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Every real matrix $A$ can be decomposed into symmetric and skew-symmetric part. Symmetric matrix has only real numbers as its eigenvalues (including $0$) and skew-symmetric matrix has only imaginary values (also including $0$).

  • Could we infer from separate calculations of eigenvalues for symmetric and skew-symmetric about eigenvalues for matrix $A$ ?
  • If so then can the same be said about eigenvectors calculated separately for these two parts of matrix and their relevance for eigenvectors of the whole matrix $A$?

Edit (after 1 day)

If the answer for the questions above is too difficult to obtain in a general case maybe it would be possible to answer for a particular case:

  • why in the case of orthogonal matrices $R$ we can write down: $EGV(sym(R))+EGV(sk(R))= EGV(R)$ where $EGV$ is here a vector obtained from respectively ordered values of eigenvalues for symmetric part of $R$, skew-sym. part of $R$ and full matrix $R$.

Example:
for 3-D rotation matrix we have ( it's hard to believe that it is just a coincidence)

$EGV(sym(R)) =\begin{bmatrix} \cos(\theta)\\ \cos(\theta)\\ 1 \end{bmatrix} , EGV(sk(R)) \begin{bmatrix} i\sin(\theta)\\ -i\sin(\theta)\\ 0 \end{bmatrix}, EGV(R)=\begin{bmatrix} \cos(\theta)+i\sin(\theta)\\ \cos(\theta)-i\sin(\theta)\\ 1 \end{bmatrix}$

Of course it would be better if it were proved starting from general properties of orthogonal matrix without calculating exact values of eigenvalues as I showed in the example above.

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  • $\begingroup$ I am afraid that there are no relations between the eigenvalues for symm and skewsymm and eigenvalues for $A$ $\endgroup$ Nov 8, 2016 at 9:16
  • $\begingroup$ More generally, knowing the eigenvalues of matrices $A$ and $B$ does not provide information on the eigenvalues of $aA+bB$. $\endgroup$
    – Jean Marie
    Nov 8, 2016 at 9:20
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    $\begingroup$ @AlessandroBlasetti Indeed, not much can be said, but some loose relations do exist, e.g. the trace of $A$ must be the sum of traces of the two decomposed parts. $\endgroup$
    – user1551
    Nov 8, 2016 at 9:22
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    $\begingroup$ You are right, it may happen that we have particular properties for this decomposition. Sometimes something can be said about singular values instead of eigenvalues (in this respect, for example, I am almost certain that the symmetrical part of a matrix $A$ is the closest symmetrical matrix to $A$ in the sense of Frobenius norm). $\endgroup$
    – Jean Marie
    Nov 8, 2016 at 9:28
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    $\begingroup$ have a look at (math.stackexchange.com/q/1941701) $\endgroup$
    – Jean Marie
    Nov 8, 2016 at 9:31

1 Answer 1

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I try to give a partial answer. As @JeanMarie said in the comments there is no relationship between the eigenvalues of two matrices, $A$ and $B$, and some linear combination $aA+bB$.

Since $0$ is an eigenvalue of both the symmetric part of $A$ and the anty-symmetric part, if $\ker(A+A^T)\cap\ker(A-A^T)\neq\emptyset$, we can easily prove that that also $A$ is not invertible.

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  • $\begingroup$ Ok. Maybe it is a very limited conclusion but gives however some information about one eigenvalue for A. Conclusions for more number of eigenvalues can be sometimes obtained from sym, and sk-sym, part of a matrix A when we have some additional information about the matrix as it is - for example - in the case of 2-d and 3-d orthogonal matrices where sym. and sk-sym. parts give clues for eigenvalues. $\endgroup$
    – Widawensen
    Nov 8, 2016 at 16:35

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