Polar decomposition of real matrices

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.