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Change of basis problem

  • #1
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Edit complete, but it doesn't seem as though I can change the title.
The latex arrows next to the 'P' aren't showing up for me but they're supposed to be left arrows


Homework Statement


Let B and C be bases of R^2. Find the change of basis matrices [itex]P_{B \leftarrow C}[/itex] and [itex]P_{C\leftarrow B}[/itex]


[itex]B={\begin{pmatrix}3\\1\end{pmatrix}, \begin{pmatrix}2\\2\end{pmatrix}}, C={\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}5\\4\end{pmatrix}}[/itex]


To find the change of basis matrix from C to B [itex]P_{B \leftarrow C}[/itex] , I followed the steps in Lay's Linear Algebra book and found the coordinates of the B vectors relative to C.

So I solved this system:

[itex]\begin{pmatrix}3\\1\end{pmatrix}=r_1\begin{pmatrix}1\\0\end{pmatrix}+r_2\begin{pmatrix}5\\4\end{pmatrix}[/itex]
[itex]\begin{pmatrix}2\\2\end{pmatrix}=s_1\begin{pmatrix}1\\0\end{pmatrix}+s_2\begin{pmatrix}5\\4\end{pmatrix}[/itex]

where r, s are real numbers.

Doing so, I got the 2x2 matrix

[7/4 -2/4]
[1/4 2/4]

(can't get matrices to work in latex)

But in the solutions, this is the change of basis matrix for going from C to B, i.e. [itex]P_{C\leftarrow B}[/itex]. Have I misinterpreted something? The steps I'm referring to are based on this theorem on page 273 of the 3rd edition of the text by Lay:

Let [itex]B=[b_1,..,b_n][/itex] and [itex]C=[c_1,..,c_n][/itex] be bases of a vector space V. Then there is a unique nxn matrix [itex]P_{B \leftarrow C}[/itex] such that [itex][x]_c=P_{B \leftarrow C}[x]_B[/itex]. The columns of [itex]P_{B \leftarrow C}[/itex] are the C-coordinate vectors of the vectors in the basis B. That is, [itex]P_{B \leftarrow C}=[[b_1]_c...[b_n]_c][/itex]
 
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  • #2
Ray Vickson
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Edit complete, but it doesn't seem as though I can change the title.
The latex arrows next to the 'P' aren't showing up for me but they're supposed to be left arrows


Homework Statement


Let B and C be bases of R^2. Find the change of basis matrices [itex]P_{B \leftarrow C}[/itex] and [itex]P_{C\leftarrow B}[/itex]


[itex]B={\begin{pmatrix}3\\1\end{pmatrix}, \begin{pmatrix}2\\2\end{pmatrix}}, C={\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}5\\4\end{pmatrix}}[/itex]


To find the change of basis matrix from C to B [itex]P_{B \leftarrow C}[/itex] , I followed the steps in Lay's Linear Algebra book and found the coordinates of the B vectors relative to C.

So I solved this system:

[itex]\begin{pmatrix}3\\1\end{pmatrix}=r_1\begin{pmatrix}1\\0\end{pmatrix}+r_2\begin{pmatrix}5\\4\end{pmatrix}[/itex]
[itex]\begin{pmatrix}2\\2\end{pmatrix}=s_1\begin{pmatrix}1\\0\end{pmatrix}+s_2\begin{pmatrix}5\\4\end{pmatrix}[/itex]

where r, s are real numbers.

Doing so, I got the 2x2 matrix

[7/4 -2/4]
[1/4 2/4]

(can't get matrices to work in latex)

But in the solutions, this is the change of basis matrix for going from C to B, i.e. [itex]P_{C\leftarrow B}[/itex]. Have I misinterpreted something? The steps I'm referring to are based on this theorem on page 273 of the 3rd edition of the text by Lay:
If you want a matrix like this:
[tex] \pmatrix{7/4 & -2/4 \\ 1/4 & 2/4}[/tex]
use the "\ p matrix" command (remove spaces!) To get the above I just entered "\ p matrix{ 7/4 & -2/4 \\ 1/4 & 2/4}". In LaTeX/TeX, the '&' separates elements in the same row and the "\\" starts a new row.

Anyway, you are encountering the classic confusion between the matrix that changes the basis vs. the matrix that changes the ccomponents. So, if {u1, u2} and {v1,v2} are two bases (each ui or vi is a two-dimensional vector), then if
[tex] \pmatrix{u_1\\u_2} = \pmatrix{a&b\\c&d} \pmatrix{v_1\\v_2} [/tex]
and if ##x = x_1 u_1 + x_2 u_2 = y_1 v_1 + y_2 v_2## we have
[tex] (y_1 \;y_2) = (x_1 \; x_2) \pmatrix{a &b\\c &d}[/tex]
 
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  • #3
D H
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Edit complete, but it doesn't seem as though I can change the title.
I can!

I'm a bit too busy right now to be able to help you with learning to solve this problem, but I'm not too busy to help with changing the title.
 
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