Show there exists $S$ which is self-adjoint such that $S^2=T^*T$ and $S$ is invertible.

Question

Let $T:V\to V$ be a linear operator such that $T$ is linear and bijective. Also $V$ is a finite dimensional inner product space over $\mathbb R$.

Show there exists $S$ which is self-adjoint such that $S^2=T^*T$ and $S$ is invertible.

I started like this:

Since $T$ is a linear operator on a finite dimensional vector space so $T$ can be represented by a matrix say $M_n$ with respect to some standard basis $\{e_1,e_2,\ldots, e_n\}$.

But I am stuck on how to construct $S:V\to V$. Also in order to show $S$ self-adjoint, it would require me to show $S$ has an orthonormal basis with real eigenvalues.

How do I come up with it?

Actually all is going messy for me. Is it possible for someone to help me. Also can someone be so kind to share some materials like books or links from where I can learn how to solve such problems with ease on my own.

Thanks a lot in advance.

Answer 1

Hint: By the spectral theorem, there exists an orthogonal transformation $U$ for which $U^*(T^*T)U$ is diagonal (or equivalently, an orthonormal basis of $V$ relative to which $T^*T$ is diagonal). If $D$ is diagonal and satisfies $D^2 = U^*(T^*T)U$, then $S = UDU^*$ satisfies $S^2 = T^*T$.

Answer 2

You can use the polar decomposition $T=US$ where $U$ is unitary and $S$ is positive. Then $T^*T= SU^*US=S^2$, which gives what you want.