anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

amistre64
  • 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
amistre64
  • 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
amistre64
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • 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 :/
anonymous
  • anonymous
ok so i'm confused, the matrix that I got for a was [1 0] [0 0]
amistre64
  • amistre64
howd you get it? that might help me know if im completely off
anonymous
  • 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]
anonymous
  • anonymous
but i could be wrong, which is exactly why i posted the question here haha.
amistre64
  • amistre64
1. 1 2. 0 = 1+0 1 0 0 = 0+0 not equal 2 so the transformation matrix doesnt tramsform v1 into v1
anonymous
  • anonymous
lmao ok then i'm wrong.
anonymous
  • anonymous
but that was my mind set for trying to get the right answer..
amistre64
  • 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?
anonymous
  • anonymous
right
amistre64
  • amistre64
lets see if it works for v2 :) \[A\vec{v_2}=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{2}{-1}=2\binom{1}{2}-1\binom{2}{4}=\binom{2-2=0}{4-4=0}/5=\binom{0}{0}\]
amistre64
  • amistre64
how to get to that thru just using the definitions of linear transformations I cant tell ...
amistre64
  • amistre64
i take a generic A matrix, and create 2 sets of equations to solve
anonymous
  • anonymous
ok, now do you know how i would find the stand basis matrix for that?
amistre64
  • amistre64
i think that is it, but im reading up on it at the moment to be sure
amistre64
  • amistre64
the part "a" is spose to get you to the A matrix that I did; but thru a different technique i think
anonymous
  • anonymous
hmm, oh. i dont think so, part a asks for the matrix in the basis B
amistre64
  • amistre64
B = 1 2 2 -1 newB = 1 0 2 0 is that part a?
amistre64
  • amistre64
if i could see your textbook im sure id be like "doh!!"
amistre64
  • 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
anonymous
  • anonymous
wahhh i wish i knew. that makes sense.
amistre64
  • amistre64
ill read up on it tonight and see what I can remember :) good luck
anonymous
  • anonymous
thanks!
anonymous
  • anonymous
did you ever end up figuring anymore of it out? i'm so stumped.
amistre64
  • 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
anonymous
  • anonymous
ok so all of the stuff you put yesterday with everything over 5, that matrix, that has nothing to do with it? haha
amistre64
  • amistre64
from what ive gathered; it was not the manner in which they are wanting this accomplished
amistre64
  • amistre64
im not sure what newB and B have to do with each other rather than writing them up as baseses for vector spaces
amistre64
  • amistre64
what is the name of the chapter that you are working on?
anonymous
  • 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.
amistre64
  • amistre64
so, part b is what: T(v1) = (1,0) ; T(v2) = (0,1)?
anonymous
  • 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.
amistre64
  • 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
anonymous
  • anonymous
lmao its all good, it's like the blind leading the blind here! thanks anyways!
amistre64
  • amistre64
yeah, all in all, good luck with it ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.