Answer Key for Exam #2

Transcription

1 Answer Key for Exam #. Use elimination on an augmented matrix: The corresponding system is x 7x + x, x + x + x, x x which we solve for the pivot variables x, x x : x +7x x x x x x x x x x x Therefore x 7 x x + x + x x x. We perform the eliminations A R. 7 A basis for the row space is the pivot rows of R, or of A. A basis for the column space is the pivot columns of A (but not of R). A basis for the nullspace can be found as in problem or by taking the negative of the upper right corner of R, putting a identity matrix below it, taking the two columns of that. So the only basis that requires more work is the left nullspace. To get it we transpose the pivot columns of A eliminate: 6. Here we can solve the corresponding system, or throw away the identity on the left, negate the rest, put a identity under it. We also have another basis for the column space in the rows of the last matrix above. In conclusion A row space basis is or

2 A null space basis is A column space basis is or A left null space basis is 6 The factored form of A that displays bases for all four is A ) ( 6 6. To see which vector to keep we start by computing all the dot products for the three vectors. If v v v then v v +, v v , v v 8 + 8, v v , v v , v v Recall that the projection of b onto a is b a a a a. If we take a v then the ratios will be 9 8 seems like a good one to keep. Then the projection of v onto v is v v v 9, v v 8 8 8, so v therefore v + e, where e is the error in the projection. We want to replace v by some multiple of e. We have e w, so we throw away v replace it by w. Next we do the same thing with v. The projection of v onto v is v v v 8, v v 8

3 therefore v + e, where e is the error in the projection, again we want to replace v by some multiple of e. We have e w, so we throw away v replace it by w. If we rename v as w, we now have w w w, where w w w w, but we probably don t have w w ; in fact, w w +, so we definitely don t have w w. To fix this we need to project one of w w onto the other one replace the projected vector by the error. We have w w 9 + w w , so it looks like keeping w might be slightly better. The projection of w onto w is therefore w w w w w w we want to replace w by a multiple of e. We have e +, + e,, which is the third vector we need. We now have that are all perpendicular to each other, we only have to fix the lengths. The dot product of the first vector with itself is, we did the other two earlier, so we finally get that an orthonormal basis for the subspace of R spanned by v, v v is. 8

4 Some other possible answers are Let A be the matrix with v v as columns. Then so A T A ( ),, 7 P, so we find that the projection matrix P onto the subspace S is We also have that P R P I

5 is the reflection matrix through S. The projection of v onto S is the reflection of v through S is P v 6 8, R v The projection is the average of the reflection v itself, this could have been used to avoid one of the last two matrix multiplications.. If P 8 then P is symmetric P P Any matrix P for which P P P T is a projection matrix. The trace of P is ( ), so the subspace T that P projects onto is -dimensional. Therefore any two rows or columns of P will be a basis for it as long as they are not multiples of each other. But since we also have to find a basis for T, which must be -dimensional, let s eliminate: P P projects onto its own column space, or its row space since P is symmetric. This gives us a nice basis for T, a basis for T comes from the null space of P : A basis for T is A basis for T is

6 6. We have the three conditions (i) P T P (P is symmetric) (ii) P P (iii) P T P P, we have to show that (i) (ii) together imply (iii), that (iii) implies both (i) (ii). That (i) (ii) together imply (iii) is easy: P T P P P by (i), which equals P by (ii). The trickiest part is to show (iii) implies (i). To see this transpose both sides of (iii) to get ( P T P ) T P T. But ( P T P ) T P T ( P T ) T P T P, so (iii) implies P P T P ( P T P ) T P T, which proves (i). So if (iii) holds then (i) does, then we can get (ii) easily from (i) (iii): P P P T P by (i), which equals P by (iii). 7. If M is a matrix which satisfies M M T, then M would be a projection matrix by the result used in problems 6 if M M M T. In other words, if M M T M is also symmetric, then M must be a projection matrix. Therefore, the question boils down to: are there any matrices M which satisfy M M T but are not symmetric? It turns out that there are exactly two such matrices. a b If M c d This gives us the four equations then M T a c b d M a b a b a + bc b(a + d) c d c d c(a + d) d + bc (i) a + bc a, (ii) b(a + d) c, (iii) c(a + d) b, (iv) d + bc d. Subtracting (iv) from (i) (iii) from (ii) we get (v) a d a d (vi) (b c)(a + d) c b. (vi) implies that either b c or a + d. If b c then M is symmetric, so to get a nonsymmetric M we must have a + d, in which case (ii) (iii) imply that b c. (v) implies that either a d or a + d, the latter is impossible since we already have a + d. So a d, since b c (i) (iv) both become + b( b), or b +. Therefore either b c, or c b. Thus the two M s which satisfy M M T but are not projection matrices are ( ) ( ). These are clockwise counterclockwise rotation matrices by, respectively. When they are squared they become rotations, which are rotations in the other direction. So the square equals the inverse for these matrices, since they are rotations the inverse also equals the transpose. Scores: The median is 8 the mean about 8.9, with a high of 96 a low of 69. Half of the scores were between 8 8.