Finding basis of w = (a,b,b) I have been asked to find the orthonormal basis for w = {(a,b,b) | a,b are Real numbers}.
I know that to find the basis of a matrix, we need to convert it into Row-Echelon-Form and choose rows from the original matrix that are Linearly Independent.
But since I have been given a vector I do not know how to do it.
Also the answer given is {(1,0,0),(0,1,1)}.
EDIT:
As pointed out by the comments, the solution given seems to be wrong.
The correct solution is {(1,0,0),(0,1/√2,1/√2)}.
 A: First you need a collection of independent vectors spanning the set $W$. For any vector of the form $(a,b,b)$, it can be written as
$$a(1,0,0)+b(0,1,1).$$
Once you show the vectors are linearly independent, you are done.
Find the rank of the matrix having these vectors as rows!!! And then you find the rank to be $2$.
Moreover, you need orthonormal. So check the inner product!! And you will find it to be zero, so orthogonal. However, to make them orthonormal, you need to normalize them. Finally, your orthonormal basis is given by
$$\{(1,0,0), \frac{1}{\sqrt 2}(0,1,1)\}.$$
I think, there is a typo in the given answer.
A: There is more than one orthonormal basis, since for instance given one you can always rotate it.
And the answer given is incorrect: the vector  $(0,1,1) $ has norm $\sqrt2$.  To rectify this multiply it by $\dfrac1{\sqrt2} $.
Since the given basis is orthogonal,  we are done.  It's clear that the space is  $2$-dimensional (two free variables).
A: After 3 months of studying Linear Algebra,I can now answer my own question:
First we need to find the basis using the variables:
a(1,0,0) + b(0,1,1).
We need to find the pivots.If number of pivots == 2 then they are independent vectors.
Also we see that the two basis vectors are orthogonal using dot product.
Hence all that is left is to divide by their magnitude to get the orthonormal vectors.
