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Let $V$ be an inner product space finitely generated over $\mathbb{C}$ and let $\alpha\in Aut(V)$. Then there exists a unique positive-definite automorphism $\theta$ of $V$ and a unique unitary automorphism $\psi$ of $V$ satisfying $\alpha=\psi\theta$.

I can show there exist $\theta$ and $\psi$ satisfying $\alpha=\psi\theta$. However, I have problem to show the uniqueness. I appreciate any help.

Assume that $\psi\theta=\psi'\theta'$ where $\psi$ and $\psi'$ are unitary automorphism of $V$ and where $\theta$ and $\theta'$ are positive definite automorphism of $V$.

As $\psi$ is unitary, we get $\psi^*=\psi^{-1}$. Hence,

$$\theta^2=\theta\psi^*\psi\theta=(\psi\theta)^*(\psi\theta)=(\psi'\theta')^*(\psi'\theta')=\theta'(\psi')^*\psi'\theta'=(\theta')^2$$

Both $\theta$ and $\theta'$ are positive definite, but I don't know how to show they are equal.

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    Because both are positive definite, you can take the "square root" of both sides. – 2017-01-19
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    @angryavian I thought $\theta^2$ is like composition of $\theta$ and $\theta$. right? – 2017-01-19
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    So I can say $\langle (\theta^2-(\theta')^2)v,v\rangle=0$, so $\langle \theta(v),\theta(v)\rangle=\langle \theta'(v),\theta'(v)\rangle$. Therefore, I get $\|\theta(v)\|=\|\theta'(v)\|$. but I just can say they have same Kernel. I don't know. – 2017-01-19

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If you first think of $\theta$ and $\theta'$ as positive definite matrices, then $\theta^2$ (which is function composition in your setup) becomes matrix multiplication.

The result I was trying to refer to is that since $\theta^2$ is positive definite, you can take a "square root" by diagonalizing it and replacing the eigenvalues with their square roots. Similarly for $(\theta')^2$. Note that this square root is unique, so they must be $\theta$ and $\theta'$ respectively. In short, $\theta^2 = (\theta')^2$ implies $\theta=\theta'$.

The argument for operators is essentially the same. Here is a post that goes through the details.