Find Transformation: Tα & Original vs. New Basis

[Shackleford]

I'm not exactly sure how to find the transformation. The professor wrote something different in class. I know [T]_α is what you multiply with the "new" basis to get the transformation of the components of the "original" basis. In this case, it's simply still alpha.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110615_211957.jpg

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110615_204346.jpg

 

[lanedance]

so guessing here, and abusing a little notation but hopefully it helps..

for a given matrix A you should able to write in the alpha basis:
A = \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} <br /> = q\vec{\alpha}_1+pq\vec{\alpha}_2+rq\vec{\alpha}_3+sq\vec{\alpha}_3 = \begin{pmatrix} p \\ q \\ r \\ s \end{pmatrix}_{\alpha}

then apply the T transform which is already written in the alpha basis

 

[lanedance]

to further understand the alpha basis, note that you could consider A expressed in the standard basis, call it s, and write
A = \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} <br /> = a\begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 0 \end{pmatrix} <br /> +b\begin{pmatrix} 0 &amp; 1 \\ 0 &amp; 0 \end{pmatrix} <br /> +c\begin{pmatrix} 0 &amp; 0 \\ 1 &amp; 0 \end{pmatrix} <br /> +d\begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}_s

 

[lanedance]

updated above

 

[Shackleford]

Oh, I see what you're doing.

 

[lanedance]

That's not the alpha basis. It's not the standard basis.

what's not the alpha basis?

you need to solve for q,p,r,s which give the components in the alpha basis

 

[Shackleford]

The components are already given.

 

[lanedance]

the way i read it (open to interp):
- the components of A in the standard basis are given
- the components of the operator T in the alpha basis is given

so i think you need to express A in the alpha basis, or express T in the standard basis