0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – rogerl Feb 9 '16 at 20:51
  • $\begingroup$ 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 $\Bbb R$. $\endgroup$ – Gregory Grant Feb 9 '16 at 20:51
1
$\begingroup$

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 $\;\{1\}\subset\Bbb R\;$ :

$$\begin{cases}\phi e_1=1\cdot1\\\phi e_2=2\cdot1\\\phi e_3=(-1)\cdot1\end{cases}\;\;\implies [\phi]=(1\;\;\;2\;\,-1\;)\in M_{1\times3}(\Bbb R)$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.