Is there exists linear algebra basis for $L^2[0,1]$ such that every element of it has length one and every two different element of it is orthogonal? We know by using axiom of choice every vector space over a division ring ( consequently any field ) has a basis like $\mathbb E$ in the meaning of linear algebra ( $\mathbb E$ is linear independent set and the set of all finite linear combination of elements of $\mathbb E$ is entire space). In infinite dimensional case this basis is not appropriate for analysis and analytical purposes because this basis has been obtained from axiom of choice and it is not constructive , so a new concept of basis has been defined as follow: 
Suppose we have an infinite dimensional vector space over $\mathbb C$ which has a norm, now we say $\mathbb M$ is basis (in new definition ) when $\mathbb M$ is linear independent set and the set of all infinite linear combination of elements of $\mathbb M$ is dense ( respect to norm ). So every basis in the meaning of linear algebra is a basis in new definition.
Consider Hilbert space $L^2[0,1]$ with usual inner product, we know it has basis respect to both two definition above. and even it has a basis in new definition such that norm of every element of it is one and every two different element of it is orthogonal ( respect to inner product ) .
Now I think $L^2[0,1]$ hasn't a basis in the meaning of linear algebra such that norm of every element of it is one and every two different element of it is orthogonal ( respect to inner product ).  
Is this true ?
 A: No such basis exists.
What you call a "linear algebra basis" is more commonly called a Hamel basis in this setting.  If $H$ is any infinite-dimensional Hilbert space, then it does not have a Hamel basis whose elements are pairwise orthogonal.
Let $B \subset H$ be an infinite set whose elements are pairwise orthogonal; we will show $B$ does not span $H$.  Without loss of generality, assume $0 \notin B$.  Choose countably many elements $x_1, x_2, \dots$ from $B$ and let $$x = \sum_{i=1}^\infty \frac{1}{i} \frac{x_i}{\|x_i\|}.$$ (It is a standard exercise to use completeness to show the series converges in $H$.)  Now let $\{y_1, \dots, y_n\}$ be any finite subset of $B$.  Choose one of the $x_i$ that is not in $\{y_1, \dots, y_n\}$.  Then $x_i$ is orthogonal to any finite linear combination of the $y_j$, but using the continuity of the inner product, $\langle x, x_i \rangle = \frac{\|x_i\|}{i} \ne 0$.  Hence $x$ is not a finite linear combination of $y_1, \dots, y_n$.  But $y_1, \dots, y_n$ were arbitrary, so $x$ is not in the linear span of $B$.
