It is a line from my textbook which says that "$S $ is a skew-symmetric matrix implies det$((I+S)^{-1})\neq 0$" which I cant justify.How to justify it?
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Maybe you could start by working on $(I+S)(I-S)$? – Vincent Guillemot Jun 22 '15 at 13:58
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$\det(AB) = \det(A)\det(B)$ and $\det(I) = 1$, so $\det(A^{-1}) = 1/\det(A) \ne 0$ whenever $A^{-1}$ exists.
The real question, I think, is why $(I + S)^{-1}$ exists if $S$ is a (real) skew-symmetric matrix. If not, there is a nonzero vector $v$ with $0 = (I+S) v = v + S v$, i.e. $Sv = -v$. But then since $S^T = -S$, $$\|v\|^2 = \|Sv\|^2 = v^T S^T S v = - v^T S^2 v = - v^T v = -\|v\|^2$$ so $v = 0$, contradiction.
Robert Israel
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