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Linear mapping: \begin{align*} F: \mathbb R^3 &\to \mathbb R^2,\\ \begin{pmatrix} x\\y\\ z \end{pmatrix} &\mapsto \begin{pmatrix} x\\y\\ \end{pmatrix} \end{align*}

I thought the check for 1-1 was to do the following:

$$F\begin{pmatrix}x_1\\y_1\\ z_1\end{pmatrix} = \begin{pmatrix} x_2\\y_2\\ z_2 \end{pmatrix}$$

where $x_1=x_2$ and $y_2 = y_1$ so it is 1-1?

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    $\begingroup$ It maps $(0,0,0)$ and $(0,0,1)$ to $(0,0)$, so it is not injective(one to one), it is however surjective(onto) since for every $(x,y) \in \mathbb{R}^2, f(x,y,0)=(x,y)$. $\endgroup$ – user45878 Feb 26 '14 at 19:08
  • $\begingroup$ By the way, instead of using "matrix", you can use "pmatrix" to get the parentheses around a matrix $\endgroup$ – M Turgeon Feb 26 '14 at 19:11
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A one-to-one function maps every "input" to one and only one "output". Consider

$f: \mathbb{R^3} \mapsto \mathbb{R^2} \rightarrow f(x,y,z) = x,y$

Then $f(x_1,y_1,z_1) = x_1, y_1$ & $f(x_1,y_1,z_2) = x_1, y_1$

So there are two (and, in this case, infinite) inputs for any single output.

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One-to-one means that no two distinct triples (in the case) get sent to the same pair. Is it possible for (a,b,c) and (x,y,z) to be sent to the same point, without being equal to each other?

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Observe that f(1,2,3) = f(1,2,4) = (1,2) but (1,2,3) is not equal to (1,2,4). So F is not one-to-one.

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Calculating the kernel of $F$, we obtain $$\ker F=\{(x,y,z): F(x,y,z)=(0,0)\}= \{(x,y,z):(x,y,z)=(0,0)\}=$$ $$=\{(0,0,z), \forall z\in \mathbb{R}\}\neq \{(0,0)\}.$$

Then, $F$ is not injective.

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