# Thread: Proving a matrix function as injective

1. ## 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?

2. $
A=\begin{pmatrix}3&5\\6&10\\-2 & -3\\\end{pmatrix}
$
Let $\overline{u}=\begin{pmatrix}
{x}\\
{y}
\end{pmatrix}$

Therefore:

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

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

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

This gives:

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

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

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

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

Hence:

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

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.

3. ## 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!