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Thread: Linear transformation relative to a basis:A question

  1. #1
    Senior Member x3bnm's Avatar
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    Linear transformation relative to a basis:A question

    My problem:
    Suppose
    1)Matrix for linear transformation $\displaystyle T$ relative to basis $\displaystyle B \text{ is}: A$
    2)Matrix for linear transformation $\displaystyle T$ relative to basis $\displaystyle B^\prime \text{ is}: A^\prime$
    3)Transition matrix from basis $\displaystyle B^\prime$ to basis $\displaystyle B$ is: $\displaystyle P$
    4)Transition matrix from basis $\displaystyle B$ to basis $\displaystyle B^\prime$ is : $\displaystyle P^{-1}$

    Now we know that:
    $\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime} $

    And
    $\displaystyle P^{-1}AP[\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime} $
    where $\displaystyle [T(\mathbf{v})]_{B^\prime}$ is coordinate matrix of $\displaystyle T(\mathbf{v})$ relative to basis $\displaystyle B^\prime$ and $\displaystyle [\mathbf{v}]_{B^\prime}$ is coordinate matrix of $\displaystyle \mathbf{v}$ relative to $\displaystyle B^\prime$

    Now the book(Elementary Linear Algebra by Larson 6th edition) I am reading states that:

    "But by the definition of the matrix of a linear transformation relative to a basis, this implies that:
    $\displaystyle A^\prime = P^{-1}AP $
    "

    My question:
    My question is: why's that?

    I know from above that:
    $\displaystyle [T(\mathbf{v})]_{B^\prime} = A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime} $

    which makes:
    $\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime} $

    But what reasoning makes the statement $\displaystyle A^\prime = P^{-1}AP$ true? How did the author deduced this to be true?
    Last edited by x3bnm; Nov 4th 2011 at 09:22 AM.
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  2. #2
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    Re: Linear transformation relative to a basis:A question

    Quote Originally Posted by x3bnm View Post
    My problem:
    Suppose
    1)Matrix for linear transformation $\displaystyle T$ relative to basis $\displaystyle B \text{ is}: A$
    2)Matrix for linear transformation $\displaystyle T$ relative to basis $\displaystyle B^\prime \text{ is}: A^\prime$
    3)Transition matrix from basis $\displaystyle B^\prime$ to basis $\displaystyle B$ is: $\displaystyle P$
    4)Transition matrix from basis $\displaystyle B$ to basis $\displaystyle B^\prime$ is : $\displaystyle P^{-1}$

    Now we know that:
    $\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime} $

    And
    $\displaystyle P^{-1}AP[\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime} $
    where $\displaystyle [T(\mathbf{v})]_{B^\prime}$ is coordinate matrix of $\displaystyle T(\mathbf{v})$ relative to basis $\displaystyle B^\prime$ and $\displaystyle [\mathbf{v}]_{B^\prime}$ is coordinate matrix of $\displaystyle \mathbf{v}$ relative to $\displaystyle B^\prime$

    Now the book(Elementary Linear Algebra by Larson 6th edition) I am reading states that:

    "But by the definition of the matrix of a linear transformation relative to a basis, this implies that:
    $\displaystyle A^\prime = P^{-1}AP $
    "

    My question:
    My question is: why's that?

    I know from above that:
    $\displaystyle [T(\mathbf{v})]_{B^\prime} = A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime} $

    which makes:
    $\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime} $

    But what reasoning makes the statement $\displaystyle A^\prime = P^{-1}AP$ true? How did the author deduced this to be true?
    if two functions are the same for every input of the domain, they are the same function.

    so if A' and $\displaystyle P^{-1}AP$ are the same for every $\displaystyle [\mathbf{v}]_{B'}$, they are the same function (and if they are both linear transformations, the same linear transformation).
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Linear transformation relative to a basis:A question

    Quote Originally Posted by Deveno View Post
    if two functions are the same for every input of the domain, they are the same function.

    so if A' and $\displaystyle P^{-1}AP$ are the same for every $\displaystyle [\mathbf{v}]_{B'}$, they are the same function (and if they are both linear transformations, the same linear transformation).
    Thanks for reply. But suppose three matrices and $\displaystyle [x]$ are such that:
    $\displaystyle ABx = CBx$ where $\displaystyle AB = CB$ for all $\displaystyle [x]$

    Then by the properties of matrix:
    $\displaystyle A$ is not necessarily equal to $\displaystyle C$

    So for $\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime}$

    why my reasoning above is not a problem for this situation?

    Why still $\displaystyle A^\prime = P^{-1}AP $?
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  4. #4
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    Re: Linear transformation relative to a basis:A question

    there's no cancellation going on, we're saying A' and P^-1AP are the same function (the fact that they have the same matrix relative to the basis B' is a big clue).
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  5. #5
    Senior Member x3bnm's Avatar
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    Re: Linear transformation relative to a basis:A question

    Quote Originally Posted by Deveno View Post
    there's no cancellation going on, we're saying A' and P^-1AP are the same function (the fact that they have the same matrix relative to the basis B' is a big clue).
    Thanks a lot. That I understand.
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