In the book(archive.org) of Linear Algebra by Greub, at page 226 it is asked that:
Note: $E$ is a $n$-dimensional real vector space.
Prove that a regular linear transformation $\phi$ of a Euclidean space can be uniquely written in the form $$\phi = \sigma \circ \tau,$$ where $\sigma$ is a positive self-adjoint transformation and $\tau$ is a rotation. Hint: Use problem 5 and 10.(This is essentlally the unitary trick of Weyl.)
Question 5:
A selfadjoint transformation $\phi$ is called positive, if $$(x,\phi(x)) \geq 0$$ for every $x\in E$. Given a positive selfadjoint transformation $\phi$, prove that there exists exactly one positive selfadjoint transformation $\psi$ such that $\psi^2 = \phi $.
Question 10:
Note: $\bar \phi$ is the adjoint mapping of $\phi$, which is defined as $$(x, \phi (y)) = (\bar \phi(x), y) \quad x,y \in E.$$
Let $\phi$ be any linear transformation of an inner product space $E$. Prove that $\phi \circ \bar \phi$ is a positive self-adjoint mapping. Prove that $$(x, \phi \circ \bar \phi(x)) \geq 0 \quad x\not = 0$$ with equality only if $x \in ker(\phi)$
Even though the author gives some hints how to prove this result, I have stuck at this problem for, I guess, a month, and still couldn't figure out how to prove it. Then I have asked this to one of my assistants, but she couldn't figure out either, so my main question is that how can we prove this result, but any kind of help, some more hints, or general idea of the proof, anykind, is also very welcomed.
Note that, I have put a link of the book directly copied from archive.org, so you can check it out yourself also.
If a linear transformation is positive, any eigenvalue must be non-negative:
$$0\le (Ax,x)=(\lambda x,x)=\lambda (x,x)=\lambda ||x||^2$$
Section 8.9 in your book shows that a self-adjoint operator has an orthonormal basis of eigenvectors $e_j$ with eigenvalues $\lambda_j$. So if $A$ is positive and self adjoint and invertible, its effect on a vector $\sum v^je_j$ can be written
$$Av=A\sum v^je_j= \sum v^jAe_j=\sum v^j\lambda_j e_j$$
where the $\lambda_j$ are positive.
Given such an $A$, Consider the linear transformation $Bv=\sum v^j\sqrt{\lambda_j} e_j$. Then a straightforward computation shows that $B^2=BB=A$ (thus $B$ is a "square root" of A), and that B also is positive and self-adjoint and invertible.
$B$ is also the unique positive self-adjoint square root of $A$. To see this, note that since $B$ is positive and self-adjoint and invertible it also has an orthonormal basis of eigenvectors $f_j$ with positive eigenvalues $c_j$. Since $Af_j=BBf_j=Bc_jf_j=c_j^2f_j$, each $f_j$ is also an eigenvector of $A$ and the corresponding $c_j$ equals $\sqrt{\lambda_i}$ for some $i$. In other words, the effect of $B$ is to scalar multiply each $\lambda_j$ eigenspace of $A$ by $\sqrt{\lambda_j}$. This determines $B$ uniquely since the eigenspaces of $A$ span $E$.
Now we can use all of the above to prove the original claim. Given an invertible linear transformation $\phi$, consider $A=\phi\phi^T$. Then $A$ is positive and self adjoint: $$(x,Ax)=(x,\phi\phi^T x)=(\phi^T x, \phi^T x)=||\phi^T x||^2\ge0$$ $$(x,Ax)=(x,\phi\phi^T x)=(\phi^T x,\phi^T x)=(\phi\phi^T x,x)=(Ax,x)$$ and it is invertible, so it has a unique positive self-adjoint square root $B$ which is also invertible.
Now suppose $\phi=BU$ for some rotation $U$. Then we must have $U=B^{-1}\phi$. So we'll define U by this formula and prove it is a rotation, which is to say it preserves the inner product
$$(Uv,Uv)=(B^{-1}\phi v,B^{-1}\phi v)=(\phi^TB^{-1}B^{-1}\phi v,v)=(\phi^T A^{-1}\phi v,v)$$
$$=(\phi^T (\phi\phi^T)^{-1}\phi v,v)=(\phi^T (\phi^T)^{-1}\phi^{-1}\phi v,v)=(v,v)$$
So $U$ is a rotation. To prove uniqueness, suppose also that $\phi=CV$ for a positive self adjoint $C$ and a rotation $V$. Then
$$\phi\phi^T=CVV^TC=C^2$$
so that $C$ is also a positive self-adjoint square root of $\phi\phi^T$. But uniqueness of $B$ was proved above, so $C=B$. Then $V=C^{-1}\phi=B^{-1}\phi=U$, so the decomposition is unique.