217 Orthonormal Basis Inquiry

Home , Orthonormality, Orthonormal basis

n 1. Deﬁne what it means that set of vectors {~v1, . . . ,~vm} in R is orthonormal. Why do n n orthonormal vectors in R always form a basis? n Solution note: Vectors {~v1, . . . ,~vm} in R are orthonormal if ~vi · ~vj = 0 for i 6= j and ~vi · ~vi = 1 for all i = 1, . . . m. A theorem in the book says that orthonormal vectors are linearly independent. n n Hence if there are n of them, they will be a basis for R , since R has dimension n. 3 2 5 2. Find an orthonormal basis of R that includes the vector 4 . Draw a picture. 5 3 −4 5 5 Solution note: 4 , 3 is one of two possible answers. The other multiplies 5 5 the second vector by −1.

 √  0 − 2 √2 3. Find an orthonormal basis for 3 that includes the vectors 0 ,  2 . R    2  1 0 √  2  √2 Solution note: Just add in  2  .  2  0

4. One advantage of an orthonormal basis B is that it is easy to ﬁnd the B-coordinates of a vector

~x. Why so? [Hint: if ~x = c1~v1 + ··· + cn~vn, what happens if you dot with ~vi? Does this work for non-orthonormal bases?]

Solution note: ci = ~x · ~vi.

5. Prove that if the set {~v1, . . . ,~vm} is orthonormal, then it is linearly independent. [Hint: There is a standard technique to start a proof of linear independence...]

Solution note: Suppose we have a relation c1~v1 + ··· + cm~vm = 0. Dot both sides with ~vi. This gives (c1~v1 + ··· + cm~vm) · ~vi = 0 · ~vi = 0

c1(~v1 · ~vi) + ··· + cm(~vm · ~vi) = 0

so using the fact that the ~vi are orthnormal to eachother, we have ci = 0, for each i. QED. 6. Find an orthonormal basis B for the plane V define by x + y − z = 0. After that, take some vector in the plane which is not parallel to either of your basis elements, and find its coordinates in B. √ √ T Solution note: We can pick any unit vector ~u1 = [ 2/2 0 2/2] in V for the first element in the basis. Now we need to find just one more element in the plane V which is perpendicular to this one (and then scale it down to be a unit vector). We need to T find ~w = [a b c] such that ~w · ~u1 = 0. Clearly this means that a = −c. But it also T has to lie on V , so b must be −2a. For example, ~w = [1 − 2 − 1] is in V , so ~u1, ~w is a basis for V . This is not quite orthonormal: the vectors are perpendicular, but the second one is not a unit vector. We√ fix this√ by dividing√ the second√ vector√ by its length. So an orthonormal basis is {[ 2/2 0 2/2]T , [1/ 6 − 2/ 6 − 1/ 6]T }.

Inquiry: What is the dimension of the orthogonal complement to a subspace?

No solutions for the rest of the worksheet because it is on the next worksheet.