# Determine the singular values of a compact operator in terms of the eigenvalues of an alternating tensor product of operators

Let $$H$$ be a $$\mathbb R$$-Hilbert space, $$A\in\mathfrak L(H)$$ be compact and $$|A|:=\sqrt{A^\ast A}$$ denote the square-root of $$A$$. By definition, the $$k$$th largest singular value $$\sigma_k(A)$$ of $$A$$ is equal to the $$k$$th largest eigenvalue $$\lambda_k(|A|)$$ of $$A$$.

Now, if $$S,T:H\to H$$ are linear, let $$S\otimes T$$ denote the linearization of $$H^2\ni(x,y)\mapsto Sx\otimes Ty\tag1$$ and $$S^{\otimes 2}:=S\otimes S$$. I would like to endow $$H^{\otimes2}$$ with an inner product so that $$S\otimes T$$ is bounded, whenever $$S,T$$ are bounded, and for its completion $$H^{\hat\otimes2}$$ the operator $$S\otimes T$$ has a unique bounded linear extension $$S\hat\otimes T$$.

Moreover, it should hold that the largest singular value $$\sigma_1(A^{\hat\otimes k})$$ of $$A^{\hat\otimes k}$$ is given by $$\prod_{i=1}^k\sigma_i(A)$$. Note that this is a property which is fulfilled by the exterior product $$A^{\wedge k}$$.

It's trivial to see that if $$S:H\to H$$ is linear, $$\lambda_i\in\mathbb R$$ and $$e_i\in\mathcal N(\lambda_i-S)$$, then $$S^{\otimes k}\underbrace{\left(\bigotimes_{i=1}^ke_i\right)}_{=:\:e}=\underbrace{\prod_{i=1}^k\lambda_i}_{=:\:\lambda}\bigotimes_{i=1}^ke_i\tag2$$ and hence $$e\in\mathcal N(\lambda-S^{\otimes k})$$.

The problem with this is that the largest eigenvalue of $$|A|^{\otimes k}$$ is clearly $$\lambda_1(|A|)^k$$; not $$\prod_{i=1}^k\lambda_i(|A|)$$ as I would like. The crucial difference to what I remarked about the exterior product before is that the exterior product consists of alternating multilinear forms (and hence $$e_1\wedge\cdots\wedge e_1$$ cannot be an eigenvector of $$A^{\wedge k}$$).

So, we would need some kind of alternating tensor product. Is such a construction possible?

Remark: Note that the natural choice for the inner product on $$H^{\otimes2}$$ is given by the Hilbert-Schmidt tensor product; maybe we can consider some kind of closed subspace of alternating tensors of it.

EDIT: What I'm trying to find is an analogue of Proposition 3.2.7 in Random Dynamical Systems: • The proposition you quote is only stated in finite dimensions. Is it important to you to have a generalization that works in infinite dimensions, or is your question more about the algebraic construction (full tensor powers of a vector spaces versus its exterior powers)? – Yemon Choi Jun 4 at 13:22

If $$\sigma_1(B) = 0$$ then $$B = 0$$. Now pick your favorite $$2\times 2$$ matrix $$A$$ with $$\sigma_1(A) = 1$$, $$\sigma_2(A) = 0$$ and get a counterexample ($$A\otimes A \neq 0$$, but you want $$\sigma_1(A\otimes A) = 1*0 = 0$$).
• @0xbadf00d perhaps I do not understand what you are asking for, sorry. If you want the norm on $H\otimes H$ so that your conditions on the singular values are satisfied, then I gave an example showing that is not possible. If on the other hand you are OK with using some other space, then why $\bigwedge^2 H$ is not suitable for you? – Aleksei Kulikov May 5 at 12:43
• $\wedge^2H$ is suitable, but I wondered whether the same construction is possible using tensor products. It should be possible to construct an "alternating" tensor product as the subspace of $H^{\otimes2}$ spanned by the thensor $(x\otimes y-x\otimes y\rangle/2$. – 0xbadf00d May 5 at 13:16