Diagonalizing Martrix (S) question

[astropi]

Homework Statement

First off, this is not for a course, I'm reviewing material. This also *should* be straightforward! I think I'm forgetting something simple, so if someone could point it out to me, I would be able to sleep easy tonight :)
OK! the question:

Given a matrix $$T = \begin{pmatrix}<br /> cosx & -sinx\\<br /> sinx & cosx\\<br /> \end{pmatrix}$$

we want to find the inverse of $${\bf S}^{-1}={\bf S}$$ and then take $$STS^{-1}$$.

Homework Equations

So first I find the eigenvalues which are cosx +/- isinx
Next I calculated the eigenvectors and got a(1) = (1,-i) [column vector]
and a(2) = (1,i) [column vector]
If you normalize the two eigenvectors you get a constant 1/sqrt[2] for both.
That gives me
$$S^{-1} = \frac{1}{\sqrt{2}}\begin{pmatrix}<br /> 1 & 1\\<br /> -i & i\\<br /> \end{pmatrix}$$

and when I solve I get $$S = \frac{1}{\sqrt{2}}\begin{pmatrix}<br /> 1/2 & i/2\\<br /> 1/2 & -i/2\\<br /> \end{pmatrix}$$

The Attempt at a Solution

If you look above, you should see that I did most everything correctly (I believe, let me know if I made an error)! However, clearly $$STS^{-1}$$ should give me back a matrix with my eigenvalues on the diagonal. However, I get an extra coefficient in front of the matrix which should cancel out. Where is the mistake? Also, along those lines, I read somewhere that $$S=(S^{-1})^{\dag}$$ is this true? I was always under the impression that S is simply the inverse and you do not need to take the adjoint?

 

[jdwood983]

I get:

$$S=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\-i&i\end{array}\right)\rightarrow S^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)$$

The latter term is where it seems you encounter your extra factor. My result does lead to the expected solution:

$$S^{-1}TS=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\cdot\left(\begin{array}{cc}\cos(x)&-\sin(x)\\ \sin(x)&\cos(x)\end{array}\right)\cdot\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\-i&i\end{array}\right)=\left(\begin{array}{cc}\cos(x)+i\sin(x)&0\\0&\cos(x)-i\sin(x)\end{array}\right)$$

 

[astropi]

I get:

$$S=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\-i&i\end{array}\right)\rightarrow S^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)$$

The latter term is where it seems you encounter your extra factor. My result does lead to the expected solution:

$$S^{-1}TS=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\cdot\left(\begin{array}{cc}\cos(x)&-\sin(x)\\ \sin(x)&\cos(x)\end{array}\right)\cdot\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\-i&i\end{array}\right)=\left(\begin{array}{cc}\cos(x)+i\sin(x)&0\\0&\cos(x)-i\sin(x)\end{array}\right)$$

Yes, thank you. I made an error in the analysis of $$S$$ and that lead to the errors :)