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Math Help - how to find a matrix of a linear transformation

  1. #1
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    how to find a matrix of a linear transformation

    Let T: V---->W be the linear transformation that has the matrix:

    1 -1 1
    2 1 1

    relative to the bases{(1,2,0),(1,1,1),(1,1,0)} of V and {(1,1),(1,-1)} of W
    Find the matrix of T relative to the bases{(2,3,0),(1,1,1),(2,3,1)} of V and {(3,-1),(1,-1)} of W
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  2. #2
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    Re: how to find a matrix of a linear transformation

    you will need two "change of basis" matrices:

    P, which changes vectors with coordinates corresponding to the basis {(2,3,0),(1,1,1),(2,3,1)} to vectors with coordinates corresponding to the basis {(1,2,0),(1,1,1),(1,1,0)}.

    Q which changes vectors with coordinates corresponding to the basis {(1,1),(1,-1)} to vectors with coordinates corresponding to the basis {(3,-1),(1,-1)}.

    if A is your given matrix, then the matrix you want is:

    QAP.

    what have you done so far?
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  3. #3
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    Re: how to find a matrix of a linear transformation

    I missed last week class cuz i was sick so I didn't learn change of basis yet
    and this is the only problem I have trouble with on my homework
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  4. #4
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    Re: how to find a matrix of a linear transformation

    *sigh*

    i will show you how to find P. you can try figuring out Q yourself.

    if we let B = {(2,3,0),(1,1,1),(2,3,1)} then since (2,3,0) = 1(2,3,0) + 0(1,1,1) + 0(2,3,1) in the basis B, (2,3,0) has coordinates [1,0,0]B.

    what we want to know is: what is [1,0,0]B in B'-coordinates?

    that is, we need to know how to express (2,3,0) as a linear combination of (1,2,0),(1,1,1) and (1,1,0).

    well, if (2,3,0) = a(1,2,0) + b(1,1,1) + c(1,1,0) = (a+b+c,2a+b+c,b) this means:

    a+b+c = 2
    2a+b+c = 3
    b = 0

    we catch a break on that third equation, so now we just need to solve:

    a+c = 2
    2a+c = 3

    subtract the top from the bottom, leaving a = 1. thus c = 1. so:

    (2,3,0) = 1(1,2,0) + 0(1,1,1) + 1(1,1,0) = [1,0,1]B'.

    now, what matrix takes (1,0,0) to (1,0,1)? well, we can't answer entirely, but we know its first column is the image of (1,0,0), and we know that: it's (1,0,1). thus, so far, we have:

    P = \begin{bmatrix}1&*&*\\0&*&*\\1&*&* \end{bmatrix}

    in B-coordinates, (1,1,1) is [0,1,0]B (remember this means: (1,1,1) = 0(2,3,0) + 1(1,1,1) + 0(2,3,1)).

    again, we need to know how to express (1,1,1) in B'-coordinates, that is, as a linear combination of (1,2,0), (1,1,1), and (1,1,0).

    here, we catch another break: it's obvious that (1,1,1) = 0(1,2,0) + 1(1,1,1) + 0(1,1,0), so in B'-coordinates, we have (1,1,1) = [0,1,0]B'.

    this tells us the second column of P is (0,1,0), so:

    P = \begin{bmatrix}1&0&*\\0&1&*\\1&0&* \end{bmatrix}

    we do the same thing to find the 3rd column, we need to find what (2,3,1) is as a linear combination of (1,2,0), (1,1,1) and (1,1,0).

    in fact, it's pretty clear that (2,3,1) = 1(1,2,0) + 1(1,1,1) + 0(1,1,0), so its B'-coordinates are [1,1,0]B'.

    now we know P, it is:

    P = \begin{bmatrix}1&0&1\\0&1&1\\1&0&0 \end{bmatrix}.

    so this changes B-coordinates into B'-coordinates. this is good, because the matrix A we were given acts on B'-coordinates.

    if we want to, we can calculate AP:

    AP = \begin{bmatrix} 1&-1&1\\2&1&1 \end{bmatrix} \begin{bmatrix}1&0&1\\0&1&1\\1&0&0  \end{bmatrix} = \begin{bmatrix}2&-1&0\\3&1&3 \end{bmatrix}

    this is the matrix of T relative to the bases B = {(2,3,0),(1,1,1),(2,3,1)} and C' = {(1,1),(1,-1)}, so we're closer (we've matched one of the bases to our target pair).

    so Q is going to be a 2x2 matrix which changes C'-coordinates to the C-coordinates we want. to get Q, you need to figure out how to write:

    (1,1) as a linear combination of (3,-1) and (1,-1)

    (1,-1) as a linear combination of (3,-1) and (1,-1). can you do this?
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    Re: how to find a matrix of a linear transformation

    (1 , 1) = a(3 , -1) + b(1, -1) = [1, -2]

    (1, -1) = a(3, -1) + b(1, -1) = [0, 1]
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  6. #6
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    Re: how to find a matrix of a linear transformation

    so Q = 1 0 ?
    -2 1
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  7. #7
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    Re: how to find a matrix of a linear transformation

    yep, that's the Q i got. now, put it all together to get your desired matrix, which is..?
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  8. #8
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    Re: how to find a matrix of a linear transformation

    2 -1 0
    -1 3 3
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  9. #9
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    Re: how to find a matrix of a linear transformation

    thank you so much
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