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Thread: Proving a matrix function as injective

  1. #1
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    Proving a matrix function as injective

    I'm having trouble on a question proving that a function, multiplying a matrix by a column is injective.

    The question is:

    $\displaystyle A=\begin{pmatrix}3&5\\6&10\\-2 & -3\\\end{pmatrix} $

    and $\displaystyle T_A(\bar{u}) =A\bar{u} $

    is $\displaystyle T_A $ injective? And why?

    Can anyone give me any help?
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  2. #2
    Super Member Showcase_22's Avatar
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    $\displaystyle
    A=\begin{pmatrix}3&5\\6&10\\-2 & -3\\\end{pmatrix}
    $
    Let $\displaystyle \overline{u}=\begin{pmatrix}
    {x}\\
    {y}
    \end{pmatrix}$

    Therefore:

    $\displaystyle \begin{pmatrix}3&5\\6&10\\-2 & -3\\\end{pmatrix}\begin{pmatrix}
    {x}\\
    {y}
    \end{pmatrix}=\begin{pmatrix}
    {3x+5y}\\
    {6x+10y}\\
    {-2x-3y}
    \end{pmatrix}=T_A(\overline{u})$

    Suppose two values of $\displaystyle \overline{u}$ are mapped to the same point (ie. $\displaystyle T_A$ is not injective).

    Let $\displaystyle \overline{u_1}=\begin{pmatrix}
    {x_1}\\
    {y_1}
    \end{pmatrix}$ and let $\displaystyle \overline{u_2}=\begin{pmatrix}
    {x_2}\\
    {y_2}
    \end{pmatrix}$.

    This gives:

    $\displaystyle \begin{pmatrix}
    {3x_1+5y_1}\\
    {6x_1+10y_1}\\
    {-2x_1-3y_1}
    \end{pmatrix}=T_A(\overline{u_1})$

    $\displaystyle \begin{pmatrix}
    {3x_2+5y_2}\\
    {6x_2+10y_2}\\
    {-2x_2-3y_2}
    \end{pmatrix}=T_A(\overline{u_2})$

    Since they are mapped to the same point $\displaystyle T_A(\overline{u_1})=T_A(\overline{u_2})$.

    Therefore: $\displaystyle \begin{pmatrix}
    {3x_1+5y_1}\\
    {6x_1+10y_1}\\
    {-2x_1-3y_1}
    \end{pmatrix}=\begin{pmatrix}
    {3x_2+5y_2}\\
    {6x_2+10y_2}\\
    {-2x_2-3y_2}
    \end{pmatrix}$

    Hence:

    From the first (and second rows, i'm cutting some working here) $\displaystyle 3(x_1-x_2)+5(y_1-y_2)=0$.

    From the third row $\displaystyle 2(x_1-x_2)+3(y_1-y_2)=0$.

    We want to cancel out the y's.
    $\displaystyle 5+3n=0 \Rightarrow \ n=\frac{-5}{3}$.

    Therefore:

    $\displaystyle 3(x_1-x_2)-\frac{10}{3}(x_1-x_2)=0$

    $\displaystyle 9(x_1-x_2)=10(x_1-x_2)$

    $\displaystyle 9x_1-9x_2=10x_1-10x_2$

    $\displaystyle -x_1+x_2=0 \Rightarrow \ x_1=x_2$.

    We also have: $\displaystyle 2(x_1-x_2)+3(y_1-y_2)=0$.
    Since we already know that $\displaystyle x_1=x_2$, this implies that $\displaystyle y_1=y_2$.

    Hence $\displaystyle \overline{u_1}=\overline{u_2}$ so the function is injective.
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  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    4

    Thanks

    Thank you so much for that! After I saw that, I kicked myself for not thinking of it.

    I managed to do it all by myself now, thanks again!
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