Polar decomposition of real matrices Can we decompose real matrices in a way that is analogous to polar decomposition in the complex case?
What I mean is: given an invertible real matrix $M$, can we always write:
$$
M = OP,
$$
maybe uniquely, where $O$ is orthogonal, and $P$ is symmetric positive-definite?
Thanks.
 A: The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix.  Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = \sqrt{M^TM}$. Now, note that $P$ has the property $\|Px\| = \|Mx\|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$.  We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
\|MP^{-1}y\| = \|y\|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined.  Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.

Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
A: A derivation of the polar decomposition for a 2x2 matrix can be found on Polar Decomposition
I also elaborate the relation with SVD.
In short, shuffling around with the SVD decompostion:
$A= U \Sigma V^T$ = $(UV^T) (V \Sigma V^T) $= $R_{\theta} S_{AT}$
$S_{AT}$ scales along the principal axes of the ellipse defined by $A^T(unitcircle)$
$R_{\theta}$ rotates over the angle between the principal axes of $A$ and $A^T$
$A^{-1}=\left({\ \ U\ \mathrm{\Sigma}}^{-1}U^{-1}\right)^{-1}R_\theta$ = $S_A R_\theta$
$R_{\theta}$ rotates over the angle between the principal axes of $A$ and $A^T$
$S_{A}$ scales along the principal axes of the ellipse defined by $A(unitcircle)$
