Is (I-S) Nonsingular for Skew-Hermitian S?

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SUMMARY

The discussion centers on proving that the matrix expression I - S is nonsingular when S is a skew-Hermitian matrix, defined by the property S* = -S. Participants suggest that directly assuming det(I - S) ≠ 0 is not valid, as this is the conclusion to be proven. Instead, they recommend demonstrating that det(I - S) = 0 leads to a contradiction, leveraging the properties of skew-Hermitian matrices. A key approach involves showing that the inner product <(I - S)x, (I - S)x> is nonzero for any nonzero vector x.

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Homework Statement



S \in C^{mxm} is skew-hermitian. S^{*}=-S
I need to show that I-S is nonsingular.


Homework Equations





The Attempt at a Solution


There are two things that I thought. First was start with det(I-A)\neq0 since (I-A) is nonsingular.
Other attempt was to try to show (I-S)(I-S)^{-1}=(I-S)^{-1}(I-S)=I.

Unfortunately, I could get nowhere. Thanks in advance for your help...
 
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You can't start with det(I - S) \neq 0, since that's essentially what you need to prove. Also, by writing (I - S)(I - S)-1, you are assuming the existence of the inverse of I - S, which is equivalent to what you want to prove.

I think your best bet is to show (not assume) that det(I - S) \neq 0. Alternatively, you might assume that det(I - S) = 0 and see if you can arrive at a contradiction with your other assumption that S* = -S.
 
You want to show I-S has kernel {0}, hence show (I-S)x is nonzero if x is nonzero. Can you show the inner product <(I-S)x,(I-S)x> is nonzero?
 

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