Thread: how to find a matrix of a linear transformation

Let T: V---->W be the linear transformation that has the matrix:

1 -1 1
2 1 1

relative to the bases{(1,2,0),(1,1,1),(1,1,0)} of V and {(1,1),(1,-1)} of W
Find the matrix of T relative to the bases{(2,3,0),(1,1,1),(2,3,1)} of V and {(3,-1),(1,-1)} of W

you will need two "change of basis" matrices:

P, which changes vectors with coordinates corresponding to the basis {(2,3,0),(1,1,1),(2,3,1)} to vectors with coordinates corresponding to the basis {(1,2,0),(1,1,1),(1,1,0)}.

Q which changes vectors with coordinates corresponding to the basis {(1,1),(1,-1)} to vectors with coordinates corresponding to the basis {(3,-1),(1,-1)}.

if A is your given matrix, then the matrix you want is:

QAP.

what have you done so far?

I missed last week class cuz i was sick so I didn't learn change of basis yet
and this is the only problem I have trouble with on my homework

*sigh*

i will show you how to find P. you can try figuring out Q yourself.

if we let B = {(2,3,0),(1,1,1),(2,3,1)} then since (2,3,0) = 1(2,3,0) + 0(1,1,1) + 0(2,3,1) in the basis B, (2,3,0) has coordinates [1,0,0]_B.

what we want to know is: what is [1,0,0]_B in B'-coordinates?

that is, we need to know how to express (2,3,0) as a linear combination of (1,2,0),(1,1,1) and (1,1,0).

well, if (2,3,0) = a(1,2,0) + b(1,1,1) + c(1,1,0) = (a+b+c,2a+b+c,b) this means:

a+b+c = 2
2a+b+c = 3
b = 0

we catch a break on that third equation, so now we just need to solve:

a+c = 2
2a+c = 3

subtract the top from the bottom, leaving a = 1. thus c = 1. so:

(2,3,0) = 1(1,2,0) + 0(1,1,1) + 1(1,1,0) = [1,0,1]_B'.

now, what matrix takes (1,0,0) to (1,0,1)? well, we can't answer entirely, but we know its first column is the image of (1,0,0), and we know that: it's (1,0,1). thus, so far, we have:



in B-coordinates, (1,1,1) is [0,1,0]_B (remember this means: (1,1,1) = 0(2,3,0) + 1(1,1,1) + 0(2,3,1)).

again, we need to know how to express (1,1,1) in B'-coordinates, that is, as a linear combination of (1,2,0), (1,1,1), and (1,1,0).

here, we catch another break: it's obvious that (1,1,1) = 0(1,2,0) + 1(1,1,1) + 0(1,1,0), so in B'-coordinates, we have (1,1,1) = [0,1,0]_B'.

this tells us the second column of P is (0,1,0), so:



we do the same thing to find the 3rd column, we need to find what (2,3,1) is as a linear combination of (1,2,0), (1,1,1) and (1,1,0).

in fact, it's pretty clear that (2,3,1) = 1(1,2,0) + 1(1,1,1) + 0(1,1,0), so its B'-coordinates are [1,1,0]_B'.

now we know P, it is:

.

so this changes B-coordinates into B'-coordinates. this is good, because the matrix A we were given acts on B'-coordinates.

if we want to, we can calculate AP:



this is the matrix of T relative to the bases B = {(2,3,0),(1,1,1),(2,3,1)} and C' = {(1,1),(1,-1)}, so we're closer (we've matched one of the bases to our target pair).

so Q is going to be a 2x2 matrix which changes C'-coordinates to the C-coordinates we want. to get Q, you need to figure out how to write:

(1,1) as a linear combination of (3,-1) and (1,-1)

(1,-1) as a linear combination of (3,-1) and (1,-1). can you do this?

(1 , 1) = a(3 , -1) + b(1, -1) = [1, -2]

(1, -1) = a(3, -1) + b(1, -1) = [0, 1]

so Q = 1 0 ?
-2 1

yep, that's the Q i got. now, put it all together to get your desired matrix, which is..?

2 -1 0
-1 3 3

thank you so much