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Linear Algebra - Change of Bases

  • Thread starter Mumba
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  • #1
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Hi, again another problem:

Let B1 = {( [tex] \stackrel{1}{3}[/tex]),([tex] \stackrel{1}{2}[/tex])} and


[tex]B_{2} = [ \frac{1}{\sqrt{2}}( \stackrel{1}{1}), \frac{1}{\sqrt{2}} (\stackrel{-1}{1}) ] [/tex]


Determine the representing matrix [tex]T = K_{B_{2},B_{1}} \in \Re^{2\times2}[/tex] for the change from B1 coordinates to B2 coordinates.

I have no idea what i should do here. I ve found how to calculate the representing matrix from a domain to a codomain.
Is this the same way? Can you give me atleast a hint, please ^^.

Thx Mumba

PS. Sorry, it looks really strange. I dont know how to formate this better.
 

Answers and Replies

  • #2
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This reminds me of quantum mechanics 1, a course that crushed m average and of which I remember very little.

I tried a couple things and the answer i got was 1/sqrt(2) ( 2 5 )
( 0 1 )
for the matrix T. But as i said again, thats just ag uess
 
  • #3
jbunniii
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If S denotes the standard basis for [itex]\mathbb{R}^2[/itex], do you know how to find

[tex]K_{S,B_1}[/tex] and [tex]K_{S,B_2}[/tex]?

If so, then observe that

[tex]K_{B_2,B_1} = K_{B_2,S} K_{S,B_1} = K_{S,B_2}^{-1}K_{S,B_1}[/tex]
 
  • #4
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No, i dont even know what K is supposed to be...
 
  • #5
jbunniii
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No, i dont even know what K is supposed to be...
I am using your notation.

[tex]K_{S,B_1}[/tex]

is the change-of-basis matrix that transforms [itex]B_1[/itex] coordinates to [itex]S[/itex] coordinates. I suggested doing it this way because you already know how to express [itex]B_1[/itex] and [itex]B_2[/itex] in [itex]S[/itex] coordinates: that is what you are given.
 
  • #6
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So i should calculate the change of base matrix for B1 to S and the inverse of the change of base matrix for B2 to S coordinates?
And multiply this to get my result?

Ok i ll try to find out how to calculate the change of base matrix ^^
Thx
 
  • #7
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Is [text]K_{S,B1}[/tex] not the same as B1?
And the same for B2? So i calculate the inverste and multply them and then im finished?
 
  • #8
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the i get as matrix:

4/sqrt(2) 3/sqrt(2)
2/sqrt(2) 1/sqrt(2)

correct?
 
  • #9
jbunniii
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Is [text]K_{S,B1}[/tex] not the same as B1?
And the same for B2? So i calculate the inverste and multply them and then im finished?
Well, [itex]B_1[/itex] is a basis (set of vectors), not a matrix, so what you said is not exactly correct. However, I think what you are trying to say is this:

[tex]K_{S,B_1}[/tex] is the matrix whose columns are the basis vectors from [itex]B_1[/itex] expressed in [itex]S[/itex] coordinates, namely

[tex]K_{S,B_1} = \left[\begin{array}{cc}1 & 1 \\ 3 & 2\end{array}\right][/tex]

and similarly for [tex]K_{S,B_2}[/tex].

So yes, now you can find [tex]K_{B_2,B_1}[/tex] as I described earlier.
 
  • #10
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Coool thx a lot, easy this way.
:D
 
  • #11
jbunniii
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the i get as matrix:

4/sqrt(2) 3/sqrt(2)
2/sqrt(2) 1/sqrt(2)

correct?
I get the same answer.
 
  • #12
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Cool :)
Thx jbunniii
 

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