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How can I find a linear map $f: \mathbb R^3 \to \mathbb R^3 $ such that $$image(f)=\{(x_1,x_2,x_3)\in\mathbb R^3: 3x_1-2x_2+3x_3=0\}?$$

Not looking for a solution to this specific problem, but more of a general approach when having to find a linear map given the kernel or image. Thanks in advance.

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  • $\begingroup$ hint try the matrix transformation [300 0-20 003] $\endgroup$ – shai horowitz Jul 9 '17 at 9:06
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    $\begingroup$ Had a chance to read the answers that have been provided, Tesla? $\endgroup$ – Gerry Myerson Jul 19 '17 at 9:29
  • $\begingroup$ Earth to Tesla – come in, please. $\endgroup$ – Gerry Myerson Jul 20 '17 at 13:41
  • $\begingroup$ Hi, I am preparing for another exam at the moment, will check the answers after that for sure. Thanks in advance. $\endgroup$ – Tesla Jul 21 '17 at 12:59
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You have a basis $(e_1,e_2,e_3)$ for $\mathbb{R}^3$. To find a linear map $f\in\mathcal{L}(\mathbb{R}^3)$ such that $\text{Im}f=S$ where $S$ is a subspace of $\mathbb{R}^3$, notice that $\text{Im}f=\text{span}(f(e_1),f(e_2),f(e_3))$. Notice also that $f$ is entirely determined if you know $f(e_1)$, $f(e_2)$ and $f(e_3)$. Hence simply find a basis for $S$ and try to choose $f(e_1)$, $f(e_2)$ and $f(e_3)$ in an appropriate way.

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You want a linear transformation $f:V\to W$ with given kernel $X$. You can find a basis for $X$, extend it to a basis for $V$, take the basis for $X$ to zero, and take the other basis vectors of $V$ to linearly independent elements of $W$ (by first finding a basis for $W$). I'm assuming here that all your spaces are finite-dimensional, and that the dimensions are such that this kind of kernel is possible.

For a given range $Y$, you can do something similar. Find a basis for $Y$, take basis vectors of $V$ to these basis vectors of $Y$, making sure that each basis vector of $Y$ is the image of at least one of the basis vectors of $V$.

In both cases, once you've worked out where the basis vectors go, extend the map to all of $V$ by linearity.

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