Let $H_i$ be a $\mathbb R$-Hilbert space, $A\in\mathfrak L(H_1,H_2)$ be compact, $|A|:=\sqrt{A^\ast A}$ and $\sigma\in\mathbb R$.
How would we describe the singular value decomposition of $A$ in the language of operator theory? (Assuming $\dim H_i\in\mathbb N$, if necessary.)
To fix terminology, say that $\sigma>0$ is a singular value of $A$ if $\sigma$ is an eigenvalue of $|A|$, i.e. $\mathcal N(\sigma-|A|)\ne\{0\}$. This definition is equivalent to claim that there are $x_i\in H_i$ with $\left\|x_i\right\|_{H_i}=1$ and $$Ax_1\sigma x_2\text{ and }A^\ast x_2=\sigma x_1\tag1.$$
By the Courant-Rayleigh minimax principle, we may enumerate the singular values of $A$ in nonincreasing order. So, let $\sigma_i(A)$ denote the $i$th largest singular value of $A$ for $i\in\mathbb N$. (If there are only $k$ different singular values, $\sigma_i(A)=0$ for all $i>k$.)
Now we may mimic some parts of the spectral theorem for compact self-adjoint operators: Let \begin{align}E_i&:=\mathcal N(\sigma_i(A)-|A|),\\d_i&:=\dim E_i\end{align} and $\left(e^{(i)}_1,\ldots,e^{(i)}_{d_i}\right)$ be an orthonormal basis of $E_i$ for $i\in\mathbb N$ and \begin{align}(\sigma_i)_{i\in\mathbb N}:=(\underbrace{\sigma_1(A),\ldots,\sigma_1(A)}_{=:\:d_1\text{ times}},\underbrace{\sigma_2(A),\ldots,\sigma_2(A)}_{=:\:d_2\text{ times}},\ldots),\\(e_i)_{i\in\mathbb N}:=\left(e^{(1)}_1,\ldots,e^{(1)}_{d_1},e^{(2)}_1,\ldots,e^{(2)}_{d_2},\ldots\right).\end{align} Then $(e_i)_{i\in\mathbb N}$ is an orthonormal basis of $\mathcal N(A)^\perp$ (since $\mathcal N(A)=\mathcal N(|A|)$) and $$|A|x_1=\sum_{i\in\mathbb N}\sigma_i\langle x_1,e_i\rangle_{H_1}e_i\tag2.$$
How do we need to proceed? And how is this related to the polar decomposition$^1$ of $A$?
$^1$ There is an unique partial isometry $U$ from $H_1$ to $H_2$ with $\mathcal N(U)=\mathcal N(A)$ and $A=U|A|$.
Best Answer
The singular value decomposition is obtained from the polar decomposition, together with the spectral theorem.
The polar decomposition gives you $A=V|A|$, where $V$ is a partial isometry such that $\operatorname{ran}V^*V=\overline{\operatorname{ran}A^*}$, and $|A|=(A^*A)^{1/2}$. Since $|A|\in B(H_1)$ and positive and compact, we apply the Spectral Theorem to obtain $$\tag1 |A|=\sum_{j=1}^\infty\sigma_j\,P_j, $$ where $\sigma_1\geq\sigma_2\geq\cdots\geq0$ and each $P_j$ is a rank-one projection. We can rewrite $(1)$ as $$\tag2 |A|=U^*DU, $$ where $U$ is a unitary and the is the diagonal operator (in the canonical basis, say) with diagonal $\sigma_1,\sigma_2,\ldots$
Then $$\tag3 A=VU^*DU=WDU $$ where $D$ is as above, $W$ is a partial isometry, and $U$ is unitary.
An often more useful way of writing this is choosing unit vectors $e_j$ with $P_je_j=e_j$ (so they form an orthonormal basis of the range of $|A|$) and write $(1)$ as $$\tag4 |A|=\sum_{k=1}^\infty\sigma_k\,\langle\cdot,e_k\rangle \,e_k. $$ Then $$\tag5 A=V|A|=\sum_{k=1}^\infty\sigma_k\,\langle\cdot,e_k\rangle \,Ve_k. $$ As $V$ is an isometry on $\operatorname{ran}|A|$, we get that $\{Ve_k\}$ is orthonormal. So the Singular Value Decomposition can be restated as saying