Invertible matrix equals to unitary matrix times positive definite Let $V$ be a finite-dimensional inner product space, and $T:V\to V$ be a invertible linear operator. Prove that there exists a unitary operator $U$ and a positive operator $P$ on $V$ such that $T=U\circ P$.
How may I prove this theorem? Or its equivalent theorem in matrix form?
 A: As noted by user1551 in the comments, this is the polar decomposition. Also, as noted by Aaron in the comments, it can be proven with the SVD. 
Here is another approach. To get the unitary matrix we orthogonalize the columns of $T$ by multiplying with $(T^*T)^{-1/2}$:
$$T = \underbrace{\left(T (T^*T)^{-1/2}\right)}_{U}\underbrace{(T^*T)^{1/2}}_{P}.$$
Of course, $P:=(T^*T)^{1/2}$ is SPD since $T^*T$ is. To see that $U:=T (T^*T)^{-1/2}$ is unitary, we compute the angle between vectors after applying it, and see that it is the same as the angle before:
$$\langle T (T^*T)^{-1/2} u, T (T^*T)^{-1/2} v \rangle = \langle u, (T^*T)^{-1/2}T^*T (T^*T)^{-1/2} v \rangle = \langle u, v \rangle.$$
And we are done.

Also of interest numerically:
One can apply the dense unitary matrix $U$ to a vector $w$ without actually constructing it or doing any dense linear algebra, by 


*

*Applying $(T^*T)^{-1/2}w$ with a preconditioned rational Krylov method, or rational matrix function approximation method, then

*Applying $T$ to the result.
All this requires is a good preconditioner for $T^*T$, which is often available (e.g., algebraic multigrid).
