## anonymous 3 years ago question:

1. anonymous

Let L: $R ^{2} -> R ^{2}$ be the linear transformation defined by $L[\left(\begin{matrix}a \\ b\end{matrix}\right)]= \left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right] \left(\begin{matrix}a \\ b\end{matrix}\right)$ a)find ker L b) find a set of vectors spanning range L.

2. anonymous

just explain the procedure how to do it. @lalaly

3. UnkleRhaukus

what does ker mean

4. anonymous

i guess its just L . ker is a printing mistake sorry for that uncle

5. UnkleRhaukus

?

6. TuringTest

kernel

7. anonymous

maybe i m sure about kernel its only written ker L

8. anonymous

i just need the way to solve it

9. TuringTest

I'm looking over my notes, I always forget this stuff

10. anonymous

no problem

11. TuringTest

oh, kernel is the same thing as null space for a linear transformation i.e. the set of all $$\vec x$$ in $$A\vec x=\vec 0$$ is the kernel, so just solve that matrix to find part a)

12. anonymous

ok

13. TuringTest

that is, solve$\left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right] \left(\begin{matrix}a \\ b\end{matrix}\right)=0$

14. anonymous

then we'll get two eqs right ???

15. TuringTest

yes, from which we will get the two vectors that will make up the basis of our kernel

16. TuringTest

at least I think that is what they want

17. anonymous

ok

18. anonymous

19. TuringTest

consulting notes again...

20. anonymous

ok

21. TuringTest

hm... I'm perhaps a bit confused because my notes don't seem to agree with wikipedia. I think the dimension of the null space, i.e. the nullity is the kernal in that case, the answer to part a) would be the number of vectors that make up the basis for the null space of your transformation matrix the answer to part b) must then be the basis itself I guess, which is the set of vectors that span $$A\vec x=\vec0$$

22. anonymous

ok thank you i got it :) i just need some practice now

23. TuringTest

yeah, me too :) welcome, good luck