## anonymous 4 years ago 2. Suppose that vectors v1 = (1, 2), v2 = (2,−1), and that the basis B is B=v1, v2 . (this is a list) Let T be the linear transformation from R2 to R2 given by T(v1) = v1 and T(v2) = 0. (a) Write down the matrix for T in the new basis B. (You should be able to do this directly from the definition of T). (b) Use this to write down the matrix for T in the standard basis.

1. amistre64

1 2 2 -1 is B T(x) = Ax a c b d = A 1(a,b) + 2(c,d) = (1,2) 1a + 2c = 1 1b + 2d = 2 2(a,b)-1(c,d) = (0,0) 2a -1c = 0 2b - 1d = 0 Of course this inst according to the definition of a linear transform, but its how i do it

2. amistre64

1a + 2c = 1 2a - 1c = 0 *2 1a + 2c = 1 4a - 2c = 0 ---------- 5a = 1 ; a = 1/5 1a + 2c = 1 *-2 2a - 1c = 0 -2a -4c = -2 2a - 1c = 0 ----------- -5c = -2; c = 2/5

3. amistre64

1b + 2d = 2 2b - 1d = 0 -2b -4d = -4 2b - 1d = 0 ------------- -5d = -4 ; d=4/5 1b + 2d = 2 4b - 2d = 0 ------------- 5b = 2; b = 2/5 $A=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}$ maybe might have to leave the 1/5 in there to work tho

4. amistre64

B = v1, v2 newB = T(v1), T(v2) $T\binom{1}{2}+ T\binom{2}{-1}$ $1\binom{1}{2}+2\binom{1}{-1/2}$ ugh ... i just dont have the practice at doing it htis way :/

5. anonymous

ok so i'm confused, the matrix that I got for a was [1 0] [0 0]

6. amistre64

howd you get it? that might help me know if im completely off

7. anonymous

I got it because if T(v1) is (1, 2) and T(v2) is (0, 0), then the matrix that takes the vectors to that one is [1 0] [0 0]

8. anonymous

but i could be wrong, which is exactly why i posted the question here haha.

9. amistre64

1. 1 2. 0 = 1+0 1 0 0 = 0+0 not equal 2 so the transformation matrix doesnt tramsform v1 into v1

10. anonymous

lmao ok then i'm wrong.

11. anonymous

but that was my mind set for trying to get the right answer..

12. amistre64

$A=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{1}{2}=1\binom{1}{2}+2\binom{2}{4}=\binom{1+4=5}{2+8=10}/5=\binom{1}{2}$ right?

13. anonymous

right

14. amistre64

lets see if it works for v2 :) $A\vec{v_2}=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{2}{-1}=2\binom{1}{2}-1\binom{2}{4}=\binom{2-2=0}{4-4=0}/5=\binom{0}{0}$

15. amistre64

how to get to that thru just using the definitions of linear transformations I cant tell ...

16. amistre64

i take a generic A matrix, and create 2 sets of equations to solve

17. anonymous

ok, now do you know how i would find the stand basis matrix for that?

18. amistre64

i think that is it, but im reading up on it at the moment to be sure

19. amistre64

the part "a" is spose to get you to the A matrix that I did; but thru a different technique i think

20. anonymous

hmm, oh. i dont think so, part a asks for the matrix in the basis B

21. amistre64

B = 1 2 2 -1 newB = 1 0 2 0 is that part a?

22. amistre64

if i could see your textbook im sure id be like "doh!!"

23. amistre64

1.1 +0. 0 = 1 2 0 0 0.1 +1. 0 = 0 2 0 0 which is why you came up with the : 1 0 0 0 to begin with

24. anonymous

wahhh i wish i knew. that makes sense.

25. amistre64

ill read up on it tonight and see what I can remember :) good luck

26. anonymous

thanks!

27. anonymous

did you ever end up figuring anymore of it out? i'm so stumped.

28. amistre64

nothing really that makes any sense to me .... B = v1 v2 B = 1 2 2 -1 .......................................... newB = T(v1) T(v2) newB = 1 0 2 0 seems to be the way its looking from what I can gather. that should be part a its the part b thats got me frazzled

29. anonymous

ok so all of the stuff you put yesterday with everything over 5, that matrix, that has nothing to do with it? haha

30. amistre64

from what ive gathered; it was not the manner in which they are wanting this accomplished

31. amistre64

im not sure what newB and B have to do with each other rather than writing them up as baseses for vector spaces

32. amistre64

what is the name of the chapter that you are working on?

33. anonymous

it's all about changing basis... idk if you gathered that. so basically its just asking for the matrix of the transformation with respect to basis b and then in part b its asking for that same matrix only with respect to the standard basis. they're homework questions assigned and not in the textbook.

34. amistre64

so, part b is what: T(v1) = (1,0) ; T(v2) = (0,1)?

35. anonymous

i'm not sure if that's how you write it ... because i dont know if it's neccessarily going to the standard matrix, but what would the standard basis matrix be of the transformation .. if that makes any sense.

36. amistre64

i aint got a clue yet :) B = v1 v2 $B\vec{x} =x_1v_1+x_2v_2$ $B\vec{x} =x_1\binom 12+x_2\binom 2{-1}$ $B' = T(\vec{v_1})\ T(\vec{v_2})$ $B'\ \vec{x} =x_1T(v_1)+x_2T(v_2)$ $B'\ \vec{x} =x_1\binom 12+x_2\binom 00$ what this has to do with anything I cant tell

37. anonymous

lmao its all good, it's like the blind leading the blind here! thanks anyways!

38. amistre64

yeah, all in all, good luck with it ;)