1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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} $

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.