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Thread: e2 Show that S and T are both linear transformations

  1. #1
    Super Member bigwave's Avatar
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    e2 Show that S and T are both linear transformations

    $S:\mathbb{R}^2\to \mathbb{R}^2$ and $T:\mathbb{R}^2 \to \mathbb{R}^2$ be transformations defined by
    $S\begin{bmatrix}x\\y \end{bmatrix}=
    \begin{bmatrix} 2x+y \\x-y \end{bmatrix},
    \quad T
    \begin{bmatrix}x\\y \end{bmatrix}=
    \begin{bmatrix}x-4y\\3x\end{bmatrix}$
    (a) Show that S and T are both linear transformations

    ok my first attempt would be to see if S preserves addition so

    $S
    \left(\begin{bmatrix}x_1\\y_1 \end{bmatrix}
    +\begin{bmatrix}x_2\\y_2 \end{bmatrix}\right)=
    \begin{bmatrix}x_1+x_2 \\y_1+y_2 \end{bmatrix}
    =\begin{bmatrix} 2(x_1+x_2)+(y_1+y_2) \\(x_1+y_1)-(y_1+y_2) \end{bmatrix}$
    and
    $S\begin{bmatrix}x_1\\y_1 \end{bmatrix}+S\begin{bmatrix}x_2\\y_2 \end{bmatrix}
    =\begin{bmatrix} 2x_1+y_1 \\x_1-y_1 \end{bmatrix}+\begin{bmatrix} 2x_2+y_2 \\x_2-y_2 \end{bmatrix}
    =​\begin{bmatrix} 2(x_1+x_2)+(y_1+y_2) \\(x_1+y_1)-(y_1+y_2) \end{bmatrix}$

    maybe typos but think steps were ok for S still have T and there is
    (b) Find $ST
    \begin{bmatrix} x\\y
    \end{bmatrix}
    \textit{ and } T^2
    \begin{bmatrix} x\\y
    \end{bmatrix}$
    (c) Find the matrices of S and T with respect to the standard basis for $\mathbb{R}^2$.
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  2. #2
    Member Walagaster's Avatar
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    Re: e2 Show that S and T are both linear transformations

    It's OK as far as it goes. Don't forget for linearity you also need $T(ax) = aT(x)$.
    Thanks from topsquark and bigwave
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  3. #3
    Super Member bigwave's Avatar
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    Re: e2 Show that S and T are both linear transformations

    ok thanks I thot addition was enough ....
    got a take a break then back in the saddle
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    Re: e2 Show that S and T are both linear transformations

    Quote Originally Posted by bigwave View Post
    $S:\mathbb{R}^2\to \mathbb{R}^2$ and $T:\mathbb{R}^2 \to \mathbb{R}^2$ be transformations defined by
    To show that a mapping $S$ is a linear transformation does $S(X+\alpha\cdot Y)=S(X)+\alpha\cdot S(Y)~?$
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  5. #5
    Super Member bigwave's Avatar
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    Re: e2 Show that S and T are both linear transformations

    I would have to follow an example to see how to do that
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    Re: e2 Show that S and T are both linear transformations

    Quote Originally Posted by Plato View Post
    To show that a mapping $S$ is a linear transformation does $S(X+\alpha\cdot Y)=S(X)+\alpha\cdot S(Y)~?$
    Quote Originally Posted by bigwave View Post
    I would have to follow an example to see how to do that
    Why not at least try it?
    $T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$

    ALSO
    ${T^2}\left( {\left[ {\begin{array}{*{20}{c}}
    x\\
    y
    \end{array}} \right]} \right) = T \circ T\left( {\left[ {\begin{array}{*{20}{c}}
    x\\
    y
    \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}
    {(x - 4y) - 12x}\\
    {3x - 12y}
    \end{array}} \right]$ WHY & HOW?
    Post your work so others may see .
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  7. #7
    Super Member bigwave's Avatar
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    Re: e2 Show that S and T are both linear transformations

    Quote Originally Posted by Plato View Post
    Why not at least try it?
    $T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$

    ALSO
    ${T^2}\left( {\left[ {\begin{array}{*{20}{c}}
    x\\
    y
    \end{array}} \right]} \right) = T \circ T\left( {\left[ {\begin{array}{*{20}{c}}
    x\\
    y
    \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}
    {(x - 4y) - 12x}\\
    {3x - 12y}
    \end{array}} \right]$ WHY & HOW?
    Post your work so others may see .
    $T\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right)
    = T\left( {\left[ {\begin{array}{*{20}{c}}{\alpha x}\\{\alpha y}\end{array}} \right]} \right)
    = \left[ {\begin{array}{*{20}{c}}{\alpha x - 4\alpha y}\\{3\alpha x}\end{array}} \right]$\\
    $\text{so then}\\$
    $S\left( {\alpha \left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]} \right)
    = S\left( {\left[ {\begin{array}{*{20}{c}}
    {\alpha x}\\{\alpha y}\end{array}} \right]} \right)
    = \left[ {\begin{array}{*{20}{c}}
    {2\alpha x+\alpha y}\\{\alpha x-\alpha y}\end{array}} \right]$
    Last edited by bigwave; Mar 27th 2019 at 12:45 PM.
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  8. #8
    Super Member bigwave's Avatar
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    Re: e2 Show that S and T are both linear transformations

    ok finally Find $ST\begin{bmatrix}x\\y\end{bmatrix}
    =S\left(\begin{bmatrix}x-4y\\3x\end{bmatrix}\right)
    =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$
    and
    $T^2\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
    =T\left(T\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)\right)
    =T\left(\left[\begin{array}{c}x-4y\\3x \end{array}\right]\right)
    =\left[\begin{array}{c}x-4y-4(3x) \\ 3(x-4y) \end{array}\right]$
    hopefully
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  9. #9
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    Re: e2 Show that S and T are both linear transformations

    Quote Originally Posted by bigwave View Post
    ok finally Find $ST\begin{bmatrix}x\\y\end{bmatrix}
    =S\left(\begin{bmatrix}x-4y\\3x\end{bmatrix}\right)
    =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$
    and
    $T^2\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
    =T\left(T\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)\right)
    =T\left(\left[\begin{array}{c}x-4y\\3x \end{array}\right]\right)
    =\left[\begin{array}{c}x-4y-4(3x) \\ 3(x-4y) \end{array}\right]$
    hopefully
    Good Job.
    However, it should be $S{\circ}T\begin{bmatrix}x\\y\end{bmatrix}$ because $ST$ indicates multiplication and not function composition.
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  10. #10
    Super Member bigwave's Avatar
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    Re: e2 Show that S and T are both linear transformations

    Quote Originally Posted by Plato View Post
    Good Job.
    However, it should be $S{\circ}T\begin{bmatrix}x\\y\end{bmatrix}$ because $ST$ indicates multiplication and not function composition.
    $$ST\begin{bmatrix}x\\y\end{bmatrix}
    =S\left(T\begin{bmatrix}x-4y\\3x\end{bmatrix}\right)
    =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$$

    do you mean this?
    Last edited by bigwave; Mar 30th 2019 at 01:31 PM.
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  11. #11
    Forum Admin topsquark's Avatar
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    Re: e2 Show that S and T are both linear transformations

    Quote Originally Posted by bigwave View Post
    $$ST\begin{bmatrix}x\\y\end{bmatrix}
    =S\left(T\begin{bmatrix}x-4y\\3x\end{bmatrix}\right)
    =\left[\begin{array}{c}2(x-4y)+3x \\ x-4y-3x \end{array}\right]$$

    do you mean this?
    Yes. The term ST is, in fact, a multiplication as Plato pointed out. You really do need to write it as $\displaystyle S \circ T$.

    On the other hand I've never seen the composition symbol used in any of the Physics classes I've taken. So if you are doing Math do it as Plato (or your instructor) says. Otherwise ask your instructor.

    -Dan
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