Determine $a \in \mathbb{R}$ for which F is an isomorphism The linear mapping $F: \mathbb{R^2} \rightarrow \mathbb{R^2}$ with $F(x_1,x_2) := (x_1+2x_2, \ 2x_1 + ax_2) \ , \  a \in \mathbb{R} \ $ is given. 
I want to know all $a \in \mathbb{R}$ for which $F$ is an isomorphism. 
Generally I would do the following: 
First, show that the given mapping is linear, in other words, that the conditions additivity and homogeneity are satisfied. 
Second, I would show that the given mapping is bijective.
In this case I don't have to show that it is a linear mapping, because this is already given. 
Bijectivity means: injectivity and surjectivity. 
If someone would put a gun to my head I would say, that you can show injectivity by showing that $\text{ker}F=\{(0,0)\}$, which in this case would be fulfilled if $ a \neq 4 $. (I solved this here.) 
I wouldn't know how to show surjectivity except of repeating the words I read in another example which – applied to this example – would be : "$dim(\mathbb{R^2})= dim(\mathbb{R^2})=2$, therefore surjectivity."
I don't see how showing that the two vector spaces in our mapping have the same dimension means we know that the linear mapping is surjective.
I'm grateful for every hint.
 A: Since $f$ is a linear map, $\text{im}(f)$ is a subspace of $\mathbb{R}^2$.

Since $\ker(f) = 0$, it follows (by rank-nullity) that $\text{im}(f)$ has dimension $2$.

But for any finite-dimensional vector space $V$ of dimension $n$ say, any subspace $W$ of $V$ with dimension less than $n$ must be a proper subspace.  Why? Suppose $W$ has dimension $m < n$. Then a basis for $W$ has $m$ elements, so can't generate $V$ since any basis for $V$ must have $n$ elements.

Thus, in the context of the current problem, since $\text{im}(f)$ has dimension $2$, it can't be a proper subspace of $\mathbb{R}^2$, hence we must have $\text{im}(f) = \mathbb{R}^2$. Therefore $f$ is surjective.

For this problem, as an alternative, more elementary approach, surjectivity can be proved as follows . . .

Let $(y_1,y_2) \in \mathbb{R}^2$, and actually find a point $(x_1,x_2) \in \mathbb{R}^2$ such that $f(x_1,x_2) = (y_1,y_2)$.

In other words, solve the equation $f(x_1,x_2) = (y_1,y_2)$ for $x_1,x_2$ in terms of $y_1,y_2$.

With all other variables except $x_1,x_2$ regarded as constant, the equation
$f(x_1,x_2)=(y_1,y_2)$ is just a system of two linear equations in two unknowns (so elementary algebra is all that's needed).
A: To show surjectivity, take any point $(x_1,y_1)$ and show that there exists $(x,y)$ such that $x+2y = x_1, 2x + ay = y_1$. If $a \neq 4$ then you always get such x and y, in fact $y = \frac{2x_1-y_1}{4-a}, x = \frac{2y_1-ax_1}{4-a}$. 
