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Homework Help: Identity minus Skew-Hermitian

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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!
  2. jcsd
  3. Jan 18, 2009 #2


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    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.
  4. Jan 18, 2009 #3
    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>
    =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)
    => I = S^2
    Now, I have to show this is a false statement...right? Can we show this is false for all cases?
  5. Jan 18, 2009 #4
    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!
  6. Jan 18, 2009 #5


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    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?
  7. Jan 19, 2009 #6
    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..?
  8. Jan 19, 2009 #7


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    <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.
  9. Jan 19, 2009 #8
    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!
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