Calculate C*-subalgebra generated by $A$, $A^*$ and $\mathbb 1$ 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
 A: 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 $M_2(\mathbb C)\oplus\mathbb C$.
