Consider an operator $\Gamma$ acting on $\mathcal{A}\otimes\mathcal{B}$, the tensor product of two finite-dimensional Hilbert spaces. Then decompose $\Gamma$ with the operator singular value decomposition: $$\Gamma = \sum_k\sigma_k\, A_k\otimes B_k$$ where $\sigma_k\geq0 \ \forall k$ and $\mathrm{tr}(A_j^\dagger A_k)=\mathrm{tr}(B_j^\dagger B_k)=\delta_{jk}$.
I'm interested only in $A_0$ (the operator on $\mathcal{A}$ corresponding to the largest singular value). Is there a way of extracting $A_0$ from $\Gamma$ without having to compute the whole SVD? If not, how about approximating $A_0$?
Let $vec(A)$ be the vectorization of the matrix $A$.
Prove $tr(A^\dagger B)=Vec(A)^\dagger Vec(B)$ for rank one matrices first, then use additivity for arbitrary matrices.
Define $R(\Gamma)=\sum_{k}\sigma_k vec(A_k)vec(B_k)^t$. Notice that $R(\Gamma)R(\Gamma)^\dagger=\sum_{k}\sigma_k^2 vec(A_k)vec(A_k)^\dagger$.
Now the problem is to find a norm 1 eigenvector, $v$, associated to the biggest eigenvalue $\sigma_0^2$ of the positive semidefinite Hermitian matrix $R(\Gamma)R(\Gamma)^\dagger$. Let $A_0=vec^{-1}(v)$.