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Thread: Change of basis

  1. #1
    MHF Contributor arbolis's Avatar
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    Change of basis

    In a given basis $\displaystyle \{ \vec e _i \}$ of a vector space, a linear transformation and a given vector of this space are determinate by $\displaystyle \begin{bmatrix} 2 & 1 & 0 \\ 1&2&0 \\ 0 & 0 & 5 \end{bmatrix}$ and $\displaystyle \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ (should be a column vector).
    Find the matrix representation of the transformation and of the vector in a new basis such that the old one is represented by $\displaystyle \vec e _1 = \begin{bmatrix} 1&1&0 \end{bmatrix}$, $\displaystyle \vec e _2 = \begin{bmatrix} 1&-1&0 \end{bmatrix}$,$\displaystyle \vec e _3 = \begin{bmatrix} 0&0&1 \end{bmatrix}$ (they should be column vectors).

    My attempt: I formed a matrix whose columns are the $\displaystyle \vec e_1$ ,$\displaystyle \vec e_2$ ,$\displaystyle \vec e_3$ and I found its inverse. Call them $\displaystyle S$ and $\displaystyle S^{-1}$.
    Then I multiplied $\displaystyle S^{-1}AS$ where $\displaystyle A$ is the first matrix I wrote. As result, I obtained the following matrix $\displaystyle A'= \begin{bmatrix} 3&0&0 \\ 0&1&0 \\ 0&0&5 \end{bmatrix}$ which would be the matrix they asked for.

    For the vector, I wasn't sure at all... I just multipled $\displaystyle A'$ by $\displaystyle \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ and I obtained $\displaystyle \begin{bmatrix} 3 & 12 & 15 \end{bmatrix}$ which would be the vector they ask for.
    Am I right?
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    Quote Originally Posted by arbolis View Post
    In a given basis $\displaystyle \{ \vec e _i \}$ of a vector space, a linear transformation and a given vector of this space are determinate by $\displaystyle \begin{bmatrix} 2 & 1 & 0 \\ 1&2&0 \\ 0 & 0 & 5 \end{bmatrix}$ and $\displaystyle \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ (should be a column vector).
    Find the matrix representation of the transformation and of the vector in a new basis such that the old one is represented by $\displaystyle \vec e _1 = \begin{bmatrix} 1&1&0 \end{bmatrix}$, $\displaystyle \vec e _2 = \begin{bmatrix} 1&-1&0 \end{bmatrix}$,$\displaystyle \vec e _3 = \begin{bmatrix} 0&0&1 \end{bmatrix}$ (they should be column vectors).

    My attempt: I formed a matrix whose columns are the $\displaystyle \vec e_1$ ,$\displaystyle \vec e_2$ ,$\displaystyle \vec e_3$ and I found its inverse. Call them $\displaystyle S$ and $\displaystyle S^{-1}$.
    Then I multiplied $\displaystyle S^{-1}AS$ where $\displaystyle A$ is the first matrix I wrote. As result, I obtained the following matrix $\displaystyle A'= \begin{bmatrix} 3&0&0 \\ 0&1&0 \\ 0&0&5 \end{bmatrix}$ which would be the matrix they asked for.

    For the vector, I wasn't sure at all... I just multipled $\displaystyle A'$ by $\displaystyle \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ and I obtained $\displaystyle \begin{bmatrix} 3 & 12 & 15 \end{bmatrix}$ which would be the vector they ask for.
    Am I right?
    The first part, A', is correct but the question did not ask for the vector <1 2 3> multiplied by A'.

    Suppose your problem were Av= u where [tex]A= \begin{barray}2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 5\end{bmatrix} and [tex]v= \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}.

    To change A to A' you would, as you say, write $\displaystyle S^{-1}AS$ with the S matrix you give. Now think how we would apply that to the equation Av= u. If we multiply both sides, on the left, by $\displaystyle S^{-1}$, we would have $\displaystyle S^{-1}Av= S^{-1}u$. But that does not have the "S" multiplying A on the right. Just multiplying the equation on the right won't help because that would give $\displaystyle S^{-1}AvS= S^{-1}uS$, not what we want.

    Instead, use the fact that $\displaystyle I= SS^{-1}$ to write the equation as $\displaystyle S^{-1}A(SS^{-1})v= S^{-1}u$ or $\displaystyle (S^{-1}AS)(S^{-1}v)= S^{-1}u$- that is, $\displaystyle A'v'= u'$ with $\displaystyle A'= S^{-1}AS$ and $\displaystyle v'= S^{-1}v$, not A'v.
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