TITLE: Injective, surjective and bijective for linear maps
QUESTION [9 upvotes]: 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 $v = \phi(u_1, u_2, u_3) \in \mathbb{R}^3$?

REPLY [2 votes]: 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  $\ker T = \{0\}$.