Vectors Question (Orthogonality)

AI Thread Summary
The discussion revolves around the application of the Gram-Schmidt process to the vectors u1, u2, and u3 to create an orthogonal basis for R3. The first vector v1 is correctly identified as [2, -2, 2], and v2 is calculated as [-1, 1, 2]. However, the attempt to calculate v3 resulted in an incorrect vector of [1, 2/3, 4/3], which did not satisfy the orthogonality condition with the previously calculated vectors. Participants emphasize the importance of checking the inner products to ensure orthogonality and suggest reviewing the calculations for potential errors. The discussion highlights the need for careful application of the Gram-Schmidt formulas to achieve the correct orthogonal basis.
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Homework Statement



Let u1 = [2 -2 2 ] u2 = [-2 2 1], u3 = [0 1 2]Use the Gram-Schmidt process to u1, u2, u3, in this order. The resulting vectors are:

v1 = [___ ___ ___], v2 = [___ ___ ___], v3 = [___ ___ ___]

And ß = {v1, v2, v3} is an orthongal basis for R3.

Homework Equations



v1 = u1

v2 = u2 - u2 . v1 v1
------------ v1 . v1
v3= u3 - u3 . v1 v1 - u3 . v2 v2
---------------v1 . v1 v2 . v2

The Attempt at a Solution



Well, the first one is easy.

v1 = [2 -2 2]

I solved v2 and got [-1, 1, 2] which was also right.

I tried v3 based on the equation and got [1, 2/3, 4/3] which was apparently incorrect.
 
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Why do you say "apparently" incorrect? You should have checked yourself that <2, -2, 2].[1, 2/3, 4/3]= 2- 4/3+ 8/3= 2+ 4/3 and [-1, 1, 2].[1, 2/3, 4/3]= -1+ 2/3+ 8/3= -1+ 10/3, neither of which is 0.

If you show what you did, that is show use exactly how you used that formula, we might be able to point out an error.
 

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