Simplifying A^{-1} · diag(A · X · A^{-1}) · A

[tom08]

I encounter a problem when simplifying the following equation, can anyone give a hint:)

Let A denote an orthonormal matrix, X be a symmetric matrix.

diag(A) is an operator that creates a diagonal matrix of A.

Then, my problem is how to simplify the equation:

$$A^{-1} \cdot diag(A \cdot X \cdot A^{-1}) \cdot A$$

 

[HallsofIvy]

"diag(A) is an operator that creates a diagonal matrix of A."

That doesn't tell us enough. There are many ways to make a diagonal matrix out of A, for example, just setting all non-diagonal values equal to 0. Do you mean "diag(A) is a diagonal matrix similar to A"? In that case, are we to assume that A is diagonalizable?

 

[tom08]

"diag(A) is an operator that creates a diagonal matrix of A."

That doesn't tell us enough. There are many ways to make a diagonal matrix out of A, for example, just setting all non-diagonal values equal to 0. Do you mean "diag(A) is a diagonal matrix similar to A"? In that case, are we to assume that A is diagonalizable?

Sorry, I'll clarify the defitioon of diag (A).

diag(A) is a square matrix in which the entries outside the diagonal are all zero, and its diagonal entries are the diagonal of A.