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Prove that if $A^T = -A$ is any skew -symmetric matrix, then $Q = (I-A)^{-1}(I+A)$ is an orthogonal matrix. Can you prove that (I - A) is always invertible?

How do I go on to prove this? Is it similar to proving that det(Q) = $^+_-$ 1 or that $A^T = A^{-1}$?

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Um. No, not determinant. Individual eigenvalues. The ordinary dot product of two column vectors $v,w$ is given by the matrix product $v^T w = w^T v$ because the transpose of a 1 by 1 matrix is itself. So suppose your $A$ has a real eigenvalue $\lambda,$ with an eigenvector $v.$ We have $Av = \lambda v,$ and $$ \lambda v^T v = v^T (\lambda v) = v^T (Av) = (Av)^T v = v^T A^T v = -v^T A v = - \lambda v^T v. $$ Now $v \neq 0,$ so $v^T v \neq 0.$ Thus $$ \lambda v^T v = - \lambda v^T v $$ means $\lambda = 0.$

So, the only possible real eigenvalue is $0.$ In particular, $1$ is never an eigenvalue, we always have $Av \neq v,$ and $(I-A)v \neq 0.$ Put more simply, $0$ is not an eigenvalue of $(I-A),$ which is thus nonsingular.

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Thank you Will, but we haven't learned eigenvalues yet. Is there any other way to prove it without eigenvalues ? –  diimension Oct 22 '12 at 2:23
    
@diimension, just write $Av=v,$ then $v^T Av = v^T v$ and see what goes wrong. That gives a proof that $Av \neq v.$ –  Will Jagy Oct 22 '12 at 2:27
    
alright, I will give it a shot. –  diimension Oct 22 '12 at 2:32
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@diimension, note this refers to real nonzero vectors, so $v \neq 0$ and $v \cdot v \neq 0$ and $v^T v \neq 0$ –  Will Jagy Oct 22 '12 at 2:38
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