Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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 :/

Answer 5

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

Answer 6

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

Answer 7

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]

Answer 8

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

Answer 9

1. 1 2. 0 = 1+0 1
0 0 = 0+0 not equal 2

so the transformation matrix doesnt tramsform v1 into v1

Answer 10

lmao ok then i'm wrong.

Answer 11

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

Answer 12

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

Answer 13

right

Answer 14

lets see if it works for v2 :)

\[A\vec{v_2}=\frac{1}{5}\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{2}{-1}=2\binom{1}{2}-1\binom{2}{4}=\binom{2-2=0}{4-4=0}/5=\binom{0}{0}\]

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

B = 1 2
2 -1

newB = 1 0
2 0 is that part a?

Answer 22

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

Answer 23

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

Answer 24

wahhh i wish i knew.
that makes sense.

Answer 25

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

good luck

Answer 26

thanks!

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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.

Answer 34

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

Answer 35

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.

Answer 36

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

Answer 37

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

Answer 38

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