Can I prove that I-S is a nonsingular matrix?

[Codezion]

Homework Statement

Given S is a Skew-Hermitian (S*=-S), Prove that I - S is a nonsingular matrix

Homework Equations

If a matrix A is nonsingular, for Ax=0, x={0}

The Attempt at a Solution

(I-S)x=0, and I have been trying to show that the solution for x is always zero. Is this the correctly direction, because I have been trying to do this since last night and I seem to be unable to do this. I would really appreciate it if you have a better suggestion. Thanks!

 

[Dick]

If (I-S) is singular, then there is a nonzero vector x such that <(I-S)x,(I-S)x>=0. (<> is the inner product). Play around with the consequences of that for a while.

 

[Codezion]

Thanks Dick:

After playing with it, I arrived at the concolusion I = S^2, but cannot see how to contradict that.

<(I-S)x, (I-S)x>
=[(I-S)x]*[(I-S)x]
=[Ix*-S*x*][Ix-Sx]
=[Ix*+Sx*][Ix-Sx]
=Ix*x+ISx*x - Ix*sx - Sx*Sx
=Ix*x - Sx*Sx (middle two cancel out)
=b(I) - b(S^2) (let x*x = b (some constant)
=0
=> I = S^2
Now, I have to show this is a false statement...right? Can we show this is false for all cases?

 

[Codezion]

Taking the square root of each side...Sqrt(I) = S, however, the square root of an identity matrix is by no means a skew Hermitian, hence the original assumption is erroneous!

Thanks Dick!

 

[Dick]

Taking the square root of each side...Sqrt(I) = S, however, the square root of an identity matrix is by no means a skew Hermitian, hence the original assumption is erroneous!

Thanks Dick!

No, no, no. You can't do any such thing. I think you have a stray sign. What I wound up with is <x,x>+<Sx,Sx>=0. What's wrong with that?

 

[Codezion]

Hmm...I played with it overnight and cannot figure out what is wrong with that statement. The only thing I can see is that x*x cannot be zero since x is a non zero vector..?

 

[Dick]

<x,x> is positive since x is nonzero and the dot product is positive definite. <Sx,Sx> is nonnegative. The sum can't be zero.

 

[Codezion]

Wow! Thank you. I did not see that <x,x> is always positive, I was just focusing on it being nonzero!

Thank you so much!