Injective, surjective and bijective for linear maps I have problems showing that $\phi$ is surjective. My understanding is, that I have to show for every $u \in \mathbb{R}^3$ that there exists a $v \in \mathbb{R}^3$ but I am not sure how.

Let $a,b,c \in \mathbb{R}$. Let's examine the $\mathbb{R}$-linear map $\phi:\mathbb{R}^3\rightarrow\mathbb{R}^3$ defined through:
  $\phi(e_1)=\begin{bmatrix}b \\ -c \\ 1\end{bmatrix}, \phi(e_2)=\begin{bmatrix}a \\ 1 \\ 0\end{bmatrix}, \phi(e_3)=\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$ with $e_1, e_2, e_3$ as standard basis of $\mathbb{R}^3$.
  
  
*
  
*Is $\phi$ injective, surjective, bijective?
  
*If $\phi$ is bijective, find the inverse function $\phi^{-1}$.

My first step was to determine the linear map itself by:

$\phi(x,y,z) = x\phi(e_1) + y\phi(e_2) + z\phi(e_3) = x\begin{pmatrix}b \\ -c \\1\end{pmatrix} + y\begin{pmatrix}a \\ 1 \\0\end{pmatrix} + z\begin{pmatrix}1 \\ 0 \\0\end{pmatrix}$

Now I want to show, that $\phi$ is injective:

Since $\ker \phi$ must be $\{0\}$ I have following linear system of equations:
  $\begin{align} bx+ay+z = 0 \\ -cx + y = 0\\x = 0\end{align}$
So it follows $x = y = z = 0$. Therefore $\phi$ must be injective.

Now if I want to show that $\phi$ is surjective can I just say, if $u \in \mathbb{R}^3$ then there is obviously a $v = \phi(u_1, u_2, u_3) \in \mathbb{R}^3$?
 A: There are a number of ways of doing this.


*

*One is the brute force approach. Given any $(v_1,v_2,v_3)\in\mathbb{R^3}$, you need to show that you can find $x$, $y$, and $z$ such that 
$$\begin{array}{rcccccl}
bx & + & ay & + & z & = & v_1\\
-cx & + & y &   &   & = & v_2\\
x  & & & & &=&v_3
\end{array}$$
no matter what $v_1$, $v_2$, and $v_3$ are. This is not too difficult to do. This shows explicitly that the function is surjective.

*Another method is to finesse it using some theory. If you already know the Rank-nullity Theorem, then you know that the dimension of the image plus the dimension of the kernel must equal the dimension of the domain:
$$\dim(V) = \dim(\mathrm{ker}\phi) + \dim(\mathrm{Im}\phi).$$ 
Here, the dimension of the domain is $3$, and the dimension of the kernel is $0$, so the dimension of the image must be $3$. Since the codomain is $\mathbb{R}^3$, of dimension $3$, then the image must equal the entire codomain; that is, the function must be surjective.
In fact, as an easy consequence of the Rank-Nullity Theorem we conclude that if $V$ if finite dimensional, then for linear transformation $T\colon V\to V$, the three conditions (injectivity, surjectivity, and bijectivity) are in fact equivalent. Any one implies the other two. 

*Other bits of theory that may come in handy are results you may know about systems of linear equations. For example, you probably know that a homogeneous system with more unknowns than equations always has nontrivial solutions. You may also know that a system of the form $A\mathbf{x}=\mathbf{b}$, where $A$ is an $n\times n$ matrix, has a solution for all $\mathbf{b}$ if and only if the only solution to $A\mathbf{x}=\mathbf{0}$ is the trivial solution. Again, this amounts to using some theory to finesse the solution.
Whether or not your grader would accept "it's obvious" depends on the level of the course. My guess is that it would not be sufficient for full marks, given the level of the problem. 
A: Also there is a result: Consider linear transformation  $T:U\to V$, where $\dim(U)= \dim(V)$.
Then $T$ invertible (i.e injective and surjective) iff  $\ker T = \{0\}$. 
