# Using Least Squares to find Orthogonal Projection

• ver_mathstats
In summary, the student is trying to solve a problem involving vectors and matrices, but is having difficulty. They took their vectors and set up a matrix, but are now needing to solve for the orthogonal projection. They are able to do this by using QR factorization and solving for the projectoin vector.f

#### ver_mathstats

Homework Statement
Use least squares to find the orthogonal projection of u onto the subspace of R4 spanned by the vectors v1, v2, and v3.
Relevant Equations
u = (0,5,4,0) v1 = (6,0,0,1) v2 = (0,1,-1,0) v3 = (1,1,0,-6)
I'm a little confused how to do this homework problem, I can't seem to obtain the correct answer. I took my vectors v1, v2, and v3 and set up a matrix. So I made my matrix:

V = [ (6,0,0,1)T, (0,1,-1,0)T, (1,1,0,-6)T ] and then I had
u = [ (0,5,4,0) T ].

I then went to solve using least squares. So I ended up doing VTV = VTu.

I took that, obtained my augmented matrix:

[ (37,0,0)T, (0,2,1)T, (0,1,38)T, (0,1,5)T ].

I solved it and everything worked fine, I got the answers: x1 = 0, x2 = 11/25, x3 = 3/25.

I noticed to that I could use QR factorization of V then solve QTu and the least squares solution would be x' satisfies Rx = QTu, we can then obtain our answer, but again this solution only has three components, why is the solution asking for four?

However, the solution requires four components. I am unsure of where I am going wrong with this problem, any help would be appreciated.

Homework Statement:: Use least squares to find the orthogonal projection of u onto the subspace of R4 spanned by the vectors v1, v2, and v3.
Relevant Equations:: u = (0,5,4,0) v1 = (6,0,0,1) v2 = (0,1,-1,0) v3 = (1,1,0,-6)

I'm a little confused how to do this homework problem, I can't seem to obtain the correct answer. I took my vectors v1, v2, and v3 and set up a matrix. So I made my matrix:

V = [ (6,0,0,1)T, (0,1,-1,0)T, (1,1,0,-6)T ] and then I had
u = [ (0,5,4,0) T ].

I then went to solve using least squares. So I ended up doing VTV = VTu.

I took that, obtained my augmented matrix:

[ (37,0,0)T, (0,2,1)T, (0,1,38)T, (0,1,5)T ].

I solved it and everything worked fine, I got the answers: x1 = 0, x2 = 11/25, x3 = 3/25.

I noticed to that I could use QR factorization of V then solve QTu and the least squares solution would be x' satisfies Rx = QTu, we can then obtain our answer, but again this solution only has three components, why is the solution asking for four?

However, the solution requires four components. I am unsure of where I am going wrong with this problem, any help would be appreciated.
You have to find the orthogonal projection of u onto the subspace of R4 spanned by the vectors v1, v2, and v3. x1, x2, x3 you got are the componets of this projection vector P with respect to the basis v1, v2, v3, that is
P= x1 v1 +x2 v2+x3 v3 , a linear combination of three four-dimensional vectors .

• andrewkirk