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Math Help - 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:

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

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

    is  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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    <br />
A=\begin{pmatrix}3&5\\6&10\\-2 & -3\\\end{pmatrix}<br />
    Let \overline{u}=\begin{pmatrix}<br />
{x}\\ <br />
{y}<br />
\end{pmatrix}

    Therefore:

    \begin{pmatrix}3&5\\6&10\\-2 & -3\\\end{pmatrix}\begin{pmatrix}<br />
{x}\\ <br />
{y}<br />
\end{pmatrix}=\begin{pmatrix}<br />
{3x+5y}\\ <br />
{6x+10y}\\<br />
{-2x-3y}<br />
\end{pmatrix}=T_A(\overline{u})

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

    Let \overline{u_1}=\begin{pmatrix}<br />
{x_1}\\ <br />
{y_1}<br />
\end{pmatrix} and let \overline{u_2}=\begin{pmatrix}<br />
{x_2}\\ <br />
{y_2}<br />
\end{pmatrix}.

    This gives:

    \begin{pmatrix}<br />
{3x_1+5y_1}\\ <br />
{6x_1+10y_1}\\ <br />
{-2x_1-3y_1}<br />
\end{pmatrix}=T_A(\overline{u_1})

    \begin{pmatrix}<br />
{3x_2+5y_2}\\ <br />
{6x_2+10y_2}\\ <br />
{-2x_2-3y_2}<br />
\end{pmatrix}=T_A(\overline{u_2})

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

    Therefore: \begin{pmatrix}<br />
{3x_1+5y_1}\\ <br />
{6x_1+10y_1}\\ <br />
{-2x_1-3y_1}<br />
\end{pmatrix}=\begin{pmatrix}<br />
{3x_2+5y_2}\\ <br />
{6x_2+10y_2}\\ <br />
{-2x_2-3y_2}<br />
\end{pmatrix}

    Hence:

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

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

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

    Therefore:

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

    9(x_1-x_2)=10(x_1-x_2)

    9x_1-9x_2=10x_1-10x_2

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

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

    Hence \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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