# If [x,z] = 0 $\implies$ [x,y] = 0, then y = $\alpha$z. True for infinite dimensional vector space?

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.

• It's true. See this. – David Mitra Dec 25 '14 at 13:46
• @DavidMitra Thanks for the reference: it will take me some time to work through it. – Tom Collinge Dec 25 '14 at 13:57
• 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. – David Mitra Dec 25 '14 at 14:09
• @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 ? – Tom Collinge Dec 26 '14 at 8:46

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.