Finding linear map when given two eigen vectors and the corresponding eigen values

Question

Let $ f : ℝ^2 \rightarrow ℝ^2$ be a linear map. I am told (1,1) is the eigenvector associated with the eigenvalue 2, and that (2,0) is the eigenvector associated with the eigenvalue -1. How do I find f(5,3)?

I apologize for any miscommunication, I am not at all used to debating maths in English since it's not my native language. Also, I understand that this might be very easy, it's just that I've been stuck on it for too long and want to understand. Thank you in advance.

Answer 1

Note that $(5,3)=3(1,1)+(2,0)$. By linearity of $f$, $f(5,3)=f(3(1,1)+(2,0)) = 3f(1,1)+f(2,0) = 3\cdot 2 (1,1)+(-1)(2,0) = (4,6)$.

Answer 2

The domain of $f$ must be $\mathbb R^{2}$ (You have typed it as $\mathbb R$).

Any vector $(x,y)$ can be written as $t(1,1)+s(2,0)$. In fact we get $t=y$ and $s=\frac {x-y} 2$ from this. By linearity we get $f(x,y)=tf(1,1)+sf(2,0)=2t(1,1)-s(2,0)$. Now just put $t=y=3$ and $s=\frac {x-y} 2=1$. This gives $f(5,3)=(4,6)$.

Answer 3

These two facts DO tell you everything about the xform. You know

1. $f(\pmatrix{1\\1}) = \pmatrix{2\\2}$.

2. $f(\pmatrix{2\\0}) = \pmatrix{-2\\0}$.

3. $f$ is linear.

You'd like to find $$ f(\pmatrix{5\\3}). $$ If you write $$ \pmatrix{5\\3} = 3 \pmatrix{1\\1} + \pmatrix{2\\0}$$ you can use linearity of $f$ to solve your problem.