Orthagonal Sets Homework: Gram-Schmidt Algorithm in R4

[theRukus]

Homework Statement

Use the Gram-Schmidt algorithm to convert the set S={x₁, x₂, x₃} to an orthagonal set, given x1 = [1 1 1 1]^T, x₂ = [6 0 0 2]^T, x₃ = [-1 -1 2 4]^T.

Homework Equations

The Attempt at a Solution

I've used the algorithm to come up with the set of vectors {[1 1 1 1]^T, [-2 -2 1 3]^T, [10/3 -8/3 -5/3 3]^T. I've triple checked that I have executed the algorithm correctly.

My first two vectors are orthagonal; their dot product is zero. The dot product of third vector with either of the other two vectors is non-zero. Is this an orthagonal set? By definition I'm assuming that it's not, but is there some way that a set of 3 vectors in R4 can be orthagonal without all three vectors themselves being orthagonal... ? I highly doubt it.. but can someone provide some insight?


Thank you

 

[micromass]

(Hint: the correct word is orthogonal, not orthagonal)

All three vecters should be pairswise orthogonal. I'm guessing that the last coordinate of the third vector should be 1...
Can you present how you found the third vector? Maybe we can find the mistake...

 

[theRukus]

When finding the third vector, I already have set U = { [1 1 1 1]^T, [-2 -2 1 3]^T } as my SO FAR orthagonal set.

Use vector [6 0 0 2] from original set, and subtract from it its projection on the vectors in U:

[6 0 0 2]^T - proj_{[1 1 1 1]T}( [6 0 0 2]^T ) - proj_{[-2 -2 1 3]}( [6 0 0 2]^T )

= [6 0 0 2]^T - (8/4)[1 1 1 1]^T - (-6/18)[-2 -2 1 3]^T

= [6 0 0 2]^T - [2 2 2 2]^T - [2/3 2/3 -1/3 -3]^T

= [10/3 -8/3 -5/3 3]^T


Can anyone find an error here?

Thank you for your help

 

[theRukus]

Haha. Uhm. Cancel that, I've found my error. It's quite obvious as well. I'll be more careful in my checking next time. Thank you for the help micromass!