if A^T (transpose of A) =-A, then how to prove that I+A is invertible?
Suppose A is an nxn matrix
- 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:
I hope this helps!