Comparing polar- and Cartan-decomposition, can we conclude that every positive definite symmetric matrix $A\in\text{Gl}(n,\mathbb{R})$ can be written as $A=\text{diag}(\lambda_1,..,\lambda_n)\cdot P$ with $P\in\text{O}(n)$ and $\lambda_i \gt 0$ ?

Conversely are all such products symmetric and positive definite?


Converse: false (Take $\Lambda=\begin{pmatrix} 1&0\\0&1\end{pmatrix}$ and $P=\begin{pmatrix}\cos\psi&\sin\psi\\-sin\psi&\cos\psi\end{pmatrix}$, say, for $\psi=\pi/3$).

The first conclusion is false as well. Indeed, take $n=2$ and let $A=A^T=\Lambda P$ with $P\in O(2)$ and $\Lambda=\operatorname{diag}(\lambda_1,\lambda_2)$. Then $\Lambda P=(\Lambda P)^T=P^T\Lambda$. An arbitrary $P\in O(2)$ has the form $P=\pm\begin{pmatrix}\cos\psi&\sin\psi\\-sin\psi&\cos\psi\end{pmatrix}$ with $P^T=\pm\begin{pmatrix}\cos\psi&-\sin\psi\\sin\psi&\cos\psi\end{pmatrix}$. Now by writing out the condition $\Lambda P=P^T\Lambda$ explicitly you will see that it holds only if $\sin\psi=0$. But then the matrix $\Lambda P$ is necessarily diagonal, which shows that you cannot obtain nondiagonal symmetric matrices in this way.

  • $\begingroup$ I see, why $AP$ is not positive definite, thanks for the counterexample. But how can I prove that the first conclusion is false as well? $\endgroup$ – sj134 Jun 8 '14 at 9:36
  • $\begingroup$ Edited post; see above. $\endgroup$ – Vladimir Jun 8 '14 at 9:48

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