# Thread: Linear Algebra: Finding a specific linear transformation issue

1. ## Linear Algebra: Finding a specific linear transformation issue

Hello, this is my first time posting. I've been wracking my brain over the following problem for two days now and without any sort of examples in my book, study guide or notes I'm starting to go crazy.

Here's the problem: Let L: $R^2$ $\rightarrow$ $R^3$ such that:
L(1 -1) = (2 1 3) AND L(-2 3) = (-1 0 -1)

a) Determine Lx
b) Determine if L is one-to-one.

Now, I know if I can find what the transformation is, with the given information, I can tell if it's one-to-one without any problems but I, quite honestly, don't know where to begin on figuring out what the transformation is.

Thank you so much in advance!

2. ## Re: Linear Algebra: Finding a specific linear transformation issue

Originally Posted by goatmafioso
... I'm starting to go crazy.
That is not a good idea.

Here's the problem: Let L: $R^2$ $\rightarrow$ $R^3$ such that: L(1 -1) = (2 1 3) AND L(-2 3) = (-1 0 -1) a) Determine Lx
$B=\{(1,-1),(-2,3)\}$ is a basis of $\mathbb{R}^2$ so, we can express uniquely $x=(x_1,x_2)=\lambda_1(1,-1)+\lambda_2(-2,3)$ . Solving, you'll obtain $(x_1,x_2)=(3x_1+2x_2)(1,-1)+(x_1+x_2)(-2,3)$ . Now, use the linearity of $L$

b) Determine if L is one-to-one.
Use the characterization: $L$ is one to one iff $\ker f=\{0\}$ .