1. Limited time only! Sign up for a free 30min personal 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!

Determinant of matrix of linear transformation

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Linear transformation T: P2->P2
    T(f) = -5f + 8f'
    Need to find detA (A is a matrix of T)

    2. Relevant equations

    T(f) = Af

    3. The attempt at a solution
    The basis of P2 is B={1, x, x2}. Some polynomial f with respect to B looks like this in general:

    (a, b, c)T

    right? So, T(f) = -(a, b, c)T + 8 (0, b, 2c)T = (-a, 7b, c)T.

    So, A*(a, b, c)T = (-a, 7b, c)T, therefore A =

    1 0 0
    0 7 0
    0 0 1

    And detA = -7. This is how I was thinking about it, but the answer is wrong and I am confused :(
     
  2. jcsd
  3. Mar 18, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    What happened to the factor of 5 from T(f)=-5f+8f'?
     
  4. Mar 18, 2010 #3
    Oh!
    Thanks!
    It always happens to me.

    So, fixing that mistake, T(f)=

    -5a
    3b
    11c

    And A becomes:

    -5 0 0
    0 3 0
    0 0 11

    and detA = -165 and it is still wrong :(
     
  5. Mar 18, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Well, if [itex]f(x)=a+bx+cx^2[/itex], then [itex]f'(x)=b+2cx[/itex]

    So, if you are representing your basis functions as

    [tex]x^0\to\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \;\;\;\;\text{and}\;\;\;\; x^1\to\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix} \;\;\;\text{and}\;\;\;\; x^2\to\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}[/tex]

    Then you have [tex]f(x)\to\begin{pmatrix}a \\ b \\ c \end{pmatrix}[/tex] and [tex]f'(x)\to\begin{pmatrix}b \\ 2c \\ 0 \end{pmatrix}[/tex]
     
  6. Mar 18, 2010 #5
    Okay, thank you again!
    Now I have:

    -5(a, b, c)T + 8 (b, 2c, 0)T =

    -5a+8b
    -5a+16c
    -5a

    And:

    A(a, b, c)T = (-5a+8b, -5a+16c, -5a)T, so A=

    -5 8 0
    -5 0 16
    -5 0 0

    detA = -640 and it is still wrong :(((
     
  7. Mar 18, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Errm...

    [tex]-5\begin{pmatrix}a \\ b \\ c \end{pmatrix}+8\begin{pmatrix}b \\ 2c \\ 0 \end{pmatrix}=\begin{pmatrix}-5a+8b \\ -5b+16c \\ -5c \end{pmatrix}[/tex]
     
  8. Mar 18, 2010 #7
    Oh, I'm dumb! Thank you very much!
     
  9. Mar 18, 2010 #8
    There is a similar problem I cannot figure out..

    T(f(t)) = f(2t) - 3f(t)

    Again need to find detA.

    In this case I don't even know how to start...
     
  10. Mar 18, 2010 #9

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Is T again a mapping from P2 to P2?

    If so, do it in the same way, just with [itex]t[/itex] as the variable instead of [itex]x[/itex]... if [itex]f(t)=a+bt+ct^2[/itex], what is [itex]f(2t)[/itex]?
     
  11. Mar 18, 2010 #10
    f(2t) = a + b2t = c2t2?
     
  12. Mar 18, 2010 #11

    Mark44

    Staff: Mentor

    No. If f(t) = a + bt + ct2, then f(2t) = a + b*2t + c(2t)2. This can be simplified.
     
  13. Mar 18, 2010 #12
    Thank you very much!
     
  14. Mar 18, 2010 #13
    Sorry, I've faced one more problem...
    There is T: R2->R2, s.t. T(V1)=8V2, and T(V2)=-4V1.

    Need to find detA.

    This is how I started thinking:
    AV1=8V2 => A=8V2 / V1
    AV2=-4V1 => A=-4V1 / V2

    Don't know how to think further...
     
  15. Mar 18, 2010 #14

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Okay, and what are [itex]V_1[/itex] and [itex]V_2[/itex]?
     
  16. Mar 18, 2010 #15
    They are just some vectors in R2, they're not given.
     
  17. Mar 18, 2010 #16

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    I don't believe that. If this was true for all [itex]V_1[/itex] and [itex]V_2[/itex] in [itex]\mathbf{R}^2[/itex], then the matrix [itex]A[/itex] would have to be full of zeroes. You'd need only look at some case where [itex]V_2=V_1[/itex] to prove that to yourself.

    I suspect that they are supposed to represent the basis vectors of [itex]\mathbf{R}^2[/itex]. In which case, you might as well look at two orthogonal basis vectors

    [tex]V_1\to\begin{pmatrix}1 \\ 0 \end{pmatrix} \;\;\;\;\text{and}\;\;\;\; V_2\to\begin{pmatrix}0 \\ 1 \end{pmatrix}[/tex]
     
  18. Mar 18, 2010 #17
  19. Mar 19, 2010 #18

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Okay, so they are some specific vectors, you just aren't told exactly what they are.

    In that case, just say

    [tex]V_1\to\begin{pmatrix}a_1 \\ b_1\end{pmatrix} \;\;\;\;\text{and}\;\;\;\; V_2\to\begin{pmatrix}a_2 \\ b_2\end{pmatrix}[/tex]

    and go from there.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determinant of matrix of linear transformation
Loading...