Finding norm of orthonormal basis? I'm sorry i'm new here. I uploaded a pictures in order to make things simpler. 
I have three linearly independent vectors:
v1= (1,1,0,0) v2=(1,-1,0,0) and v4=(0,2,0,0).
As you may see from the picture, the result for u3 is equal to the vector (0,0,1,0). 
It says that since the norm of u3 is 1, which i found by square rooting the entries in this matrix, then the set {u1, u2, u3} is an orthonormal
basis of Span(v1, v2, v4).
I do not understand this statement.
Can somebody please explain why this is so? I do not understand the relation between this result, 1, and how it implies that {u1, u2, u3} is an orthonormal basis. 
Thanks for the help, and sorry again for uploading the pic.
(please give me a quite simple explanation . I have a form dyscalculia and it takes me a while to understand these processes.
 A: Asserting that $(u_1,u_2,u_3)$ is an orthonormal basis of that space consists in checking several things:


*

*each $u_k$ has norm $1$;

*each two distinct vectors are orthogonal.


Whover wrote this is not claming that just because $\lVert u_3\rVert=1$, then $(u_1,u_2,u_3)$ is an orthonormal basis. That was just the last thing that had to be checked.
A: It is an application of G-S process. Since $v_1$ and $v_2$ are orthogonal, we have already $2$ vectors for the orthonormal basis.
Then we proceed with the orthogonalization of $v_4=(0,2,1,0)$ with respect to $u_1$ and $u_2$.
By the construction the span of the two sets of vectors are identical.
Refer also to the related


*

*Gram Schmidt-The arts behind it

*Find orthonormal vectors for given vectors using Gram-Schmidt

*Is there a more convenient method for converting a base to be orthogonal than Gram Schmidt?
A: As noted in the other answers and their discussions,
to show that a set of vectors is an orthonormal basis you have to show that each pair of vectors is orthogonal and each vector has unit length.
A useful fact about the Gram-Schmidt process, which the quoted material employed,
is that if you give it any set of $n$ vectors as input, there are only two possible
outcomes:
either at some point in the process one of the $\mathbf u'$ vectors turns out to be the zero vector,
or you get a set of $n$ vectors that is an orthonormal basis of the space spanned by your input vectors.
Unless you encounter a zero vector,
each pair of vectors is automatically orthogonal due to the formulas that produced each vector,
and of course scaling each vector $\mathbf u'$ by $1/\lVert\mathbf u'\rVert$
guarantees that each vector in the result has unit length.
In short, it is mathematically impossible for Gram-Schmidt to turn $n$ input vectors into $n$ non-zero output vectors that are not an orthonormal basis.
In the quoted material, however, there are a couple of places where serendipity simplifies the calculations.
One such place is when computing $\mathbf u_2',$
where it turns out that $\mathbf v_2$ is already orthogonal to $\mathbf u_1'.$
The other place is where it turns out that $\mathbf u_3'$ already has unit length.
But if we ignored these two facts and just blindly applied Gram-Schmidt,
we would already know that all the checks for orthogonality and unit length would be satisfied.
