Proving $$(P^T P^T) \Lambda P P \equiv \Lambda$$ where $P$ is an orthogonal matrix, $\Lambda$ is diagonal matrix. All matrices have dimensions $n \times n$.

Since this is the last step of the proof shown in $\chi^2$ for dependent Gaussian distributions It is known that all diagonal elements of $\lambda_i \geq 0$

  1. Multiplied orthogonal matrices give another orthogonal matrix

Proof: $$ P \cdot P^T = I\\ Q := P \cdot P\\ P^{-1} = P^T\\ PP \cdot (PP)^T = PP \cdot P^T P^T = P I P^T = P \cdot P^T = I $$

So $Q$ is orthogonal as well.

  1. How can I now prove that $Q \Lambda Q^T = \Lambda$?

For a full rank $\Lambda$ with equal diagonal elements and otherwise zero this can proven: $Q \Lambda Q^T = Q \lambda \cdot I Q^T = \lambda Q \cdot Q^T = \lambda \cdot I = \Lambda$

How can I prove this for the general case with differing diagonal elements?

  • $\begingroup$ I think this comes from a typo in my post referred to above. Will check this and correct it soon. $\endgroup$ – kjetil b halvorsen Feb 13 '15 at 12:28

I don't think this is true. Counter example: $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} $ $\begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix} $ $\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} $= $\begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix} $

  • $\begingroup$ Thanks. I also started playing around with octave and came to the same conclusion. $\endgroup$ – drahnr Feb 8 '15 at 12:48

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