Linear Algebra (Similar Matrices)

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Homework Help Overview

The discussion revolves around the representation of a linear transformation T in the context of linear algebra, specifically focusing on the standard matrix representation and the concept of similar matrices. The original poster is attempting to clarify the dimensions of the matrix associated with T, which operates on polynomials.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the dimensionality of the matrix representation of T, questioning whether T maps polynomials of degree 3 or less to itself or to polynomials of degree 2 or less. There is also discussion about the basis vectors and their implications for the matrix size.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the transformation and its representation. Some guidance has been offered regarding the interpretation of the transformation and its matrix form, but no consensus has been reached on the correct dimensions or representation.

Contextual Notes

There is an ongoing debate about the nature of the transformation T and the corresponding matrix size, which may be influenced by the definitions of the polynomial spaces involved. The original poster expresses uncertainty about the correctness of their initial matrix representation.

DanielFaraday
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Homework Statement



Write the standard matrix representation TE for T.

Homework Equations



[tex] T\left(ax^3+bx^2+cx+d\right)=(a-b)x^2+(c-d)x+(a+b-c)[/tex]

The Attempt at a Solution



[tex] [T]_E=\left(<br /> \begin{array}{c}<br /> a+b-c \\<br /> c-d \\<br /> a-b \\<br /> 0<br /> \end{array}<br /> \right)[/tex]

I think this is right, but a subsequent problem asks me to show that this matrix is similar to another matrix. Similarity only makes sense for an n x n matrix, so this must be wrong.

Any ideas?
 
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The basis 'vectors' are x^3, x^2, x and 1. So ax^3+bx^2+cx+d=(a,b,c,d).This is exactly the same as other exercises you have already done, it's T(a,b,c,d)=(0,a-b,c-d,a+b-c). Sure, it's an 4x4 matrix.
 
Is T here a function from the space of polynomials of degree 3 or less to itself or to the space of polynomials of degree 2 or less? If the former, the matrix is 4 by 4. If the latter, it is 4 by 3 (but the first row is all "0"s).
For example, taking x3 as the first "basis vector", T(x3)= ax2+ 0x+ a as a polynomial of degree 2 or less or 0x3+ ax2+ 0 x+ a as a polynomial of degree 3 or less. That is:
[tex]T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ 0 \\ a\end{bmatrix}[/tex]
in the first case or
[tex]T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ a \\ 0 \\ a\end{bmatrix}[/tex]
in the second.

Do you see that those give you the first column of the matrix?
 
Ah yes, I see. I was thinking of T as a vector instead of a transformation. Thanks!
 

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