That question came to me, by reading a proof, where the author shows that a semisimple Lie-algebra $\mathfrak{g} \subset \mathfrak{sl}(n,\mathbb{C})$ contains the full subalgebra of diagonal matrices of trace $0$. He does this by showing, that $[E_{1j}, E_{j1}]= E_{11}-E_{jj} \in \mathfrak{g}$ for $j=2,...,n$. He than states, that a semisimple Lie-subalgebra of $\mathfrak{sl}(n,\mathbb{C})$ of rank $n-1$ is equal to $\mathfrak{sl}(n,\mathbb{C})$ which is, what i want to verify.
What i understand as the rank of a finite-dimensional semisimple Lie algebra (over an algebraically closed field of characteristic zero) is the (unique) dimension of a Cartan subalgebra. Equivalently, it is defined to be the dimension of the maximal abelian subalgebra, or in the context of subalgebras of $\mathfrak{sl}(n,\mathbb{C})$, the largest number of (linear combinations of) generators which commute with each other.
I know that for $\mathfrak{sl} (n,\mathbb{C} )$, the Cartan subalgebra may be taken to be the diagonal subalgebra of $\mathfrak{sl} (n,\mathbb{C} )$, consisting of diagonal matrices whose diagonal entries sum to zero. So if this subalgebra has dimension $n-1$ then $\mathfrak{sl} (n,\mathbb{C} )$ has rank $n-1$.
So basically the two questions i have are:
Q1: Why has the Subalgebra of $\mathfrak{g} \subset \mathfrak{sl}(n,\mathbb{C})$ of diagonal trace $0$ matrices, dimension $n-1$? (answered in the comments)
Q2: Why is a semisimple Lie-subalgebra of $\mathfrak{sl}(n,\mathbb{C})$ of rank $n-1$ equal to $\mathfrak{sl}(n,\mathbb{C})$?