Equality of dimensions of two space Let $T:H_1 \to H_2$ be a bijective bounded linear operator between Hilbert spaces $H_1$ and $H_2$. Do the algebraic dimensions of  $H_1$ and $H_2$ coincide? I think the answer is positive but I can not prove it. Please give me any regard. Also, please introduce any references about dimension of spaces.
 A: If $H_1,H_2$ are finite dimensional this is a well known result in linear algebra.
If they're infinite dimensional then you need to define those notions. The most "natural" way to do that is to ask whether there is a bijection between a basis of $H_1$ and a basis of $H_2$.
The answer is yes: Let $\{e_i:i\in I\}$ be an algebraic basis for $H_1$ (which exists using the axiom of choice), we claim that $\{T(e_i):i\in I\}$ is a basis of $H_2$.
Proof: let $h_2\in H_2$ as $T$ is a bijection there exists $h_1\in H$ such that $T(h_1)=h_2$. As $\{e_i:i\in I\}$ an algebraic basis for $H_1$ there exists finitely many indices $e_{i_1},...,e_{i_n}$ such that $h_1=\sum_{k=1}^n c_{i_k}e_{i_k}$ and therefore $T(h_1) = \sum_{k=1}^n c_{i_k}T(e_{i_k})$ we conclude that $\{T(e_i):i\in I\}$ is a generating set.
It is clearly minimal, as if we can remove $T(e_i)$ for some $i$ then $T(e_i)$ is a linear combination of some $T(e_{j_1}),...,T(e_{j_m})$ but then since $T$ is injective we have that $e_i$ a linear combination of $e_{j_1},...,e_{j_m}$ which contradicts the fact that $\{e_i:i\in I\}$ is minimal.
We conclude that $\{T(e_i):i\in I\}$  is a minimal generating set, hence a basis.
