math54 - quiz 10- solns

# math54 - quiz 10- solns - 1 √ 2-1 √ 2 1 √ 2 1 √ 2 1...

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Math 54 Quiz 10 Solutions 1. (3 pts) Find the orthogonal projection of h 1 0 1 i onto the subspace of R 3 spanned by h 1 1 0 i , and h 0 0 1 i . Notice that the basis of the subspace given is orthogonal. So, the projection onto the subspace is the sum of the projections onto each orthogonal basis element: h 1 0 1 i · h 1 1 0 i h 1 1 0 i · h 1 1 0 i 1 1 0 + h 1 0 1 i · h 0 0 1 i h 0 0 1 i · h 0 0 1 i 0 0 1 = 1 / 2 1 / 2 1 2. (4 pts) Find an orthonormal basis for the subspace of R 4 spanned by v 1 = 0 1 - 1 0 , v 2 = 1 2 1 1 , v 3 = 0 1 1 0 Noticing that v 1 and v 3 are already orthogonal, let’s use them as the ﬁrst two vectors in our orthogonal basis. Next we ﬁnd the third vector by Gram Schmidt; projecting away the parts of v 2 that point along v 1 and v 3 . u 3 = v 2 - v 1 · v 2 v 1 · v 1 v 1 - v 3 · v 2 v 3 · v 3 v 3 = 1 2 1 1 - 1 2 0 1 - 1 0 - 3 2 0 1 1 0 = 1 0 0 1

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Normalizing our vectors, we get
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Unformatted text preview: 1 / √ 2-1 / √ 2 , 1 / √ 2 1 / √ 2 , 1 / √ 2 1 / √ 2 3. (3 pts) Suppose that ~v is an eigenvector of some matrix A with eigen-value λ . Suppose also that A~v is orthogonal to ~v . What can you say about λ ? We have that A~v = λ~v , and that h A~v,~v i = 0. Putting these together, we get that 0 = h A~v,~v i = h λ~v,~v i = λ h ~v,~v i Now, since ~v is an eigenvector, it is nonzero, so its length is nonzero. Above, we showed that λ times the length squared is zero, so λ must be zero....
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## This note was uploaded on 09/17/2011 for the course MATH 54 taught by Professor Chorin during the Spring '08 term at Berkeley.

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math54 - quiz 10- solns - 1 √ 2-1 √ 2 1 √ 2 1 √ 2 1...

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