Find a transition matrix from bases problem

  • Thread starter proctortom
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  • #1
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Homework Statement



I have 2 bases, a = {1, x, x^2} and b = {-2 - 2x + 3x^2 , 1 + 2x - x^2 , -1 - x + 2x^2} of P2.

Find the transition matrix Pab.

How is this done?!?!


Homework Equations


Since this is Linear Algebra, there aren't really any relevant "Equations" as such. More logic based. Right?


The Attempt at a Solution



I am quite muddled. Best I could get was to make [v]s = [1; 1; 1] (Thats a vertical matrix of 1s)

Not quite sure where to go from here.
 

Answers and Replies

  • #2
HallsofIvy
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You are mapping form a 3 dimensional space to a 3 dimensional space (actually P2 to itself but using different bases so you are treating P2 as two different spaces) so you want a 3 by 3 matrix.

Any vector in P2 is of the form ux^2+ vx+ w so can be written as u(1)+ v(x)+ w(x^2) where I have put parentheses to show the basis vectors. Changing to basis b would mean finding numbers p, q, and r so that ux^2+ vx+ w= p(-2 - 2x + 3x^2)+ q(1 + 2x - x^2)+ r(-1 - x + 2x^2).

The simplest way to find the transition matrix is to determine how to write each of the vectors in basis a in terms of basis b and use those coefficients as columns in the matrix.
For example, the first vector in basis a is "1" so we want to find p, q, and r such that
1= p(-2 - 2x + 3x^2)+ q(1 + 2x - x^2)+ r(-1 - x + 2x^2). That is the same as
1= -2p- 3px+ 3px^2+ q+ 2qx- qx^2- r- rx+ 2rx^2= (-2p +q- r)+ (-3p+ 2q- r)x+ (3p- q+ 2r)x^2. For that to be true for all x we must have -2p+ q- r= 1, -3p+ 2q- r= 0, and 3p- q+ 2r= 0. Solve those three equations for p, q, and r.
 

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