Proving Linearity Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ Prove that
$f$ is linear $\iff f(\alpha x+\beta y) = \alpha f(x) + \beta f(y)$ for all $x,y \in \mathbb{R}^n$ and all $\alpha , \beta \in \mathbb{R}$.
I know that this statement is one of the definitions of linearity but I don't know how to prove it. Does it have something to do with the corresponding $m$x$n$ matrix that exists for this function?
 A: Seeing as the question posed is an "if and only if" situation, we need to use each fact to confirm the other. If we have some linear function $ f:\mathbb{R}^n →\mathbb{R}^m $, then...
$$ (1): f(x+y)=f(x) + f(y)  $$
$$(2): f( \alpha x)=\alpha f(x) $$
If we have $f(\alpha x +\beta y)$ and $f$ is linear, using the first property (1):
$$ f(\alpha x+\beta y)=f(\alpha x) + f(\beta y). $$
Then using the second property (2):
$$f(\alpha x) = \alpha f(x) \ \ \ and  \ \ \ f(\beta y)=\beta f(y), $$
$$ f(\alpha x) + f(\beta y) = \alpha f(x) + \beta f(y).$$
hence, 
$$f(\alpha x + \beta y) = \alpha f(x) + \beta f(y).$$
Therefore $f$ is linear $ → f(\alpha x + \beta y) = \alpha f(x) + \beta f(y). $ It seems a bit trivial, I'm not sure if it's sufficient or not.
I'm also not sure how to prove that $ f(\alpha x + \beta y) = \alpha f(x) + \beta f(y) \rightarrow $ $f$ is linear. 
A: I am not sure whether this is what you want or not,  but I'm giving it a shot here.
If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is linear, that means $f(x)$ can be expressed as $f(x) = Ax$ where $A \in \mathbb{R}^{m\times n}$. Then,
$f(\alpha x + \beta y) = A(\alpha x + \beta y)$. By the linearity properties of a matrix (here is the catch, I'm not sure whether this is even allowed in your problem because it is kinda the same thing as your definition to me), $f(\alpha x + \beta y) = \alpha Ax + \beta Ay = \alpha f(x) + \beta f(y)$.
If the linearity of matrix is not allowed then I am out of ideas for now. I will come back and edit if I come up with anything else.
A: It is defined that way -- you don't need to prove it.
You can then use this definition to show that some function is linear.
Let's also take a step back to when you probably recently learned about dimensions of a vector space = the number of linearly independent vectors that span the vector space.  This is how dimension is defined -- you don't have to prove it.  Now, you use this definition of dimension to show that a vector space has dimension, $k$ - with computations and whatnot.
