# Define $\phi:\mathbb{R}^3 \rightarrow \mathbb{R}$ by $\phi(e_1) = 1$, $\phi(e_2) = 2$, $\phi(e_3)=-1$. Determine ker$\phi$ and im$\phi$

Let {$e_1,e_2,e_3$} be the standard basis for $\mathbb{R}^3$ and define $\phi:\mathbb{R}^3 \rightarrow \mathbb{R}$ by $\phi(e_1) = 1$, $\phi(e_2) = 2$, $\phi(e_3)=-1$. Determine the subspaces ker$\phi$ and im$\phi$, and verify the rank theorem in this case.

I know how to find kernel and image given a matrix and I also know the Rank Theorem

Kernel is the nullspace and Image is the span of vectors of the linear transformation

The Rank Theorem is dim$V$ = dim(ker$\phi$) + dim(im$\phi$)

but I'm confused as to how to find them with the information I am given.

• Can you write down a matrix for $\phi$? If you can, you should be able to use what you know to find the kernel and image of the map. To write down a matrix for $\phi$, you need to know what it does to an arbitrary element $(x,y,z)\in\mathbb{R}^3$; since it is linear and you know what it does to basis elements, this should be easy. – rogerl Feb 9 '16 at 20:51
• Basically $(x,y,z)\mapsto x+2y-z$. So just find all $(x,y,z)$ such that $x+2y-z=0$. That gets you the kernel. The image is obviously all of $\Bbb R$. – Gregory Grant Feb 9 '16 at 20:51

You already have everything you need to write a matrix with respect of basis $\;\{e_1,e_2,e_3\}\subset\Bbb R^3\;$ and $\;\{1\}\subset\Bbb R\;$ :
$$\begin{cases}\phi e_1=1\cdot1\\\phi e_2=2\cdot1\\\phi e_3=(-1)\cdot1\end{cases}\;\;\implies [\phi]=(1\;\;\;2\;\,-1\;)\in M_{1\times3}(\Bbb R)$$