[Math] Eigenvalues for symmetric and skew-symmetric part of a matrix

eigenvalues-eigenvectorslinear algebramatricesrotations

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.

Best Answer

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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