I've given a matrix $A=\left( \begin{array}{ccc} 1-3 \cos (2 \lambda ) & 3 i \sin (2 \lambda ) & 2 i \sin (\lambda ) \\ -3 i \sin (2 \lambda ) & 3 \cos (2 \lambda )+1 & 2 \cos (\lambda ) \\ 0 & 0 & 4 \\ \end{array} \right)$.
I have to calculate the subalgebra generated by $A,A^*$ and $\mathbb{1}$.
How should I proceed? Is it useful to compute the Jordan normal form? Is there some trick?
Cheers, Peter
Consider $$ A-I=\begin{bmatrix} -3 \cos (2 \lambda ) & 3 i \sin (2 \lambda ) & 2 i \sin (\lambda ) \\ -3 i \sin (2 \lambda ) & 3 \cos (2 \lambda ) & 2 \cos (\lambda ) \\ 0 & 0 & 3 \\ \end{bmatrix} $$ If you square it you get $$ \begin{bmatrix} 9&0&12i\,\sin\lambda\\ 0&9&12\,\cos\lambda\\ 0&0&9 \end{bmatrix}. $$ So $$ B=(A-I)^2-9I=\begin{bmatrix} 0&0&12i\,\sin\lambda\\ 0&0&12\,\cos\lambda\\ 0&0&0 \end{bmatrix}. $$ Now $$E_{33}=\frac1{144}B^*B=\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}\in C^*(A).$$
If you look at $$ C=\frac1{144}\,BB^*=\begin{bmatrix}\sin^2\lambda&i\cos\lambda\,\sin\lambda&0\\ -i\cos\lambda\,\sin\lambda&\cos^2\lambda&0\\0&0&0\end{bmatrix}, $$ its eigenvalues are $0$ and $1$, so it is a projection.
Also, $$ X=\frac13\,\left[(A-I)-\frac16\,B-3E_{33}\right]=\begin{bmatrix} - \cos (2 \lambda ) & i \sin (2 \lambda ) & 0\\ - i \sin (2 \lambda ) & \cos (2 \lambda ) &0 \\ 0 & 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} - 1+2\sin ^2 \lambda & 2i \sin \lambda \cos\lambda & 0\\ -2i \sin \lambda \cos\lambda & 1-2\sin^ 2 \lambda &0 \\ 0 & 0 & 0 \end{bmatrix} =\begin{bmatrix} -1&0&0\\0&-1&0\\0&0&0 \end{bmatrix} +2C $$
We have $$A=3X+\frac16\,B+I+3E_{33},$$ with all the summands in $C^*(A)$. I claim that $$ C^*(A)=\text{span}\,\{X,B,B^*,I,E_{33}\}. $$ They are clearly linearly independent. And $X^2=I-E_{33}$, $B^2=(B^*)^2=0$, $XB=B$, $BX=0$, $XE_{33}=E_{33}X=0$, $B^*B=144E_{33}$, $BB^*=(X-X^2)/2$, $BE_{33}=B$, $E_{33}B=0$. These relations are enough to show that the span of $X,B,B^*,I,E_{33}$ is a C$^*$-algebra.
Thus $C^*(A)$ has dimension $5$, which means it is isomorphic to $M_2(\mathbb C)\oplus\mathbb C$.