= c. = c. c 2. We can find apply our general formula to find the inverse of the 2 2 matrix A: A 1 5 4

Transcription

1 . In each part, a basis B of R is given (you don t need to show B is a basis). Find he B-coordinate of the vector v. (a) B {, }, v Solution.(5 points) We have: + Therefore, the B-coordinate of v is equal to. (You would lose point if you don t give as the final answer or give answer.) (b) B {, }, v 5 Solution.( points) We need to find c and c such that: c + c 5 as the final () That is we want to solve the following system: } 5 {{ } A c c () We can find apply our general formula to find the inverse of the matrix A: A 5 Consequently: 5 c }{{} A c We can find apply our general formula to find the inverse of the matrix A: A 5 Therefore, the B-coordinate of v is equal to (5 points for forming the system () or (). 5 points for solving the system.).

2 . Find bases for the image and the kernel of the linear transformation defined by the following matrix: A 5 7 Solution.(5 points) We firstly find the RREF of A. (I) (I) (II) (II) Because there are leading ones in the first two columns of the RREF of A, the first two columns of A form a basis for the image of A. That is to say, a basis for the image of A is given by the following vectors: 5 We can also use the RREF of A to find any vector in the kernel. Any such vector has the following form: t t t t Therefore, the following vector gives a basis for the kernel of A; (5 points for finding the RREF, 5 points for finding a basis for the image and 5 points for finding a basis for the kernel. You would lose points if you make an arithmetic mistake in computing RREF. You would lose points if you give the first two columns of the RREF of A (instead of the first two columns of A) as a basis for the image of A.)

3 . Determine whether each statement below is correct or not. You don t need to justify your answer; just give a yes or no answer. (a) Let C(R,R) be the vector space that consists of continuous functions f : R R. Define W as: Is W a vector subspace of C(R,R)? W {f C(R,R) f() } Solution.( points) No, because the function f(x) is not an element of W. (b) Let W R be defined as: Is W a vector subspace of R? W {(x,y,z) R x y z} Solution.( points) No, because, for example, (,,) is in W while (,, ) (,,) is not an element of W. (c) Let P be the vector space of degree at most polynomials and B be the basis for P that consists of, x, and x. Is: T : P R an isomorphism? T (Q) Q B Solution.( points) Yes. In general given a basis B for a vector basis, the B-coordinate map is an isomorphism. (d) Let l be the linear space of all infinite sequences of numbers where addition and scalar multiplication is defined as: (x, x, x,... ) + (y, y, y,... ) : (x + y, x + y, x + y,... ) Consider the map T : l l : Is the vector (,,,,... ) in the image of T? k(x,x, x,... ) : (kx, kx, kx,... ) T (x, x, x,... ) : (, x, x, x,... ) Solution.( points) Yes. For example T (,,,,... ) (,,,,... ). (e) Let T be defined as in the previous part. Is the vector (,,,,... ) in the kernel of T? Solution.( points) No. T (,,,,... ) is equal to (,,,,... ) which is not zero.

4 . In each part, is the given set of vectors a basis for R? If not, is it possible to find a subset of the given vectors to form a basis? Justify your answer; if you think it is not possible explain why. If you think it is possible, find a subset which is a basis. (a) v v 7 Solution.(5 points) No, this set doesn t form a basis. Because a basis for R has three elements. Obviously removing one or two elements from this set of vector also would leave us less than three vectors. Therefore, we cannot form a basis by removing one or two of these vectors. ( points for explaining that v and v is not a basis. points for showing that there is not a subset which is a basis. You would lose point if you say v and v do not span R but you don t explain why.) (b) v v 5 v Solution.(5 points) This set has elements which is equal to the dimension of R. Thus to show that it gives us a basis, it suffices to show that these vectors are linearly independent: c + c + c c v + c v + c v 5c + c c (c) The last equation implies that c. Then the second equation asserts that c also has to be zero. Finally, we can use the first equation to show that c is also zero. Therefore these three vectors are linearly independent and they form a basis. ( points for showing that the three vectors are linearly independent. points for showing they span R or mentioning that there are the same number of vectors as the dimension of R.) v v v 5 v Solution.( points) No, this set doesn t form a basis. Because a basis for R has three elements. However there is a chance to come up with a basis after removing one element of this set of vectors. To do that, it s helpful to form a matrix which has v, v, v and v as its column and then find the RREF of the resulting matrix: (I) (I) (II) (II) (III) (III)

5 The first, the second and the last row of the RREF have a leading one. Therefore, v, v, v form a basis for R. ( points for arguing that v, v, v and v is not a basis for R. 5 points for finding RREF. points for the subset which is a basis. You would lose points if you make an arithmetic mistake in computing RREF.) 5

6 5. Let A be the following matrix: A and the basis B {v, v } consists of the following vectors: 7 v v Find the matrix B of the linear transformation T ( x) A x with respect to the basis B. Solution.(5 points) We can find the matrix B either column-by-column or by using the formula B S AS where S is the matrix whose columns are v and v 7 S We use the second method. Therefore, we have to find the inverse of S: S Consequently: B S AS (If you use B S AS, then points for the formula B S AS, points for forming S, points for finding S, and points for computing the product. If you find B column by column, then points for the general formula for the columns of B and points for finding each column.). Different parts of this question are not necessarily related to each other. So if you can t do one part, don t give up on the other parts. (a) Show that the following set determines a basis for R : (Hint: You can use the fact that dim(r ).) { } B : f,f,f,f Solution.(5 points) The dimension of R is equal to and there are vectors in B. Therefore, it suffices to show that the elements of B are linearly independent: c c +c +c +c c + c c c c This implies that c, c, c +c and c c. By adding up the last two equations we can conclude that c. The difference of those two equations assert that c. Therefore, c c c c. That is to say, the elements of B are linearly independent. (If you just say these matrices are linearly independent without justification, you would lose points)

7 (b) Find the B-coordinate of the following matrix: 9 5 Solution.(5points) c c c + c c c c + c c + c c c 5 c + c c c 9 Therefore, the B-coordinate of the given matrix is equal to: c c 5 c c c c ( points for forming the equation, points for solving the equation and point for giving the final answer as a vector in R.) (c) Show T : R R, defined as follows, is a linear transformation. T (M) M M Solution.(5points) We have two check the two properties of a linear transformation. First: T (M + M ) (M + M ) (M + M ) M + M M M ( ) ( ) M M + M M T (M ) + T (M ) and then: T (cm) (cm) (cm) c(m M) ct (M) ( points for checking the first property and points for checking the second property.) (d) Is T an isomorphism? Justify your answer. Solution.(5points) Consider the identity matrix: id Then: T (id) id id Therefore, the non-zero matrix id is in the kernel of T. That implies that T is not an isomorphism. 7

8 (e) Find the B-matrix of the linear transformation T. Solution.(5points) The B-matrix is given by the following formula: T (f ) B T (f ) B T (f ) B T (f ) B where f, f, f and f are given in the first part. T (f ) 5 T (f ) 5 T (f ) T (f ) As in part (b) we can find the B-coordinate of each of these vectors: T (f ) B T (f ) B T (f ) B Therefore, the B-matrix is equal to: T (f ) B (f) (Bonus question) Find a basis for the kernel and the image of T. Solution.(points) We firstly find the RREF of the matrix from the previous part. As the first step we replace the first and the third row: +(IV) (I).5 (II) (II) (II).5.5 Therefore, a basis for the image of the B-matrix is given by the first and the third column. These two columns are T (f ) B and T (f ) B. Consequently, a basis for the image of T is given by the matrices T (f ) and T (f ). An element in the kernel of the B-matrix has the following form: s +.5t s.5t t s + t.5.5 8

9 Therefore, a basis for the kernel of the B-matrix is:.5.5 These are the B-coordinate of the following matrices: These two matrices give a basis for the kernel of T. ( points for finding the RREF, points for finding the B-coordinate of a basis for the image of T, points for finding the B-coordinate of a basis for the kernel of T, and points for fining matrices with given B-coordinates.) 9