1. ## Something more??

Hey guys!
I have a question here where I'm supposed to get the kernel and image of a linear transformation. I think I must be missing something, coz I feel like I haven't done enough!

Let T: $R^3$ $\rightarrow$ $R^2$ be the linear transformation defined by $T((x_{1},x_{2},x_{3}))=(x_{1}+x_{2},x_{2}-2x_{3})$

(a) Find the kernel of T
(b) Show that (1,0) and (0,1) are in the range of T. Determine the range of T.

For (a), I just let:
$x_{1}+x_{2}=0$,and
$x_{2}-2x_{3}=0$,
and I get $x_{1}=-2x_{3}$.
So is my answer: kerT= $(-2x_{3}+x_{2},x_{2}+x_{1})$
I'm guessing this is a stupid question, and I'm missing out loads! Help would be appreciated!
Thanks,
SimplySparklers

2. Originally Posted by simplysparklers
Hey guys!
I have a question here where I'm supposed to get the kernel and image of a linear transformation. I think I must be missing something, coz I feel like I haven't done enough!

Let T: $R^3$ $\rightarrow$ $R^2$ be the linear transformation defined by $T((x_{1},x_{2},x_{3}))=(x_{1}+x_{2},x_{2}-2x_{3})$

(a) Find the kernel of T
(b) Show that (1,0) and (0,1) are in the range of T. Determine the range of T.

For (a), I just let:
$x_{1}+x_{2}=0$,and
$x_{2}-2x_{3}=0$,
and I get $x_{1}=-2x_{3}$.
So is my answer: kerT= $(-2x_{3}+x_{2},x_{2}+x_{1})$
I'm guessing this is a stupid question, and I'm missing out loads! Help would be appreciated!
Thanks,
SimplySparklers
The Kernel of $T$ is the set of all $x \in \mathbb{R}^3$ such that $Tx=\bold{0}$, so solving the equations (that you have above) that this implies gives the Kernel is the set $\{(x,-x,-x/2): \ x\in \mathbb{R} \}$

CB