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

  • Thread starter Thread starter newreference
  • Start date Start date
  • Tags Tags
    Matrix
newreference
Messages
1
Reaction score
0

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...
 
Physics news on Phys.org
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top