= Solution
Apply the <spectral theorem for compact Hermitian operators> to positive $A=T^*T$. Its positive <eigenvalues> $\mu_i$ have <orthonormal> <eigenvectors> $u_i$ spanning $\ker A^\perp$, with a finite or countable list. In the infinite case $\mu_i\to0$. Define
$$
\lambda_i=\sqrt{\mu_i}>0,\qquad v_i=\frac{Tu_i}{\lambda_i}.
$$
Then $(v_i,v_j)=(u_i,T^*Tu_j)/(\lambda_i\lambda_j)=\delta_{ij}$, so the $v_i$ are orthonormal. Also $\ker A=\ker T$, because $(x,Ax)=\|Tx\|^2$. Decompose $x=x_0+\sum_i(u_i,x)u_i$ with $x_0\in\ker T$ and apply bounded $T$ to the norm-convergent expansion. This gives the <singular value decomposition>
$$
\boxed{Tx=\sum_{i=1}^N\lambda_i(u_i,x)v_i.}
$$
The series converges in <norm>; its squared tail <norm> is $\sum_{i>m}\mu_i|(u_i,x)|^2\to0$. No separability of the whole <Hilbert space> is necessary: the nonzero spectral subspace of a <compact operator> is separable. For $T=0$ use the empty list $N=0$.
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