# H is finite dimensional

Let $$H$$ be a Hilbert space and $$\lbrace x_ i \rbrace_ i \subset H$$ a countable set of vectors that algebraically spans $$H$$, i.e. every vector in $$H$$ is a linear combination of some finite subset of $$\lbrace x _i \rbrace_ i$$. Show that $$H$$ is finite dimensional.

My attempt:

In this set of vectors I can find a minimal spanning set, i.e basis of $$H$$, say $$\lbrace y_i\rbrace_i$$. suppose the basis is infinite. then take a non-zero element $$x\in H$$ then $$x=\sum_{i\geq1}a_iy_i$$. but $$x$$ can be written as linear combination of finitely many elements, say $$x=\sum_{k=1}^{n}b_ky_k$$. Therefore, $$0=\sum_{k=1}^{n}(a_k-b_k)y_k+\sum_{i\geq n+1}a_iy_i$$. So that implies the tail of the series converges to $$0\Longrightarrow x=0$$ (Contradiction.)

If this prove is correct then where are we really using the inner product of Hilbert space and the fact that Hilbert spaces are complete.

• Where is the contradiction coming from?
– daw
Feb 11, 2021 at 7:08
• Also you do not use any inner product (despite claiming so)
– daw
Feb 11, 2021 at 7:11

Let $$(y_i)$$ be a countable basis of $$H$$. Then using the Gram-Schmidt orthonormalization process we can construct an orthonormal basis $$(z_i)$$ of $$H$$: $$(z_i)$$ is a basis and $$\langle z_i,z_j\rangle = \delta_{i,j}$$ for all $$i,j\in \mathbb N$$.
Define $$u_n:=\sum_{k=1}^n \frac1k z_k$$. This is a Cauchy sequence in $$H$$. Hence converging to $$u$$. One can verify $$u=\sum_{k=1}^\infty \frac1kz_k$$ by looking at $$\langle u-u_n,z_k\rangle$$. Let $$v\in span(z_1\dots z_n)$$ with $$v=\sum_{k=1}^n v_kz_k$$. Then $$\|u-v\|_H^2 = \|\sum_{k=1}^n(\frac1k - v_k)z_k\|_H^2 + \|\sum_{k=n+1}^\infty \frac1kz_k\|_H^2 \ge \sum_{k=n+1}^\infty \frac1{k^2} >0.$$ Hence, $$u$$ cannot be a finite linear combination of the $$(z_i)$$'s. Contradiction.