Linear Algebra: Finding the Standard Matrix from a Function

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To find the standard matrix of the transformation T(f(t)) = f(3t-2) from P2 to P2, one must apply the transformation to the basis vectors of P2, which are {1, t, t²}. The first column of the matrix corresponds to T(1), the second to T(t), and the third to T(t²). The discussion clarifies that while Rn is a vector space, it requires a different basis consisting of n-tuples, not polynomials. Understanding the distinction between polynomial bases and vector space bases is crucial for correctly forming the transformation matrix.
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Homework Statement



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

Homework Equations



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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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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).
 
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.
 
Your question doesn't make sense. Rn is a vector space, not a basis.
 
Sorry, I was looking at my homework when I typed the last post. I meant vector space.
 
Same thing. You choose a basis and apply the transformations to the basis vectors to get the columns of the matrix representing the transformation.
 
vela said:
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)?
 
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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