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.

- anonymous

- chestercat

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- 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

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

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

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## More answers

- 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

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

- amistre64

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

- 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

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

- 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

lmao ok then i'm wrong.

- anonymous

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

- 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

right

- 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

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- 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

wahhh i wish i knew.
that makes sense.

- amistre64

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

- anonymous

thanks!

- anonymous

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

- 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

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

- amistre64

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

- 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

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

- 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

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

- 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

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

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

- amistre64

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

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