Finding T(v) relative to B and B'

  • Thread starter dzimitry
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In summary, to find T(v) using the matrix relative to B and B', we first solve for the coefficients of the basis vectors in T(x, y, z) in terms of the basis vectors in B' (a, b, c, d). Then, we form a matrix A using these coefficients as the columns. Finally, we can find T(v) by multiplying the vector v by A.
  • #1
dzimitry
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Homework Statement



find T(v) using the matrix relative to B and B'

T(x, y, z) = (2x, x + y, y + z, x + z)
v = (1, -5, 2)
B = { (2, 0, 1), (0, 2, 1), (1, 2, 1) }
B' = { (1, 0, 0, 1), (0, 1, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0) }


Homework Equations





The Attempt at a Solution



T(2, 0, 1) = (4, 2, 1, 3)
= 4(1, 0, 0, 1) + 2(0, 1, 0, 1) + 1(1, 0, 1, 0) + 3(1, 1, 0, 0)
= (8, 5, 1, 6)
T(0, 2, 1) = (0, 2, 3, 1)
= (4, 3, 3, 2)
T(1, 2, 1) = (2, 3, 3, 2)
= (7, 5, 3, 5)

A = 8 4 7
5 3 5
1 3 3
6 2 5

Av = (2, 0, -8, 6)

but if the person I am checking against is right, the answer should be (2, -4, -3, 3)

I am confused as to if I can even use the method I am using in this case.

Thanks in advance
 
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  • #2
that's a matrix A btw, everything that was indented got shifted.
 
  • #3
dzimitry said:

Homework Statement



find T(v) using the matrix relative to B and B'

T(x, y, z) = (2x, x + y, y + z, x + z)
v = (1, -5, 2)
B = { (2, 0, 1), (0, 2, 1), (1, 2, 1) }
B' = { (1, 0, 0, 1), (0, 1, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0) }


Homework Equations





The Attempt at a Solution



T(2, 0, 1) = (4, 2, 1, 3)
= 4(1, 0, 0, 1) + 2(0, 1, 0, 1) + 1(1, 0, 1, 0) + 3(1, 1, 0, 0)
No, (4, 2, 1, 3) is NOT equal to (8, 5, 1, 6)! You are doing this backwards. You want to find numbers, a, b, c, d, such that (4, 2, 1, 3)= a(1, 0, 0, 1)+ b(0, 1, 0, 1)+ c(1 , 0, 1, 0)+ d(1, 1, 0, 0). That is you jeed to solve a+ c+ d= 4, b+ d= 2, c= 1, and a+ b= 3.
Then
[tex]\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}[/tex]
will be the first column of the matrix.

= (8, 5, 1, 6)
T(0, 2, 1) = (0, 2, 3, 1)
= (4, 3, 3, 2)
T(1, 2, 1) = (2, 3, 3, 2)
= (7, 5, 3, 5)

A = 8 4 7
5 3 5
1 3 3
6 2 5

Av = (2, 0, -8, 6)

but if the person I am checking against is right, the answer should be (2, -4, -3, 3)

I am confused as to if I can even use the method I am using in this case.

Thanks in advance
 
  • #4
ok that makes sense...and for the vector v = (1, -5, 2), do I need to solve a system like
(1, -5, 2) = a(2, 0, 1) + b(0, 2, 1) + c(1, 2, 1) and use (a, b, c) as my v and multiply that by A?
 

1. What is the purpose of finding T(v) relative to B and B'?

Finding T(v) relative to B and B' allows us to determine the transformation between two coordinate systems, B and B', by finding the transformation matrix T. This is useful in various applications such as computer graphics, robotics, and physics.

2. How do I find T(v) relative to B and B'?

To find T(v) relative to B and B', we first need to identify the basis vectors for each coordinate system, B and B'. Then, we can use the formula T(v) = B^-1 * B' * v, where B^-1 is the inverse of the basis matrix for B and v is the vector we want to transform.

3. Can T(v) relative to B and B' be represented in different forms?

Yes, T(v) relative to B and B' can be represented as a matrix, a set of linear equations, or as a geometric transformation such as rotation, translation, or scaling.

4. Are there any limitations to finding T(v) relative to B and B'?

The process of finding T(v) relative to B and B' assumes that both coordinate systems, B and B', are linearly independent and have the same number of dimensions. Additionally, the basis vectors for each system must be orthogonal (perpendicular) and unit vectors (length of 1).

5. How is T(v) relative to B and B' used in real-world applications?

T(v) relative to B and B' is used in various fields such as computer graphics to transform 3D objects, in robotics to navigate and control movements, and in physics to study the motion and behavior of particles in different coordinate systems.

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