Linear Algebra - Change of Bases

[Mumba]

Hi, again another problem:

Let B₁ = {( $$\stackrel{1}{3}$$),($$\stackrel{1}{2}$$)} and


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


Determine the representing matrix $$T = K_{B_{2},B_{1}} \in \Re^{2\times2}$$ 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 don't know how to formate this better.

 

[dacruick]

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, that's just ag uess

 

[jbunniii]

If S denotes the standard basis for $\mathbb{R}^2$, do you know how to find

$$K_{S,B_1}$$ and $$K_{S,B_2}$$?

If so, then observe that

$$K_{B_2,B_1} = K_{B_2,S} K_{S,B_1} = K_{S,B_2}^{-1}K_{S,B_1}$$

 

[Mumba]

No, i don't even know what K is supposed to be...

 

[jbunniii]

No, i don't even know what K is supposed to be...

I am using your notation.

$$K_{S,B_1}$$

is the change-of-basis matrix that transforms $B_1$ coordinates to $S$ coordinates. I suggested doing it this way because you already know how to express $B_1$ and $B_2$ in $S$ coordinates: that is what you are given.

 

[Mumba]

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

 

[Mumba]

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 I am finished?

 

[Mumba]

the i get as matrix:

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

correct?

 

[jbunniii]

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 I am finished?

Well, $B_1$ 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:

$$K_{S,B_1}$$ is the matrix whose columns are the basis vectors from $B_1$ expressed in $S$ coordinates, namely

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

and similarly for $$K_{S,B_2}$$.

So yes, now you can find $$K_{B_2,B_1}$$ as I described earlier.

 

[Mumba]

Coool thanks a lot, easy this way.
:D

 

[jbunniii]

the i get as matrix:

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

correct?

I get the same answer.

 

[Mumba]

Cool :)
Thx jbunniii