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

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Suppose A is an nxn matrix

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- kbLv 78 years agoFavorite 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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