# 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:

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

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

is $\displaystyle T_A$ injective? And why?

Can anyone give me any help?

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

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!