is this true for Hilbert space direct sum of $H$ when $H$ is infinite dimensional?

Question

Let $(H_{\alpha})_{{\alpha \in I}}$ be a $I-$indexed family of Hilbert spaces over $\mathbb{F}$.

let $H=\bigoplus H_\alpha$ be their Hilbert space direct sum. Can we say $\dim H=\sum\limits_{\alpha\in I} \dim H_\alpha$ if H is infinite dimensional?

thanks for your help

H is infinite dimensional, so dimH is infinite. Assuming that ∑_(α∈I)dimH_α is finite then gives a contradiction. — Urgje, Apr 13 '14 at 09:48
if we assuming that $\sum\limits_{(α∈I)}dimH_α$ is finite then what is the contradiction?!!! — joker, Apr 13 '14 at 10:46
∑_(α∈I)dimH_α is finite implies that dimH_α must be finite for each α. Secondly the set of α's must be finite. But then the direct sum ⨁H_α has finite dimension. — Urgje, Apr 13 '14 at 10:57
Sorry but i dont underestand why the set of α's must be finite! — joker, Apr 13 '14 at 11:09
Since for each α there is a Hilbert space of dimension at least one is added. — Urgje, Apr 13 '14 at 11:56

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

Yes we can say this for all no matter what $\operatorname{hilb.dim}(H)$ is. Let $S_i$ be a set of cardinality $\operatorname{hilb.dim}(H_i)$, then $H_i=\ell_2(S_i)$. From this post we know that $$ H=\bigoplus_{i\in I} H_i=\bigoplus_{i\in I}\ell_2(S_i)=\ell_2\left(\bigsqcup_{i\in I}S_i\right) $$ Hence $$ \operatorname{hilb.dim}(H)=\operatorname{Card}\left(\bigsqcup_{i\in I}S_i\right)=\sum_{i\in I}\operatorname{Card}(S_i)=\sum_{i\in I}\operatorname{hilb.dim}(H_i) $$

is this true for Hilbert space direct sum of $H$ when $H$ is infinite dimensional?

1 Answers1