Proving Orthogonality: Simple Matrix Question with Skew-Symmetric A"

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To prove that (I - A) . inv(I + A) is orthogonal when A is skew-symmetric and (I + A) is singular, one must utilize properties of matrix operations. The key relationships include that the transpose of a product is the product of the transposes in reverse order, and that the transpose of a skew-symmetric matrix equals its negative. The orthogonality condition requires showing that the product of the matrices equals the identity matrix. The discussion highlights the realization that the inverse of a transpose equals the transpose of the inverse, which is crucial for the proof. Ultimately, understanding these matrix properties is essential for establishing the orthogonality of the given expression.
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Homework Statement



Prove that if A is skew-symmetric (i.e. A = -A') then


(I - A) . inv(I + A)

is orthogonal, assuming that (I + A) is singular. inv(X) denotes inverse matrix of X. X' denotes transpose of X.


The Attempt at a Solution



I need to prove orthogonality :

I know for square matrices : inv(XY) = inv(Y) . inv(X) and (XY)' = Y'X'

So if X and Y were orthogonal then XY would be.

But in this case X = I - A and Y = inv(I + A), but are they orthogonal?

I also know that (I - A)' = (I + A)

But can't figure it out. Do I have to resort to the defn of inv(X) = adj(X) / det(X) ?


Would appreciate any clues... Thanks in advance
 
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Aha stupid me... I now note that inv(X)' = inv(X')
 
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