Orthonormal base of the subspace

[Guest]

Considering the usual internal product of R4, find an orthonormal base of the subspace U = [(1,1,0,0), (0,2,1,0), (0,0,0,3)].

[HallsofIvy]

That's pretty straight forward "Gram-Schmidt" isn't it? [urlhttps://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process][/url]

Take any one of the vectors and divide by its length to get a unit vector:
The length of (0, 0, 0, 3) is 3 of course so a unit vector in its direction is (0, 0, 0, 1).
(That should have been obvious!)

Now find the projection of (0, 0, 0, 1) onto (1, 1, 0, 0). But that is very obviously 0- (0, 0, 0, 1) and (1, 1, 0, 0) have dot product 0*1+ 0*1+ 0*0+ 1*0= 0 so are already orthogonal. The length of (1, 1, 0, 0) is [tex]\sqrt{1^2+ 1^2+ 0^2+ 0^2}= \sqrt{2}[/tex] so [tex]\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0, 0\right)= \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0, 0 \right)[/tex] is a unit vector and (0, 0, 0 1) and [tex]\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0, 0 \right)[/tex] are two orthogonal unit vectors.

Use the third given vector, (0, 2, 1, 0), to find a vector that is orthogonal to both of those and divide by its length.