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Thread: Linear Transformation

  1. #1
    Junior Member
    Nov 2008

    Linear Transformation

    If is a linear transformation such that
    1 2 11 14 6 and 4 -1 17 2 6 ,
    what is the standard matrix of T?

    Any help is appreciated
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  2. #2
    Sep 2009
    I dont know what standard matrix is, but I am going to assume it means the matrix of
    T with respect to the standard basis.

    Now what you need to find is what T does to the standard basis

    T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)=T\l  eft(\begin{bmatrix}1\\0\end{bmatrix}+2\cdot\begin{  bmatrix}0\\1\end{bmatrix}\right)=T\left(\begin{bma  trix}1\\0\end{bmatrix}\right)+2T\left(\begin{bmatr  ix}0\\1\end{bmatrix}\right)=\begin{bmatrix}11\\14\  \6\end{bmatrix}

    And :

    T\left(\begin{bmatrix}4\\-1\end{bmatrix}\right)=T\left(4\cdot\begin{bmatrix}  1\\0\end{bmatrix}-1\cdot\begin{bmatrix}0\\1\end{bmatrix}\right)=4T\l  eft(\begin{bmatrix}1\\0\end{bmatrix}\right)-1T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)=\b  egin{bmatrix}17\\2\\6\end{bmatrix}

    Multiplying the lower formula by 2 and adding to the upper one gives:

    9T\left(\begin{bmatrix}1\\0\end{bmatrix}\right)+0=  \begin{bmatrix}11\\14\\6\end{bmatrix}+2\cdot\begin  {bmatrix}17\\2\\6\end{bmatrix}

    So that
    T\left(\begin{bmatrix}1\\0\end{bmatrix}\right)=\fr  ac{1}{9}\cdot\begin{bmatrix}54\\18\\18\end{bmatrix  }

    Now find T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)

    And the standard matrix would then be:

    \begin{bmatrix}T\left(\begin{bmatrix}1\\0\end{bmat  rix}\right) & T\left(\begin{bmatrix}0\\1\end{bmatrix}\right)\end  {bmatrix}
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  3. #3
    Jan 2009
    As above, I repeat equations.

    T(\vec{x})=A\vec{x},A=[T(\vec{e}_1)\ \ T(\vec{e}_2)]

    T(\vec{e}_1)=\frac{1}{9}\cdot\begin{bmatrix} *45* \\18\\18\end{bmatrix}=\begin{bmatrix} 5 \\2\\2\end{bmatrix}\ ,\ T(\vec{e}_2)=\begin{bmatrix} 3 \\6\\2\end{bmatrix}\ ,\ A=\begin{bmatrix} 5&3\\2&6\\2&2\end{bmatrix}
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