If $T$ is semi-simple, then you can find a basis so that the matrix of $T$ is diagonal in the following way:
$T$ has an eigenvector $v_1 \in V$ to an eigenvalue $\lambda_1$. This gives us a $T$-invariant subspace generated by $v_1$: $U_1 := <v_1>$$U_1 :=\langle v_1\rangle$. Since $T$ is semi-simple, there is a T-invariant complement space $U_1^c$, that means $V = U_1 \oplus U_1^c$. Let $(b_1, ..., b_{n-1})$ be a basis of $U_1^c$, then $B_1 = (v_1, b_1, ..., b_{n-1})$ is a basis of $V$.
Now the important step is to make clear how the matrix of $T$ with respect to basis $B_1$ looks like. It consists of two blocks, namely a $\mathbb{C}^{1\times 1}$ submatrix (The restriction of $T$ on $U_1$) and a $\mathbb{C}^{n-1 \times n-1}$$\mathbb{C}^{(n-1)\times(n-1)}$ submatrix (The restriction of $T$ on $U_1^c$), assuming that $\mathbb{C}$ is your field. This is because both subspaces are $T$-invariant!
We have done the first step of the diagonalization. Now we take the restriction of $T$ on $U_1^c$, $T_1: U_1^c \to U_1^c$. This is given by the $(n-1)\times(n-1)$-block in the matrix. And so we continue with $T_1$ and so on until the we have found a basis in which the matrix of $T$ is diagonal.
