In Peskin Schroeder pag.696 a Higgs mechanism for an $SU(3)$ gauge theory with a scalar field $\phi$ in the adjoint representation is presented. The covariant derivative of $\phi$: $$ D_{\mu}\phi_{a} = \partial_{\mu}\phi_{a}+ g f_{abc}A^{b}_{\mu}\phi_{c}\tag{20.32}$$ defining the quantity $$\Phi=\phi_c t^c\tag{20.34}$$ We consider the expansion of $\Phi$ around two vacuum choice: $$1) \,\,\,\Phi_0 = v \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$ $$2) \,\,\,\Phi_0 = v \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ In the first case the unbroken generators are: $T_1,T_2,T_3,T_8$
In the second case the unbroken generators are: $T_3,T_8$
Where $T_i$ are Gell-Mann matrices according to the normalization which here is $T_i = \lambda_i/2$.
In the book is said that $SU(3)$ breaks spontaneously to $S U (2) \times U(1)$ in the first case and $U(1) \times U(1)$ in the second case.
I really can't understand how to prove that $T_1,T_2,T_3,T_8$ generate $S U (2) \times U(1)$ and $T_3,T_8$ generate $U(1) \times U(1)$
Starting from the simplest how a 3 x 3 matrix $$T_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ can generate $U(1)$, (ot the same for $T_8$)?
A basis for the Lie Algebra of $SU(2)$ (aka its infinitesimal generators):
$$
\left(\ \frac{1}{2} \sigma_{1}, \ \frac{1}{2} \sigma_{2}, \ \frac{1}{2} \sigma_{3} \right)
$$
I can recognize that inside the matrices that remain I can found $\sigma_{1}, \sigma_{2}, \sigma_{3}$
$$T_1 = \begin{bmatrix}
(\sigma_1)_{00} & (\sigma_1)_{01} & 0 \\
(\sigma_1)_{10} & (\sigma_1)_{11} & 0 \\
0 & 0 & 0
\end{bmatrix}
T_2 = \begin{bmatrix}
(\sigma_2)_{00} & (\sigma_2)_{01} & 0 \\
(\sigma_2)_{10} & (\sigma_2)_{11} & 0 \\
0 & 0 & 0
\end{bmatrix}
T_3 = \begin{bmatrix}
(\sigma_3)_{00} & (\sigma_3)_{01} & 0 \\
(\sigma_3)_{10} & (\sigma_3)_{11} & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
However the matrices are not $2 \times 2$ but $3 \times 3$.
Recalling the definition of $SU(2)$:
$$SU(2) \equiv \{ M \in GL(n, \mathbb{C}) | M^{\dagger} \mathbb{1}_2 M = \mathbb{1}_2, \,\, det(M)=1 \} $$
We notice that:
$$det(T_1)=det(T_2)=det(T_3)=0$$
Moreover thare is a dimensional problem in performing $T_1^{\dagger} \mathbb{1}_2 T_1,$ $\,\,$ $T_2^{\dagger} \mathbb{1}_2 T_2,$ $\,\,$ $T_3^{\dagger} \mathbb{1}_2 T_3$.
I'm quite lacking in group theory so I would need a step-by-step answer.
