MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 2016

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1 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 Copyright c 26 UC Regents/Dimitri Shlyakhtenko Copying or unauthorized distribution or transmission by any means expressly prohibited

2 (i: T MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 2 Problem (True/False, pt each) Mark your answers by filling in the appropriate box next to each question F ) Any basis for R 3 consists of 3 orthonormal vectors (False): eg (ii: T F ) If A is a 4 4 orthogonal matrix, Ae e 2, Ae 2 e 3, Ae 3 e, then Ae 4 e 4 or Ae 4 e 4 (True): if w Ae 4 then w must be perpendicular to the other columns of A, ie, to e 2, e 3 and e so be a multiple of e 4 But since its length must be, w ±e 4 (iii: T F ) There are no vectors v, w in R 7 so that v w 7 and v w 2 (True): The Cauchy-Schwartz inequality states that v w v w which would mean which cannot be true (iv: T F ) (A t x) y x (Ay) for any vectors x, y and any matrix A (here denotes the dot product) (True): if A has entries a ij, x has entries x i and y has entries y i then A t x y ij y j a ij x i ij a ij x i y j while x (Ay) ij x i a ij y j ij x i y j a ij (v: T F ) A square orthogonal matrix is invertible (True) Since A t A I, A cannot have any kernel, so by rank nullity has full rank, so is invertible (vi: T F ) If A t A I then A is an orthogonal matrix (True) if A has columns v,, v n then the i, j-th entry of A t A is exactly v i v j ; so A is orthogonal iff v i v j is zero for i j and if i j ie, A t A I (vii: T F ) Every basis for the plane x+y+z consists of two vectors (True) the dimension of the plane is 2 (viii: T F ) The columns of an orthogonal matrix are orthonormal (True) this is the definition of an orthogonal matrix (ix: T F ) The determinant of an invertible matrix is nonzero (True) det I det AA (det A) (det A ) so det A (x: T F ) For any matrix A, the kernel of A is perpendicular to the image of A t (True) see book,,

3 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 3 Problem 2 ( pts) Let P be the subspace of R 4 consisting of vectors x+y w and x+z w (a) Find a basis for P Note that by definition P consists of those vectors which satisfy which means that with Row-reducing A x+y+w, x+z+w A P ker A The last two columns correspond to free variables so that we have: x y z a + b w so that and vectors must lie in P and be linearly independent x y z w satisfying form a basis Note: other answers are possible; the two (b) Find the dimension of P Since there are two basis vectors, the dimensino is 2 (c) Find an orthonormal basis for P Using Gram-Schmidt we get: w w 2 v 2 (v 2 u )u, u 3 3 2/3 /3 /3

4 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 4 which gives the answer u 2 5 3, 5 Other answers are possible: the two vectors must lie in the plane and be orthonormal

5 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 5 Problem 3 ( pts) Let A 2 2 Diagonalize A; in other words, find a basisbin which the matrix A is diagonal There are several approaches to this problem Approach : ( Brute force ) We are given that E T E A and we seek a basis B so that BT B is diagonal Thus we want BT B B S E A E S B to be diagonal; in other words, if we write S B S E, then we seek a matrix S v v 2 a b (here v c d, v 2 is our desired basis) so that S p AS q Thus we seek a, b, c, d, p, q so that S p AS This equation can be rewritten as q p AS S q so that 2 2 giving us From this we obtain: a b a b c d c d 2a+c 2b+d a+ 2c b+2d Adding the first two equations gives 2a+c pa a+ 2c pc 2b+d qb b+2d qd p q pa qb pc qd 3a+ 3c pa+ pc p(a+c) so that either a c or p 3 If p 3, we can subtract the second equation from first to get a c p(a c) 3(a c) so a c Since S is invertible, det S, so it must be that b d Now subtract the last equation from the first to get b d q(b d) which shows that q ; adding the last two equations finally gives that 3b+3d q(b+d) b+d so that b d This gives a b S a b

6 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 6 so that our desired basis can be taken to be, (or in fact any two vectors of the a b form, with both a and b nonzero) One can directly verify that this works a b The choise a c in a similar way leads to p, q 3 and another basis,, a b (or, with a, b both nonzero) a b Some simplifications along the way: you can notice that det S AS det A which gives you immediatley pq 3; also you can rescale S by any nonzero number and thus assume eg that det S Approach 2: ( Transformations ) Let T be the transformation so that its matrix with p respect to the usual basis is A We seek a basis B v v 2 so that B T B q p By definition, this means that the B-coordinates of Tv are, so that Tv pv Similarly, Tv 2 qv 2 So we seek two vectors v, v 2 so that Tv is a multiple of v and Tv 2 is a multiple of v 2 3 At this point we can see by inspection that T 3 while T 3 so that v and v 2 work If you didn t notice these vectors right away, you can rewrite the equation Tv pv as (T pi)v ie you seek p so that T pi has a nozero kernel, ie, has zero determinant (by ranknullity) But ( ) 2 p det(t pi) det 2 p 2 p det (2 p) 2 p 2 4 4p+ p 2 p 2 4p+3 which implies that p 3 or p Now we can find v and v 2 by solving Tv 3v, Tv 2 v 2 2 or equivalently v is in the kernel of 3, eg v 2 2 and v 2 is in the kernel of, eg v 2 2,

7 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 7 Problem 4 ( pts) (a) Suppose that a, b, c are three vectors in R 4 Let p b(a c) c(a b) Show that p a We compute a p : so a p so a p a p a (b(a c) c(a b)) (a b)(a c) (a c)(a b), (b) Suppose that u, v are two vectors in R 7 so that u v v v u u 2 Prove that u v Let us compute the length of u v: u v 2 (u v) (u v) u u u v v u+v v So u v so u v Another solution: let us compute the angle between u and v: cos θ u v 2 u u v v 2 2 so that θ But since u u v v 2, u and v has the same length ( 2) and are thus equal Note: the solution would not change at all if u, v were vectors in any dimension

8 MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 26 8 Problem 5 ( pts) LetBbe the basis for R 2 given consisting of v, v (a) Find the coordinates of the vectors w and w 2 2 in the basis B 27 We write w B B S E w E ( E S B ) w E Alternatively, you can readily see that w 2v + v 2 Similarly, w 2 B / /2 Alternatively, you can readily see that w v 2 v 2 (b) Let T be the transformation of R 2 given by Te e + e 2, Te 2 e e 2, where e, e 2 are the standard basis of R 2 Find the matrix of T in the basisb Solution : We immediatley see that Te, Te 2 so that E T E BT B B S E E T E E S B ( E S B ) ET E E S B 2 Solution 2: Since v e + e 2 we get that Tv Te + Te 2 e + e 2 + e e 2 v + v 2 while Tv 2 Te Te 2 e + e 2 (e e 2 ) v v 2, which directly gives BT B