This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Hour Exam II â€“ Mathematics 33A 1. Find the matrix of 3 4 4 3 Â¶ with respect to the basis { â€¢ 4 2 â€š , â€¢ 2 4 â€š } . Solution: Since the matrix S taking coordinates in the new basis to coordinates in the standard basis is 4 2 2 4 Â¶ , this just amounts to computing 4 2 2 4 Â¶ 1 3 4 4 3 Â¶ 4 2 2 4 Â¶ = 1 20 4 2 2 4 Â¶ 3 4 4 3 Â¶ 4 2 2 4 Â¶ = 5 5 Â¶ If you happened to look at what 3 4 4 3 Â¶ does to the basis vectors, you were pleasantly surprised to find 3 4 4 3 Â¶â€¢ 4 2 â€š = 5 â€¢ 4 2 â€š and 3 4 4 3 Â¶â€¢ 2 4 â€š = 5 â€¢ 2 4 â€š That immediately says that the matrix with respect to this basis is 5 5 Â¶ . 2. (a) (5 pts.) Find the matrix of the orthogonal projection onto the orthogonal complement of the plane 2 x 1 x 2 + 2 x 3 = 0. Discussion. This question is very easy and very hard at the same time. What makes it hard is that you have probably lost track of what you once knew about planes. The plane 2 x 1 x 2 + 2 x 3 = 0 goes through (0,0,0) and has the normal vector (2,1,2). So the line (2t,t,2t) is perpendicular to the plane, and goes through (0,0,0). That line is the orthogonal complement to the plane. To get the projection on the line you need the unit vector ~u = (2 / 3 , 1 / 3 , 2 / 3) in the direction of the normal. Then the projection is 2 P ( ~x ) = ( ~x Â· ~u ) ~u = u 2 1 u 1 u 2 u 1 u 3 u 2 u 1 u 2 2 u 2 u 3 u 3 u 1 u 3 u 2 u 2 3 ~x = 4 / 9 2 / 9 4 / 9 2 / 9 1 / 9 2 / 9 4 / 9 2 / 9 4 / 9 ~x. Note that this is symmetric: matrices of orthogonal projections are always sym metric. It also has rank 1. Projections onto lines have one dimensional images, and therefore have to have rank 1....
View
Full Document
 Fall '08
 lee
 Math, Linear Algebra, Linear combination, orthogonal projection

Click to edit the document details