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Change of basis computation gone wrong...

  • #1

Homework Statement



Consider the real-vector space of polynomials (i.e. real coefficients) ##f(x)## of at most degree ##3##, let's call that space ##X##. And consider the real-vector space of polynomials (i.e. real coefficients) of at most degree ##2##, call that ##Y##. And consider the linear transformation ##A## from ##X## to ##Y## defined by the following:$$A(f) = 2f' - (x+1)f''.$$Let $$[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}$$be the matrix of ##A## with respect to the bases ##(1, x, x^2, x^3)## of ##X## and ##(1, x, x^2)## of ##Y##, let$$[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)}$$be the matrix of ##A## with respect to the bases ##(1,x+1,(x+1)^2,(x+1)^3)## of ##X## and ##(1,x+1,(x+1)^2)## of ##Y##, let ##B## be the change of basis matrix from ##(1,x,x^2,x^3)## to ##(1,x+1,(x+1)^2,(x+1)^3)##, and let ##C## be the change of basis matrix from ##(1, x, x^2)## to ##(1,x+1,(x+1)^2)##.

Problem. Show that$$[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)} = C^{-1}[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}B.$$

Homework Equations



This category is a bit silly.

The Attempt at a Solution


[/B]
Alright, let's do this, lads.

Let's first calculate$$[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}.$$Assume ##f(x)=a+bx+cx^2+dx^3##, then$$A(f)=2f' - (x+1)f''=2(b+2cx+3dx^2)-(x+1)(2c+6dx)=$$$$=2b+4cx+6dx^2-2cx-6dx^2-2c-6dx=(2b-2c)+(2c-6d)x,$$
therefore
$$[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}.=\begin{pmatrix}0&2&-2&0\\0&0&2&-6\\0&0&0&0\end{pmatrix}.$$Okay, now let's do$$[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)},$$which is similar. So we begin with ##f(x) = a + b(x+1) + c(x+1)^2 + d(x+1)^3##, then$$A(f) = 2f' - (x+1)f'' = 2(b + 2c(x+1) + 3d(x+1)^2) - (x+1)(2c + 6d(x+1)) = 2b + 2c(x+1).$$So$$[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)}.=\begin{pmatrix}0&2&0&0\\0&0&2&0\\0&0&0&0\end{pmatrix}.$$Okay, onto ##B##, ##C##, ##C^{-1}##. I calculated that$$B = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 0 & 1 & -2 & 3 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 1\end{pmatrix},$$
$$C = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{pmatrix},$$
$$C^{-1} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{pmatrix}.$$However, when I plug everything into my original equation which I want to verify, i.e.$$[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)} = C^{-1}[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}B,$$I find that the two sides are not equal.

Where did I go wrong? Thanks in advance.
 

Answers and Replies

  • #2
haruspex
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Problem. Show that$$[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)} = C^{-1}[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}B.$$
Based on the definitions, I would have expected it to be the other way around:
that$$[A]_{(1,x,x^2,x^3)}^{(1,x,x^2)}= C^{-1}[A]_{(1,x+1,(x+1)^2,(x+1)^3)}^{(1,x+1,(x+1)^2)} B.$$
Maybe try that.
 

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