## annas Group Title question: 2 years ago 2 years ago

1. annas Group Title

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. annas Group Title

just explain the procedure how to do it. @lalaly

3. UnkleRhaukus Group Title

what does ker mean

4. annas Group Title

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

5. UnkleRhaukus Group Title

?

6. TuringTest Group Title

kernel

7. annas Group Title

maybe i m sure about kernel its only written ker L

8. annas Group Title

i just need the way to solve it

9. TuringTest Group Title

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

10. annas Group Title

no problem

11. TuringTest Group Title

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. annas Group Title

ok

13. TuringTest Group Title

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

14. annas Group Title

then we'll get two eqs right ???

15. TuringTest Group Title

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

16. TuringTest Group Title

at least I think that is what they want

17. annas Group Title

ok

18. annas Group Title

19. TuringTest Group Title

consulting notes again...

20. annas Group Title

ok

21. TuringTest Group Title

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. annas Group Title

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

23. TuringTest Group Title

yeah, me too :) welcome, good luck