ChapterZero

I’m stumped– you give it a try

I’m going to add problems as I get stumped.

A bilinear functional on a hilbert space is said to be elementary if there are vectors such that $F(u,v) = \langle x, u \rangle \langle y, v \rangle.$ Suppose are linearly independent vectors, what are necessary and sufficient conditions on so that $F(u,v) = \langle x, u \rangle \langle y, v \rangle + \langle z, u \rangle \langle w, v \rangle$ is elementary?
Let $\{ A_\alpha \}$ be a family of mutually commuting operators. Then there is a common Schur basis for $\{ A_\alpha \}$ – that is, there exists a unitary such that for all $\alpha$ , $Q^\star A_\alpha Q$ is upper triangular.
The elementary tensors $x\ocirc \cdots \ocirc x$ , with all factors equal, are all in the subspace $\vee^k H$ . Do they span it?
If $\dim H = 3$ , then $\dim \otimes^3 H = 27$ , $\dim \wedge^3 H = 1$ and $\dim \vee^3 H = 10$ . In terms of an orthonormal basis of , write an element of $(\wedge^3 H \oplus \vee^3 H)^\perp$

On the 4th question, I could construct one such element, but I’m trying to get the form of a general element. I’m stuck on finding a basis for $(\wedge^3 H)^\perp$ – so far I know $e_i \otimes e_j \otimes e_k$ are in it if i,j,k aren’t distinct, and $e_i \otimes e_j \otimes e_k - e_j \otimes e_k \otimes e_i$ are in it if i,j,k are distinct–, and I imagine finding a basis for $(\vee^3 H)^\perp$ is even more painful.

Possibly relevant posts:

Inner products (6/7/2006)
Dr. Singh’s Take on the Fourier Unicity Theorem (via Invariant Subspaces) (6/17/2005)
Another sign matrix problem (3/26/2010)

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I’m stumped– you give it a try

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