# Basis for Kernel and Image of a linear map

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?

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

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