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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 $\mathbb{R}^3$.

  1. Is $\phi$ injective, surjective, bijective?

  2. 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 $v = \phi(u_1, u_2, u_3) \in \mathbb{R}^3$?

  • 4
    +1 for showing your work! Re: last question: Well, you *can* say that, but you *shouldn't* unless it is obvious to you (and you wouldn't ask this question if it were, would you?). You're almost there, in fact. Note that solving $\phi(u_1,u_2,u_3) \in \mathbb{R}^3 = v$ for $u_1,u_2,u_3$ amounts to solving a certain linear system of equations.2011-05-12
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    Since your map is a linear map from $\mathbb{R}^3$ to itself, the invertible matrix theorem applies here and in particular tells you about the relationship between injective and surjective.2011-05-12

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