I am self-studying group theory from Lie Algebras in Particle Physics by H. Georgi and I am having trouble following some of his arguments. In section 7.2 titled Weights and roots of $SU(3)$ he starts out by finding the weights of the generators \begin{equation} T_3 = \begin{pmatrix} \frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 0 \end{pmatrix} \hspace{1cm} T_{8} = \frac{\sqrt{3}}{6} \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -2 \end{pmatrix}. \end{equation} The eigenvectors and the associated weights are given to be \begin{align} \begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix} & \longrightarrow (1/2,\sqrt{3}/6)\\ \begin{pmatrix} 0\\1\\0 \end{pmatrix} &\longrightarrow (-1/2,\sqrt{3}/6)\\ \begin{pmatrix} 0\\0\\1 \end{pmatrix} &\longrightarrow (0,-\sqrt{3}/3) \end{align} So far so good. What I don't understand is how Georgi finds the roots. He says that "The roots are going to be differences of weights, because the corresponding generators take us from one weight to another." I do not understand this statement, can someone please elaborate?
Next he says "It is not hard to see that the corresponding generators are those that have only one off-diagonal entry" \begin{align} \frac{1}{\sqrt{2}}(T_1\pm i T_2) &= E_{\pm 1,0}\\ \frac{1}{\sqrt{2}}(T_4\pm i T_5) &= E_{\pm 1/2,\pm\sqrt{3}/2}\\ \frac{1}{\sqrt{2}}(T_6\pm i T_7) &= E_{\mp 1/2,\pm\sqrt{3}/2} \end{align} I thought the generators were the $T_i$'s. Can someone please explain? I find the book to be a bit terse sometimes and lacking elaborate explanations.