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I'm reading Halmos's Finite Dimensional Vector Spaces, in which he makes several references to the infinite dimensional case. In my edition this item appears as question 6 at the end of section 14.

The proof for a finite dimensional space is fairly straightforward: y and z are linear functionals and span at most one dimensional subspaces of the dual space V', say Y and Z respectively. The annihilators Y$^0$ and Z$^0$ exist in the double dual V''. For a finite dimensional space V`` is "equal" (isomorphic) to V and the condition [x,z] = 0 $\implies$ [x,y] = 0, equates to Y$^0$ is a subspace of Z$^0$, and they are both either of the same dimension as V, or one less. This means that the vectors x (if any) for which [x,y] and [x,z] are non-zero exist in the same one dimensional subspace of V, so that if x$_0$ is a basis for this (1 dimensional) space and $\alpha $ = [x$_0$,y]/[x$_0$,z] then [x$_0$,z] $\alpha $ = [x$_0$,y], and if [w,y] is non-zero then w = $\beta$x$_0$ (because w is in the 1-D space with basis x$_0$) so that [w,y] = [$\beta$x$_0$,y] = $\beta$[x$_0$,y] = $\beta$[x$_0$,z] $\alpha $ = [$\beta$x$_0$,z] $\alpha $ = [w,z] $\alpha$

So, my question is, is this true in an infinite dimensional space ? I can't prove it or find a counter-example.

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  • $\begingroup$ It's true. See this. $\endgroup$ – David Mitra Dec 25 '14 at 13:46
  • $\begingroup$ @DavidMitra Thanks for the reference: it will take me some time to work through it. $\endgroup$ – Tom Collinge Dec 25 '14 at 13:57
  • $\begingroup$ Perhaps this is a better link. You need to know the codimension of a non-zero linear functional on a vector space is $1$. For that, see this. $\endgroup$ – David Mitra Dec 25 '14 at 14:09
  • $\begingroup$ @DavidMitra. A link in your original reference took me to math.stackexchange.com/questions/158173/… The accepted answer on this seems particularly simple (relying on every (infinite) vector space has a basis). Do you see any problem with it ? $\endgroup$ – Tom Collinge Dec 26 '14 at 8:46
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I just worked on this today, and my approach was much simpler. Let $x_1$ and $x_2$ be vectors such that $[x_1,z]$ and $[x_2,z]$ are nonzero. Let $\alpha_i = [x_i,y]/[x_i,z]$ for $i=1, 2$ and $\beta = [x_2,z]/[x_1,z]$, Consideration of $[x_2 - \beta x_1,z]$ shows $\alpha_1 = \alpha_2$. I may have messed up some details here - I am writing up my recollection of my solution.

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