Linear algebra: Gram-schmidt process
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Linear algebra: Gram-schmidt process

[From: ] [author: ] [Date: 12-06-11] [Hit: ]
c2,and your third orthonormal vector is c = [1/3, 2/3,(note that you could have let c2 = c3 be anything and solved for c1.........
Find an orthonormal basis for the Euclidean space R3 that contains the vectors [2/3; -2/3; 1/3] and [2/3; 1/3; -2/3]

answer: [1/3; 2/3; 2/3]

How do you solve this question? All I know is that b and c has to be the same, which was derived from doing the dot product of the first vector with the unknown vector, and the second vector with the unknown vector. Can someone please help me?

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The two vectors you show, call them a and b are already orthonormal so all you have to do is find one more vector that is orthogonal to both of these. Use can use Gram-Schmidt but that is overkill here. Just let you vector c = [c1, c2, c3] and do the dot product with a and b and we get:
2 c1 - 2 c2 + c3 = 0
2 c1 + c2 -2 c3 = 0
(where I multiplied out the denominator 3 to make it simple)
subtracting these two equations gives:
-3 c2 + 3 c3 = 0 or c2 = c3
let c2 = c3 = 2/3
then from the first equation above we substitute c2 and c3 and we have
then c1 = (2/3) - 1/3 = 1/3
and your third orthonormal vector is c = [1/3, 2/3, 2/3]

(note that you could have let c2 = c3 be anything and solved for c1...and then normalize c)
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