1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Easy Bases Linear Algebra

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    See Attachment. Help with b) c) will be appreciated.

    3. The attempt at a solution

    For the third question, my approach is sub in values for a and b which correspond to the co-effs. of the given basis. Then assemble a matrix from them. Eg: f[(5,0);(1,3)]=[(1,1);(2;-1)] which gives me [(-17,4);(15,-3)] for c).

    Any help is appreciated.
     

    Attached Files:

  2. jcsd
  3. Nov 24, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm not sure how you got your final matrix.

    What you want to do is apply the function to the basis vectors. If you do this with the first basis vector, you get
    [tex]f(1+x) = \begin{pmatrix}5 \\ 0\end{pmatrix}[/tex]This may be, in part, what you intended to say, but what you wrote above is backward. Now you want to express the resulting vector in terms of the given basis of R2. That is, you want to find the constants a11 and a21 such that
    [tex] \begin{pmatrix}5 \\ 0\end{pmatrix} = a_{11} \begin{pmatrix}2 \\ 3\end{pmatrix} + a_{21} \begin{pmatrix}1 \\ 2\end{pmatrix}[/tex]If you work this out, you'll find a11=10 and a21=-15. These two constants will be the first column of the matrix
    [tex]A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} 10 & a_{12} \\ -15 & a_{22} \end{pmatrix}[/tex]which represents the function f relative to the two bases. Do the same thing with the second basis vector to find the second column of A.
     
  4. Nov 25, 2011 #3
    Wow! Thanks a lot, it's so much clearer now :) Would you happen to know what to do for 6b). I've tried the same approach you suggested for the other question but didn't get the right answer.
     
    Last edited: Nov 25, 2011
  5. Nov 25, 2011 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You do the same thing. Show us your work so we can see where you're the problem lies.
     
  6. Nov 27, 2011 #5
    -2[1,5]+5[1,6]=[3,-4] and 1[1,5]+2[1,6]=[3,13] Basis: {[3,-4];[3,13]}
     
  7. Nov 28, 2011 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm not sure what that has to do with the matrix for the transformation.
     
  8. Nov 30, 2011 #7
    T[(1,5);(1,6)]=[(-2,5);(1,2)].
    a1+5a2=-2;a1+6a2=1;a3+5a4=5 and a3+6a4=2 where T[(a1,a3);(a2;a4)]
    [(1,5);(1,6)|(-2,1)] and [(1,5);(1,6)|(5,2)] which gives me, T=[(-13,28);(11,-23)] which is not correct.
     
  9. Nov 30, 2011 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Please explain your steps and use standard notation. I still can't figure out what you're doing. For example, what does "T[(1,5);(1,6)]=[(-2,5);(1,2)]" mean?
     
  10. Dec 1, 2011 #9
    Sorry vela, i've attached my working on this msg.
     

    Attached Files:

    Last edited: Dec 1, 2011
  11. Dec 1, 2011 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Suppose you have two vectors [itex]\vec{x}, \vec{y} \in \mathbb{R}^2[/itex] such that [itex]\vec{y} = T(\vec{x})[/itex]. Then you can write
    \begin{align*}
    \vec{x} = x_1 \hat{e}_1 + x_2\hat{e}_2 \\
    \vec{y} = y_1 \hat{e}_1 + y_2\hat{e}_2
    \end{align*}where [itex]\hat{e}_1=\begin{pmatrix}1 \\ 0 \end{pmatrix}[/itex] and [itex]\hat{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}[/itex] are the standard basis vectors for [itex]\mathbb{R}^2[/itex]. When you say T has the matrix [itex]\begin{pmatrix} -2 & 1 \\ 5 & 2 \end{pmatrix}[/itex] with respect to this basis, it means that
    [tex]\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix}-2 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}[/tex]
    In this problem, you're given a second basis, [itex]\hat{b}_1 = \begin{pmatrix} 1 \\ 5\end{pmatrix}[/itex] and [itex]\hat{b}_2 = \begin{pmatrix} 1 \\ 6 \end{pmatrix}[/itex]. As before, you can express [itex]\vec{x}[/itex] and [itex]\vec{y}[/itex] as a linear combination of them:
    \begin{align*}
    \vec{x} = c_1 \hat{b}_1 + c_2 \hat{b}_2 \\
    \vec{y} = d_1 \hat{b}_1 + d_2 \hat{b}_2
    \end{align*}Obviously c1 and c2 aren't going to be the same numbers as x1 and x2, and d1 and d2 won't be equal to y1 and y2. However, because [itex]\vec{y} = T(\vec{x})[/itex], c1, c2, d1, and d2 will be related by a matrix equation
    [tex]\begin{pmatrix} d_1 \\ d_2 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}[/tex]It is this matrix, [itex]\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}[/itex], that you're looking for, the matrix of T relative to the basis B.

    Now consider what happens when you choose c1=1 and c2=0. If you think about it a bit, you should be able to solve for the first column of the matrix. Conceptually, you're doing exactly the same thing you did in the previous problem.
     
    Last edited: Dec 1, 2011
  12. Dec 2, 2011 #11
    Ok, thanks Vela.
     
  13. Dec 3, 2011 #12
    for 6A it says show the vectors are a basis of R^2. Since the vectors are 2-d is it enough just to show they are linearly ind then that shows they are a basis. I thought there was somthing that said

    a set of n-dimensional vectors is a basis of R^n if the set is linearly IND.

    I am not posting this as a solution I am asking as a question
     
  14. Dec 4, 2011 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, that's right, though what you said wasn't quite right. There's a theorem that says: if you have a set of n linearly independent vectors (not a set of n-dimensional vectors) from an n-dimensional space, that set of n vectors is a basis. So in this problem, it's enough to show that the vectors are linearly independent to prove they are a basis for R2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Easy Bases Linear Algebra
Loading...