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$\phi: \mathbb{R^3} \rightarrow \mathbb{R^2}, \begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix} \mapsto \mathbb{R^2}, \begin{pmatrix} -2&1&5\\3&-1&-2\end{pmatrix} \begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix} + \begin{pmatrix} x_1&x_2&x_3\\x_3&x_2&x_1\end{pmatrix}\begin{pmatrix} 4\\-1\\3\end{pmatrix}$

I calculated the transformation matrix for the standart bases:

\begin{pmatrix} 2&0&8\\6&-2&2\end{pmatrix}

Calculated the nullspace:

\begin{pmatrix} 1&0&4&0\\0&1&11&0\end{pmatrix} the basis of the Kernel($\phi$) : \begin{pmatrix} -4\\-11\\1\end{pmatrix}

Now the dimensionformula says for a lineare transformation from one vectorspace to another one $f: V \rightarrow W$ is: dim($V$) = dim(Kernel($\phi$)+dimImage($\phi$).

So dimImage($\phi$) is (also the rang of the nullspacematrix) is: $3 = 1+x \Leftrightarrow 2 = x$

Now why is the transformation surjective?

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  • $\begingroup$ $3=1+x\iff x=2$ and not $3=1+x\iff x=1.$ $\endgroup$ – MathBeginner Feb 12 '17 at 10:41
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Using rank theorem $$\dim(\mathbb R^3)=\dim\ker \phi+\dim Im(\phi)=1+\dim Im(\phi)\implies \dim Im(\phi)=2$$ Since $Im(\phi)$ is a subspace of $\mathbb R^2$, you get $Im(\phi)=\mathbb R^2$, and thus, it's surjective.

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  • $\begingroup$ Yes, just saw it $\endgroup$ – WhatAMesh Feb 12 '17 at 10:41
  • $\begingroup$ So from what I understand we have a linear map from $\mathbb{R^3} \rightarrow \mathbb{R^2}$ and because our Image also has dimension 2 (just like the vectorspace we are mapping to), every point of the image is reached by the transformation $\endgroup$ – WhatAMesh Feb 12 '17 at 10:43

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