Solution This space is just Col A where A 1 2 2 3 1 2 1 1 2 2 8 10 6 3 3 3 9 So

Solution this space is just col a where a 1 2 2 3 1 2

This preview shows page 7 - 10 out of 15 pages.

Solution. This space is just Col A , where A = 1 0 2 2 3 0 1 - 2 - 1 - 1 - 2 2 - 8 10 - 6 3 3 0 3 9 . So we just need to find a basis for Col A . This part is left to you. Ex. 4.4.27 Determine whether { 1 + 2 t 3 , 2 + t - 3 t 2 , - t + 2 t 2 - t 3 } is linearly independent or not. Solution. Say E = { 1 , t, t 2 , t 3 } . Then { [1 + 2 t 3 ] E , [2 + t - 3 t 2 ] E , [ - t + 2 t 2 - t 3 ] E } = 1 0 0 2 , 2 1 - 3 0 , 0 - 1 2 - 1 is linearly independent iff { 1 + 2 t 3 , 2 + t - 3 t 2 , - t + 2 t 2 - t 3 } is linearly independent. As 1 2 0 0 1 - 1 0 - 3 2 2 0 - 1 1 2 0 0 1 - 1 0 0 -1 0 0 0 , every column is a pivot column, the columns are linearly independent. (The matrix is not onto, hence the columns do not span R 4 .) Ex. 4.6.15 If A is a 3 × 7 matrix, find the smallest and the largest possible dimension of Nul A , Col A , Row A . Also find the minimum and the maximum rank of A . Solution. For a 3 × 7 matrix, at most it has 3 pivots and at least it has 0 pivot, hence max rank A = max dim Col A = max dim Row A = 3 (for example, A = 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 ) and min rank A = min dim Col A = min dim Row A = 0 (for example, A is the zero ma- trix). As we have rank theorem, saying rank A +dim Nul A = 7, hence max dim Nul A = 7 and min dim Nul A = 4. 7
Image of page 7
Ex. 4.7.9 Find P C←B , where B = 4 4 , 8 4 and C = 2 2 , - 2 2 . Find C -coordinate of 4 4 + 8 4 by using P C←B . Solution. 2 - 2 4 8 2 2 4 4 1 0 2 3 0 1 0 - 1 . Hence P C←B = 2 3 0 - 1 . And [ 4 4 + 8 4 ] C = P C←B [ 4 4 + 8 4 ] B = 2 3 0 - 1 1 1 = 5 - 1 Ex. 4.7.14 Find P C←B , where B = { 1 , 1 + t, 1 + t + t 2 } and C = { t 2 , t, 1 } . Solution. It is sufficient to consider 1 0 0 , 1 1 0 , 1 1 1 and 0 0 1 , 0 1 0 , 1 0 0 . Then 0 0 1 1 1 1 0 1 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 1 1 0 0 1 1 1 1 . Hence P C←B = 0 0 1 0 1 1 1 1 1 . Eigenvalues and Eigenvectors. (1) ~x is an eigenvector of A R n × n corresponding to eigenvalue λ iff ~x 6 = ~ 0 and A~x = λ~x . (2) Eigenspace corresponding to m is E m = { eigenvectors corresponding to m } ∪ n ~ 0 o . (3) Eigenvalues are roots of the characteristic equation det( A - λI ) = 0. (4) Eigenspace corresponding to m is E m = Nul( A - mI ). (5) Algebraic multiplicity of m is a if det( A - λI ) = ( m - λ ) a · · · . (6) Geometric multiplicity of m is dim E m . Properties. (1) The eigenvalues of a triangular matrix are the entries on its diagonal. (2) Eigenvectors corresponding to distinct eigenvalues are linearly independent. (3) If A is similar to B , that is, A = PBP - 1 for some invertible P , then A and B have the same characteristic polynomial, same eigenvalues with the same algebraic multiplicities and the same geometric multiplicities. 8
Image of page 8
Matrix for a linear transformation. (1) The matrix for T : V W relative to the bases B and C is A = h [ T ~ b 1 ] C · · · [ T ~ b n ] C i , where B , C are bases for V , W , respectively. It will satisfy [ T~x ] C = A [ ~x ] B and A = P - 1 C TP B . In particular T : V V is the identity map, then A = P C←B .
Image of page 9
Image of page 10

You've reached the end of your free preview.

Want to read all 15 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes