Find the matrix representations of the Differentiation Map in the Basis

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SUMMARY

The discussion focuses on finding the matrix representations of the differentiation map D: P2(R) → P2(R) with respect to two bases: the standard basis St = {1, x, x²} and the custom basis B = {x² - 1, 2x² + x - 3, 3x² + x}. It is established that B is a basis for P2(R) and that D is a linear transformation. The key steps involve determining the action of D on the basis vectors and expressing the results as linear combinations of the basis elements to derive the matrix representations DSt←St, DSt←B, and DB←B.

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  • Understanding of linear transformations in vector spaces
  • Familiarity with polynomial spaces, specifically P2(R)
  • Knowledge of matrix representation of linear transformations
  • Proficiency in expressing polynomials as linear combinations of basis elements
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  • Learn how to compute matrix representations for linear transformations
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Homework Statement



Show that B = {x2 −1,2x2 +x−3,3x2 +x} is a basis for P2(R). Show that the differentiation map D : P2(R) → P2(R) is a linear transformation. Finally, find the following matrix representations of D: DSt←St, DSt←B and DB←B.

Homework Equations





The Attempt at a Solution



I have proved that it is a linear transformation, but I'm really not sure where to begin with the second part. Any help would be appreciated. Thanks
 
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I take it St means "standard basis" I.e. {1,x,x^2}.
First see what the action of D is on the basis vectors for each basis and express these results as a linear combination of basis elements. Then see how the matrix form is supposed to do exactly the same thing to match up components.
 
I find it helpful to express expansion in a given basis by writing the row (matrix) of basis elements and the element as a matrix product. Thus e.g.

ax^2 + bx + c = \left[ 1, x, x^2 \right]\cdot \left(\begin{array}{c}c \\ b\\ a\\ \end{array}\right)

You can then solve for the components of the matrix \mathbf{M} in the equation:

D \left\{ \left[ \mathbf{basis}_1 \right] \cdot\left(\begin{array}{c}c \\ b\\ a\\ \end{array}\right) \right\}=\left[\mathbf{basis}_2\right] \cdot \left\{ \mathbf{M}\cdot \left(\begin{array}{c}c \\ b\\ a\\ \end{array}\right)\right\}
EDIT: Oops! I had my products backward, now fixed.
 

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