I have to find the basis for the kernel and image of the following linear map.

$ \phi: R^3 → R^2, ϕ \begin{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix}\end{pmatrix}= \begin{pmatrix} x -y \\z \end{pmatrix} $

For the range, I think we can express any arbitrary linear transformation as:

$ x\begin{pmatrix} 1 \\ 0 \end{pmatrix} - y\begin{pmatrix} 1 \\ 0 \end{pmatrix} + z\begin{pmatrix} 0 \\ 1 \end{pmatrix} $. So I think that a basis for the range would be $\left\{{{\begin{pmatrix} 1 \\ 0 \end{pmatrix}},{\begin{pmatrix} 0 \\ 1 \end{pmatrix}}}\right\}$

As for the kernel, we set

$\begin{pmatrix} x -y \\z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

We get $x=y$ and $z=0$. Therefore a basis for the kernel would be

$\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\right\}$

I am doing this right?

Thanks in advance

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    $\begingroup$ Yup, looking good. $\endgroup$ – Dionel Jaime Oct 1 '17 at 23:54
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    $\begingroup$ Welcome to Math SX! It's quite correct. $\endgroup$ – Bernard Oct 1 '17 at 23:54

Your calculations are correct.


Everything looks good. A slightly more elementary (but less insightful) approach that works quite well for calculations that are a little harder to "guess" as you say is to form the matrix for the linear transformation by checking the basis:

$f(1,0,0)=(1,0), \, f(0,1,0)=(-1,0) \, , f(0,0,1)=(0,1)$

So, we get the matrix

$$f:\mathbb R^3 \to \mathbb R^2 \iff\begin{pmatrix}1 & -1 & 0\\0 &0 & 1 \end{pmatrix}:\mathbb R^3 \to \mathbb R^2$$

From this, you can determine the "column space" since the first and third column vectors are linearly independent, it has to be $\mathbb R^2$ for the image, and the kernel is generated by the "null space" which amounts to solving a homogeneous system of linear equations, i.e:

$$\begin{pmatrix}1 & -1 & 0\\0 &0 & 1 \end{pmatrix} \begin{pmatrix} x \\y \\z\end{pmatrix}=\begin{pmatrix}0\\0 \end{pmatrix}$$ although this is just a restatement of your own method, it computationally allows the use of Gauss-Jordan Elimination and identifies an isomorphic subspace in $\mathbb R^3$ orthogonal to the image that is also one dimensional.


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