Showing span to be in $\Bbb R^2$ space Let say $L: \mathbb{R}^2 \to \mathbb{R}^2$ ($\mathbb{R}$ refers to set of real numbers) and $e_1$ and $e_2$ to be the standard basis of $\mathbb{R}^2$. I want to disprove that the $\text{span}(L(e_1),L(e_2)) = \mathbb{R}^2$ by counter example. Please let me know if my approach is correct here.
By definition, we know that $L(e_1)$ and $L(e_2)$ spans $\mathbb{R}^2$ if and only if $L(e_1)$ and $L(e_2)$ are linearly independent (please let me know if this definition is correct). Suppose function L is as such: $L(x,y) = (5,5)$, then $L(e_1)$ and $L(e_2)$ are not linearly independent. Therefore in this case, $ L(e_1)$ and $L(e_2)$ does not span $\mathbb{R}^2$.
Any insights and/or corrections will be helpful. Thank you.
 A: Yup you're right. See you've let the transformation to be in a way that it's not even a Linear transformation. But yes it's possible to have an example such that $span\{L(e_1),L(e_2)\} \ne \mathbb{R}^2$ because there's a theorem that says if
$ V$ and $W$ be vector spaces over a field $F$. Let $\{ \alpha_1,  \alpha_2, ......, \alpha_n \}$ be a basis of $V$ and $\{ \beta_1, \beta_2,......, \beta_n \}$ be arbitrarily chosen elements (not necessarily distinct) in $W$. Then there exists one and only one linear mapping $T :  V \to W$ such that $T(\alpha_i)= \beta_i$ for $i=1,2,.....,n$.
With the help of this theorem let's map $e_1$ and $e_2$ to same elements in $\mathbb{R}^2$ and then find the transformation. 
Here you have what you wanted.
Hope it helps.
A: Let me get this right. Here, you're defining $L(x,y) = (5,5)$, not $L(x,y) = (5x,5y)$? If the first is the case, then $L$ wouldn't even be a linear map, since its image is not a subspace (it's just a point, and one that is not at the origin), so naturally, its range is not $\mathbb R^2$.
I hope that answers your question.
A: The definition is correct. However if you assume that $L$ is constant (as you seem to do), then $L$ can not be a linear map. Anyway it is true in this case, that the span of $L(e_1)$ and $L(e_2)$ is not all of $\mathbb{R}^2$. 
A better example would be the function defined by $L(x,y) := (x,0)$. In this case we get a linear map and $L(e_1) = e_1$ and $L(e_2) = 0$, which is linearly dependent with $e_1$. Thus $\text{Span}(L(e_1),L(e_2)) \neq \mathbb{R}^2$.
I hope this clarifies things a bit! 
