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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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    @diimension, note this refers to real nonzero vectors, so $v \neq 0$ and $v \cdot v \neq 0$ and $v^T v \neq 0$2012-10-22
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I assume that $A\in M_{n}(\mathbb{R})$.

Since $A$ is real skew-symmetric we know that $\det(I+A)\neq 0$ (also, it is know in the case of real skew-symmetric matrices $det(A)\geq0$).

So, $\det(I-A)=\det(I+A^{T})=\det((I+A)^{T})=\det(I+A)\neq0$.