Find T(v) by using the standard matrix and the matrix relative to B and B'

[RET80]

Homework Statement

Basically I need to find an equation that is relative to both B and B'
T: R^2 --> R^3, T(x,y) = (x+y,x,y)
B = {(1,-1),(0,1)}
B' ={(1,1,0),(0,1,1),(1,0,1)}

Homework Equations

THE ONLY example from the book (Elementary Linear Algebra by Larson and Falvo)
6.3 Example 5

Let T: R^2 --> R^2
T(x1, x2) = (x1+x2, 2x1-x2)

Find the matrix for T relative to the bases:
B ={(1,2),(-1,1)}
B'={(1,0),(0,1)}

By the definition of T, you have:
T(v1) = T(1,2) = (3,0) = 3(w1) + 0(w2)
T(v2) = T(-1,1) = (0, -3) = 0(w1) - 3(w2)

The coordinate matrices for T(v1) and T(v2) relative to B' are
[T(v1)]B' = (3, 0)
[T(v2)]B' = (0, -3)

The matrix for T relative to B and B' is formed by using these coordinate matrices as columns to produce:
A =
[3 0]
[0 -3]

The Attempt at a Solution

B = {v1, v2}
B' = {w1, w2, w3}

T(v1) = T(1,-1) = (0,1,1) = 0(w1) + 1(w2) + 1(w3) = (-1, 1, 0)
T(v2) = T(0,1) = (1,0,1) = 1(w1) + 0(w2) + 1(w3) = (2, 1, 1)

[T(v1)]B' = (-1, 1, 0)
[T(v2)]B' = (2, 1, 1)

Which creates the matrix:
[-1 2]
[1 1]
[0 1]

Apparently this matrix is incorrect, but the example in the book (which is given in part 2) states that, this should be the process, I am very lost and confused and have spent a good four hours on this single problem, trying to understand the concept

 

[vela]

Homework Statement

Basically I need to find an equation that is relative to both B and B'
T: R^2 --> R^3, T(x,y) = (x+y,x,y)
B = {(1,-1),(0,1)}
B' ={(1,1,0),(0,1,1),(1,0,1)}

Homework Equations

THE ONLY example from the book (Elementary Linear Algebra by Larson and Falvo)
6.3 Example 5

Let T: R^2 --> R^2
T(x1, x2) = (x1+x2, 2x1-x2)

Find the matrix for T relative to the bases:
B ={(1,2),(-1,1)}
B'={(1,0),(0,1)}

By the definition of T, you have:
T(v1) = T(1,2) = (3,0) = 3(w1) + 0(w2)
T(v2) = T(-1,1) = (0, -3) = 0(w1) - 3(w2)

The coordinate matrices for T(v1) and T(v2) relative to B' are
[T(v1)]B' = (3, 0)
[T(v2)]B' = (0, -3)

The matrix for T relative to B and B' is formed by using these coordinate matrices as columns to produce:
A =
[3 0]
[0 -3]

The Attempt at a Solution

B = {v1, v2}
B' = {w1, w2, w3}

T(v1) = T(1,-1) = (0,1,1) = 0(w1) + 1(w2) + 1(w3) = (-1, 1, 0)
T(v2) = T(0,1) = (1,0,1) = 1(w1) + 0(w2) + 1(w3) = (2, 1, 1)

You made a small mistake calculating T(1,-1); it should be (0,1,-1).

The reason it's not working is because you've misinterpreted what they did in the example. The idea is to express each vector as a linear combination of w₁, w₂, and w₃. In other words, find c₁, c₂, and c₃ such that

(0,1,-1) = c₁w₁+c₂w₂+c₃w₃

and likewise for the second vector. These coefficients then become the columns of the matrix.

[T(v1)]B' = (-1, 1, 0)
[T(v2)]B' = (2, 1, 1)

Which creates the matrix:
[-1 2]
[1 1]
[0 1]

Apparently this matrix is incorrect, but the example in the book (which is given in part 2) states that, this should be the process, I am very lost and confused and have spent a good four hours on this single problem, trying to understand the concept

 

[RET80]

When I used c1w1 + c2w2 + c3w3 = v1 and v2 I got...

v1
c1 + c3 = 0
c1 + c2 = 1
c2 + c3 = -1

which then reduced down to the matrix:
[1 0 0 | 1]
[0 1 0 | 0]
[0 0 1 | -1]and v2
c1 + c3 = 1
c1 + c2 = 0
c2 + c3 = 1

which then reduced down to the matrix:
[1 0 0 | 0]
[0 1 0 | 0]
[0 0 1 | 1]

then I put both v1 and v2 together
[1 0]
[0 0]
[-1 1]

which still isn't correct...Is there something I'm missing or did I go wrong somewhere in my row reduction?
or does it have to do with [T(v)]B' ?

What I'm basically going to do after this is find T(v)
in the problem v is given to me as v=(5,4)
so I use B = (1,-1),(0,1)
and do a linear combination with it such as...

v = (5,4) = c1(1,-1) + c2(0,1)
c1 = 5
-c1 +c2 = 4

which reduced down to a matrix:
[1 0 | 5]
[0 1 | 9]

so then [v]B = (5,9)
then I multiply A[v]B to get my final answer which SHOULD be (9,5,4)

So I HAVE [v]B I just don't think I have A yet.

 

[vela]

That's the correct answer for the problem you described. Why do you think it's wrong?

 

[RET80]

That's the correct answer for the problem you described. Why do you think it's wrong?

the answer is 9, 5, 4 in the book but I'm getting 5, 0 4

if I multiply:
[1 0] [5]
[0 0] [9]
[-1 1]

I get (5,0,4) which is wrong...

 

[vela]

Your book is wrong. You can write the transformation T as

$$T\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x+y\\x\\y\end{bmatrix} = x\begin{bmatrix}1\\1\\0\end{bmatrix}+y\begin{bmatrix}1\\0\\1\end{bmatrix}$$

Note the two constant vectors are first and third vectors in basis B'. That means the image of T can't include any vector where c₂ isn't equal to 0, so (9, 5, 4) can't possibly be what (5, 9) maps to.

 

[Blissett]

I am also on the same problem. Part A asks to find T(v) using the standard matrix.

T: R^2 --> R^3, T(x,y) = (x+y,x,y), v = (5,4)

so A =
[ 1 1 ;
1 0 ;
0 1 ]

and Av = (9,5,4) which is the answer given in the back of the book. So since the problem basically asks you to find A in 2 different ways (standard vs basis), shouldn't the book be right?Wait wait never mind, you are right vela, my prof attempted to do it a couple days ago, the book is wrong.

 

[vela]

You're using a different vector than RET80. Which one is the correct one?

Your matrix A, which is 3x3, can't possibly be correct. You can't multiply it by a 2-component vector. The dimensions don't work out.

 

[Blissett]

The A in my previous post is a 3x2, multiplied by a 2x1(v)

so A =
[1 1]
[1 0]
[0 1]

v =
[5]
[4]

Sorry I was using MATLAB notation with the semicolons.

A is found by finding the standard matrix of a linear transformation from T(x,y) = (x+y,x,y). Essentially this A matrix is what you have in post #6 with the 2 columns put together.