I’m stumped– you give it a try

April 1st, 2008 ~ Posted in: Mathematics

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

  1. A bilinear functional on a hilbert space is said to be elementary if there are vectors x,y such that F(u,v) = \langle x, u \rangle \langle y, v \rangle. Suppose x,y,z are linearly independent vectors, what are necessary and sufficient conditions on w so that F(u,v) = \langle x, u \rangle \langle y, v \rangle + \langle z, u \rangle \langle w, v \rangle is elementary?
  2. 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 Q such that for all \alpha, Q^\star A_\alpha Q is upper triangular.
  3. The elementary tensors x\ocirc \cdots \ocirc x, with all factors equal, are all in the subspace \vee^k H. Do they span it?
  4. 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 H, 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.

Mathematics ~

2 Responses to “I’m stumped– you give it a try”

  • 1. George
    April 2nd, 2008 at 12:14 pm

    For the first of these, it may help to notice this:

    If (x,u)(y,v)+(z,u)(w,v)=(a,u)(b,v), then, for all u,v,

    ((a,u)b, v) = ((x,u)y + (z,u)w, v)

    and

    ((b,v)a, u) = ((y,v)x + (w,v)z, u).

    Therefore,

    (a,u)b = (x,u)y + (z,u)w

    and

    (b,v)a = (y,v)x + (w,v)z.

    If a and b are nonzero, then the first of these two shows that if u is perpendicular to both x and z, then it’s also perpendicular to a. So a=Ax+Bz. Similarly, b=Cy+Dw. It shouldn’t be too bad from here.

  • 2. Alex
    April 2nd, 2008 at 3:00 pm

    Thanks for the idea, George.

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