Change of basis (Matrix and operator)

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Ddanndt
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Change of basis (Matrix and operator)

Postby Ddanndt » Wed May 12, 2010 9:54 am UTC

Hi everybody. I've been studying vector spaces for a few weeks and we're now doing change of basis but I keep getting the matrix and operators wrong. So can someone please explain the new basis/ old basis thing clearly please? :cry: :cry:
God does not care about our mathematical difficulties — He integrates empirically.
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gorcee
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Re: Change of basis (Matrix and operator)

Postby gorcee » Wed May 12, 2010 11:18 am UTC

Let's take a 2-D example. Say you have an x-y coordinate frame. If you have a vector $(v_1, v_2)$ in this coordinate frame, then it can be written as a linear combination of the standard basis vectors, $e_1 = (1,0)$ and $e_2 = (0,1)$.

Transforming something to a new basis just means that you're using a different set of basis vectors to write the same thing. That is, you want your vector to be the same, but you're referring to it using a different coordinate system.

The key idea here is that a vector is always able to be written as a linear combination of basis vectors. If, instead of the standard basis, you had a basis $\hat{e}_1 = (1,2)$, $\hat{e}_2 = (3,-1)$, then you could write $(v_1,v_2) = c_1\hat{e}_1 + c_2\hat{e_2}$. If you write out the system fully, you'll get a linear system (hint, the first equation is $v_1 = c_1 \hat{e}_{11} + c_2 \hat{e}_{21}$) and you can compute $c_1$ and $c_2$.

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PM 2Ring
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Re: Change of basis (Matrix and operator)

Postby PM 2Ring » Wed May 12, 2010 1:06 pm UTC

Double $ signs work better:

gorcee wrote:Let's take a 2-D example. Say you have an x-y coordinate frame. If you have a vector $$(v_1, v_2)$$ in this coordinate frame, then it can be written as a linear combination of the standard basis vectors, $$e_1 = (1,0)$$ and $$e_2 = (0,1)$$.

Transforming something to a new basis just means that you're using a different set of basis vectors to write the same thing. That is, you want your vector to be the same, but you're referring to it using a different coordinate system.

The key idea here is that a vector is always able to be written as a linear combination of basis vectors. If, instead of the standard basis, you had a basis $$\hat{e}_1 = (1,2)$$, $$\hat{e}_2 = (3,-1)$$, then you could write $$(v_1,v_2) = c_1\hat{e}_1 + c_2\hat{e_2}$$. If you write out the system fully, you'll get a linear system (hint, the first equation is $$v_1 = c_1 \hat{e}_{11} + c_2 \hat{e}_{21}$$) and you can compute $$c_1$$ and $$c_2$$.

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squareroot1
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Re: Change of basis (Matrix and operator)

Postby squareroot1 » Wed May 12, 2010 3:14 pm UTC

Are you getting the matrix entirely wrong, or just "backwards" (where you would swap the old and new basis or invert the matrix to get the correct value)? I remember always writing it the wrong-way-round and getting hung up cause of that.

To work with the numbers already given, label A the change of basis matrix from [imath]e_i[/imath] to [imath]f_i[/imath]. We know that [imath]f_1=e_1+2e_2[/imath] and [imath]f_2=3e_1-e_2[/imath]. This means that A has to send (1,2) onto (1,0) and (3,-1) onto (0,1). Find an expression for A which does that (Hint: consider combining the input and output vectors into matrixes,) and you will have your change of basis matrix.

gorcee
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Re: Change of basis (Matrix and operator)

Postby gorcee » Wed May 12, 2010 3:52 pm UTC

Gah thanks PM... too much time spent on MathOverflow, not enough previewing before posting.

aleph_one
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Re: Change of basis (Matrix and operator)

Postby aleph_one » Wed May 12, 2010 4:41 pm UTC

Is there a common notation for matrices and vectors in different bases that make clear active and passive coordinate transformations?

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Yakk
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Re: Change of basis (Matrix and operator)

Postby Yakk » Wed May 12, 2010 5:24 pm UTC

$$(a,b,c,d)_B$$ for the vector in basis B?
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