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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.

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 $lVert u_3rVert=1$, then $(u_1,u_2,u_3)$ is an orthonormal basis. That was just the last thing that had to be checked.

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?

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 $1/lVertmathbf u'rVert$
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.