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 &amp; -sinx\\<br /> sinx &amp; 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 &amp; 1\\<br /> -i &amp; i\\<br /> \end{pmatrix}$

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

<br /> S=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&amp;1\\-i&amp;i\end{array}\right)\rightarrow S^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&amp;i\\1&amp;-i\end{array}\right)<br />

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

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

 

[astropi]

I get:

<br /> S=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&amp;1\\-i&amp;i\end{array}\right)\rightarrow S^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&amp;i\\1&amp;-i\end{array}\right)<br />

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

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

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