- #1

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If A is orthogonal, it is obvious that this is true, but what if A is not orthogonal?

- Thread starter thoughtgaze
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- #1

- 74

- 0

If A is orthogonal, it is obvious that this is true, but what if A is not orthogonal?

- #2

- 18

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There are some conditions on A to allow diagonal of D=A'VA be composed of eigenvalues of V. If this is the case, V and D have the same determinant and trace (first is a product of eig.values and the second is the sum). Same determinant implies

det(A'VA) = det(V),

which means det(A) =1 or det(A)=-1. Same trace implies

tr(V)=tr(A'VA)=tr(AA'V).

I don't immediately see what does this condition mean for A.

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