I came to know without proof or explanation that smallest non-trivial Hilbart space is generated by two basis vectors. What is its proof?
One example I know. Denote $a = (0 , 1)$ and $b = (1 , 0)$. then the space generated by $a , b$ with proper inner product shall be such a Hilbert space. I am also looking for another example.
Thank you for your help.
Hilbert spaces are vector spaces, i.e. modules over a field, with an inner product. So the "smallest" must have smallest possible dimension for certain, if not a proper subspace exists: i.e. this "smallest" space is just a field.
But then, the cardinalities of fields can go all the way down to $2$--if you allow finite fields--whence the answer would be $\Bbb F_2$, the field with two elements and the usual trivial absolute value for the norm. It's compact, hence complete. And the inner product is really dumb, but it works.
If you are one of those wet-blankets who demand that your complete, normed spaces be vector spaces over $\Bbb R$ or $\Bbb C$, then $\Bbb R$ with multiplication should rightly be called the smallest Hilbert space if you want just "complete, inner-product space" and $\Bbb C$ if you demand your spaces be properly Hermitian.
Note: Most definitions I've seen demand Hilbert spaces be Banach, so inherently they're real or complex spaces, but the nature of your question made me think you might just be asking about complete inner product spaces, so the second paragraph is to address that.