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 $\lambda_i \geq 0$
- 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.
- 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?
Best Answer
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} $