovFeb 26, 2016 at 17:37
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1$\begingroup$ Did you look at Theorem 6.4.14 of: ebooks.cambridge.org/ebook.jsf?bid=CBO9780511840371 $\endgroup$– SuvritFeb 27, 2016 at 17:42
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1$\begingroup$ I found the relevant page online, it's not the right result. It is basically trivial to prove the statement without the orthogonality assumption on $P$. And it is not particularly hard to prove it with the orthogonality condition, but so far I didn't see a reference. $\endgroup$– Lev BorisovFeb 27, 2016 at 18:40
1 Answer
Eventually, let me present the proof which I mentionned in my comment.
Herebelow, I use repeatedly orthogonally similarities. Mind that if $S'$ is orthogonally similar to $S$, then $S'$ satisfies the same assumption: its square is symmetric.
Define $M=[S,S^T]$, which is symmetric, hence orthogonally diagonalizable. We may assume that $M={\rm diag}(M_0,\ldots,M_r)$, where $M_0$ is a null bloc (possibly missing) and $M_i={\rm diag}(d_i,\ldots,d_i,-d_i,\ldots,-d_i)$ for some $d_1<\cdots< d_r$.
After having applied the same orthogonal similarity to $S$ as to $M$, let us write $S$ blockwise according to the same partition. Because of $S^2=(S^T)^2$, we have $$MS+SM=SS^TS-S^TS^2+S^2S^T-SS^TS=0_n.$$ In other words, $S_{ij}M_j+M_iS_{ij}=0$ for every $i,j$.
If $i\ne j$, this implies $S_{ij}=0$ because the spectra of $M_i$ and of $-M_j$ are disjoint. Therefore $S$ is block-diagonal.
If $i\ge1$, the identity $S_{ii}M_i=M_iS_{ii}$ shows that $S_{ii}$ is anti-diagonal: $$S_{ii}=\begin{pmatrix} 0_p & B_i \\ C_i & 0_q \end{pmatrix}.$$ From the definition of $M$, we have $$B_iB_i^T-C_I^TC_i=d_iI_p,\qquad C_iC_i^T-B_i^TB_i=-d_iI_q.$$ Taking the trace, we find that the sizes $p$ and $q$ of the diagonal blocks in $S_{ii}$ are equal. Thus $B_i$ and $C_i$ are square matrices.
Conjugating $S_{ii}$ by the orthogonal matrix ${\rm diag}(P,Q)$, we may replace $B_i$ by $P^TB_iQ$ (and $C_i$ by $Q^TC_iP$ meanwhile). By SUV factorization, we are led to the case where $B_i$ is diagonal with non-negative diagonal entries. Remarking that the symmetry of $S_{ii}^2$ implies that $C_i$ commutes with the diagonal matrix $B_i^2$, we deduce that $C_i$ is block-diagonal, each diagonal block corresponding to a block $bI_m$ in $B_i$.
We are thus reduced to the case of blocks $$s=\begin{pmatrix} 0_m & bI_m \\ \Sigma & 0_m \end{pmatrix},$$ where $b>0$ and $\Sigma$ is a symmetric matrix. Now, conjugating by an orthogonal matrix of the form $$\begin{pmatrix} 0_m & U \\ U & 0_m \end{pmatrix},$$ we obtain instead the block $$s'=\begin{pmatrix} 0_m & \sigma:=U^T\Sigma U \\ bI_m & 0_m \end{pmatrix}.$$ Choosing appropriately $U$, we get a diagonal $\sigma$. Rearranging the coordinates, we see that $s'$ is block-diagonal with $2\times 2$ diagonal blocks of the form $$\begin{pmatrix} 0 & c \\ b & 0 \end{pmatrix},$$
There remains the case $i=0$. This is a result about nilpotent matrices $N$ ($=S_0$). Since $N^2$ is nilpotent and symmetric, thus diagonalisable, we have $N^2=0$. Then we may use that the ambient space is the orthogonal direct sum of $R(N)$, $R(N^T)$ and $\ker N\cap \ker N^T$.