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Is the linear transformation matrix T invertible

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    T: M22 --> M22 defined by T(A) = AB where B =
    [ 3 2 ]
    [ 2 1 ]

    Is the linear transformation matrix T invertible with respect to the standard bases? If so, find it.

    2. Relevant equations

    none

    3. The attempt at a solution

    This is going to sound stupid, but I need help in finding what the transformation matrix T actually is before I can say if it is invertible or not. So far I got:

    T(A) = AB =
    [a b ] [ 3 2 ]
    [c d ] [ 2 1 ]

    =
    [ 3a+2b 2a+b ]
    [ 3c+2d 2c+d ]

    and I found T(e11) =
    [3 2 ]
    [0 0 ]

    T(e12) =
    [ 2 1 ]
    [ 0 0 ]

    T(e21) =
    [ 0 0 ]
    [ 3 2 ]

    T(e22) =
    [ 0 0 ]
    [ 2 1 ]

    and now I'm stuck. Thinking ahead a little, I know that if the kernal of T = 0, it will be invertible, or if the determinant of the matrix is not 0, it is invertible.

    P.S. is there an easier way to write matrices here?
     
  2. jcsd
  3. Mar 22, 2009 #2
    What is the inverse of T? Hint: T^-1 will look a lot like T, but with something replaced by its inverse.
     
  4. Mar 22, 2009 #3

    Mark44

    Staff: Mentor

    No, it doesn't sound stupid to me. This transformation takes a 2x2 matrix A as input, and produces another 2x2 matrix AB as its output, where B is as you showed.
    I think what is meant by asking if T is invertible with respect to the standard basis is this: Your basis vectors (the matrices eij) are linearly independent. Are the new vectors (the matrices T(eij)) also linearly independent?
    The determinant of AB is zero some of the time, and this is completely determined by the matrix A. This means that whether T is invertible depends completely on the matrix A that is used. Calculate the determinant of AB to see this.


    There's a nicer way using tex script, but I wouldn't call it easier.
     
  5. Mar 22, 2009 #4
    Yes they're also linearly independent.

    Is the inverse =
    [ -1 2 ] [ a b ]
    [ 2 -3 ] [ c d ]

    =
    [ 2c-a 2d-b ]
    [ 2a-3c 2b-3d ]
    ?
     
  6. Mar 22, 2009 #5

    Mark44

    Staff: Mentor

    It's easy enough to check. Multiply what you got by AB, and it should come out to I.

    I think I remember a theorem that says, if A and B are invertible, then (AB)^(-1) = B^(-1)A^(-1). The matrix you show on the left in your product looks like it might be B^(-1), but the one on the right definitely isn't A^(-1).
     
  7. Mar 22, 2009 #6
    It's starting to look really aweful. Is there another way to find the inverse?
     
  8. Mar 22, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You don't actually need to find the matrix corresponding to T. You just have to define it's inverse. It looks like you've already found the inverse of B. Think about S(X)=XB^(-1). Could S be T^(-1)?
     
  9. Mar 22, 2009 #8
    what's that X in S(X)=XB^(-1) ?
     
  10. Mar 22, 2009 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    X is any matrix, just like the 'A' in T(A)=AB. I could have written S(A)=AB^(-1) as well, they both mean the same thing.
     
  11. Mar 22, 2009 #10
    This is what I got after what you said:
    [ a b ] [ -1 2 ]
    [ c d ] [ 2 -3 ]
    =
    -a+2b 2a-3b
    -c+2d 2c-3d
    You hinted towards this may be being the inverse and I multiplied it out by

    [ 3a+2b 2a+b ]
    [ 3c+2d 2c+d ]

    And it doesn't look like the inverse in the end.
     
  12. Mar 22, 2009 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    S is the inverse of T if S(T(A))=A=T(S(A)). S(T(A))=S(AB)=ABB^(-1)=A. You seem to be multiplying T(A)*S(A) and trying to get A or something. That's not what the inverse means.
     
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