Fix some some double sequence $(y_{n,m})_{n,m\in \mathbb{N}}\subset \mathbb{C}$ with $\sum_{n,m}|y_{n,m}|^2<\infty.$ Let $$A = \{x\in\ell^2: \sum_{n=1}^\infty y_{n,1}x_n = 0 \,\,\wedge\,\, \sum_{n=1}^\infty y_{n,m+1}x_n = -x_m\,\,\forall m\geq 1 \},$$ $$B = \{x\in\ell^2: \sum_{n=1}^\infty y_{m,n}x_n = -x_{m+1}\,\,\forall m\geq 1\}.$$ I want to show that $\text{dim } A = \text{dim } B - 1.$ I plan to do it in two steps:
Show that $A'= \{x\in\ell^2: \sum_{n=1}^\infty y_{n,m+1}x_n = -x_m\,\,\forall m\geq 1\}$ has the same dimension as $B$ via some clever isomorphism.
Show that intersecting $A'$ with $\{x\in\ell^2: \sum_{n=1}^\infty y_{n,1}x_n = 0\}$ reduces the dimension by $1.$
I am stuck on both steps. I think I'm missing some insight on what $A'$ and $B$ look like and how they depend on the choice of $y.$ I think it might be better to replace $y$ in the definitions of $A$, $A'$ and $B$ by the complex conjugate $\bar{y},$ to use tools from Hilbert spaces, since we can then write these sets as orthogonal complements, but even then I am still stuck. Any help with either step would be much appreciated.
This is false. If $y_{n,m}=0$ for all $n,m$ then $A =\ell^{2}$ and $B$ is one dimensional.