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Linear Algebra: Finding the Standard Matrix from a Function

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Cod
#1
Nov6-10, 01:33 PM
P: 307
1. The problem statement, all variables and given/known data

Find the standard matrix of T(f(t)) = f(3t-2) from P2 to P2.

2. Relevant equations

n/a

3. The attempt at a solution

The overall question has to do with finding the determinants, so the matrix is provided; however, I want to know how the author came up with the standard matrix of T.

Any help is greatly appreciated.
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vela
#2
Nov6-10, 01:53 PM
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PF Gold
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You can find the columns of the matrix representation by applying the transformation to the basis vectors. If you're using the basis {1, t, t2}, the first column of the matrix would correspond to T(1), the second column to T(t), and the third column to T(t2).
Cod
#3
Nov7-10, 04:33 PM
P: 307
Thanks for the guidance. What happens when the basis is Rn? I realize R2 is a 3x3 matrix, R3 is a 4x4, and so on.

vela
#4
Nov7-10, 04:36 PM
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PF Gold
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Linear Algebra: Finding the Standard Matrix from a Function

Your question doesn't make sense. Rn is a vector space, not a basis.
Cod
#5
Nov7-10, 07:05 PM
P: 307
Sorry, I was looking at my homework when I typed the last post. I meant vector space.
vela
#6
Nov7-10, 08:05 PM
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Same thing. You choose a basis and apply the transformations to the basis vectors to get the columns of the matrix representing the transformation.
Cod
#7
Nov8-10, 04:50 PM
P: 307
Quote Quote by vela View Post
Same thing. You choose a basis and apply the transformations to the basis vectors to get the columns of the matrix representing the transformation.
So, I could choose a basis of 1, t, t2; 1, t, t2, t3; and so on (to tnth)?
vela
#8
Nov8-10, 05:13 PM
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PF Gold
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Not for Rn because those aren't vectors in Rn. P2 consists of polynomials of degree less than or equal to 2, and each polynomial is a linear combination of 1, t, and t2. It turns out 1, t, and t2 are also independent, so they form a basis for P2. Rn, however, is a set of n-tuples, not polynomials. You need a collection of n linearly independent n-tuples to have a basis for Rn.


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