if A^T (transpose of A) =-A, then how to prove that I+A is invertible?

Update:

Suppose A is an nxn matrix

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  • kb
    Lv 7
    8 years ago
    Favorite Answer

    A is invertible <==> λ = 0 is not an eigenvalue of A.

    So, A + I is invertible <==> λ = -1 is not an eigenvalue of A + I.

    (This comes from the fact that if λ is an eigenvalue of A <==> λ+c is an eigenvalue of A + cI.)

    This is indeed the case, because A has no nonzero real eigenvalues (because A = -A^t):

    See (5) here:

    http://dakecht.wordpress.com/2011/12/06/skew-symme...

    I hope this helps!

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