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Unformatted text preview: MIT 18.335, Fall 2006: Review Problems, Solutions 1. (Trefethen/Bau 6.5) If P is nonzero, then there is an x such that Px is nonzero. For any such x , we have that bardbl Px bardbl 2 = bardbl P 2 x bardbl 2 ≤ bardbl P bardbl 2 bardbl Px bardbl ⇒ bardbl P bardbl 2 ≥ 1 If P is an orthogonal projector, its singular values are 1 and 0. In particular, σ 1 = bardbl P bardbl 2 = 1. Finally, if bardbl P bardbl 2 = 1 we show that P must be orthogonal. Consider two vectors u ∈ range( P ) and v ∈ null( P ). Decompose u orthogonally into a component along v and a rest r : u = v * u v * v v + r Since Pr = Pu = u and v,r are orthogonal we get bardbl Pr bardbl 2 2 = bardbl u bardbl 2 2 = vextendsingle vextendsingle vextendsingle vextendsingle v * u v * v vextendsingle vextendsingle vextendsingle vextendsingle 2 bardbl v bardbl 2 2 + bardbl r bardbl 2 2 But when bardbl P bardbl 2 = 1 it must also be true that bardbl Pr bardbl 2 2 ≤ bardbl P bardbl 2 2 bardbl r bardbl 2 2 = bardbl r bardbl 2 2 which is only possible if v * u = 0, that is, u and v are orthogonal and P is an orthogonal projector. 2. (Trefethen/Bau 7.2) Because of the special structure of A , the odd columns of ˆ Q will span the same space as the odd columns of A and similarly for the even columns. Therefore, an odd column of A is a linear combination of odd columns of ˆ Q (only), and similarly for an even column. This means that ˆ R will be an upper-triangular matrix with r ij = 0 when i + j is odd.is odd....
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This note was uploaded on 09/26/2010 for the course MECHANICAL 2.001 taught by Professor Prof.carollivermore during the Spring '06 term at MIT.
- Spring '06