How can I see that the rank of $\mathrm{SU}(N)$ is $N-1$? Is there a way to understand intuitively why this must be the case? Or is there a kind of easy-to-see proof of this result? Thank you in advance.
 A: It seems easier to work on the level of the Lie algebras, the tangent to a maximal torus being precisely a maximal abelian subalgebra. The Lie algebra of $SU(n)$ is $n\times n$ traceless skew-Hermitian matrices. If they commute, they are simultaneously diagonalizable (by a version of spectral theorem), so in some basis are diagonal. That means (recalling the traceless condition) that the dimension is at most $n-1$. Of course it is then exactly $n-1$.
A: $\DeclareMathOperator\U{U}\DeclareMathOperator\diag{diag}\DeclareMathOperator\SU{SU}\DeclareMathOperator\Cent{C}\DeclareMathOperator\GL{GL}$Per @Max's suggestion, I add an approach very similar to theirs, but where things boil down slightly differently in the end.
It is clear that $T = \ker(\det : \U(1)^n \to \U(1))$ is an $(n - 1)$-dimensional torus.  (Explicitly, the isomorphism with $\U(1)^{n - 1}$ is given by $(z_1, \dotsc, z_n) \mapsto (z_1, \dotsc, z_{n - 1})$.)  It is straightforward to show that $\Cent_{\SU(n)}(T) = T$, but I'll put an explanation in the next paragraph anyway.  Granted this, we know that a maximal torus in $\SU(n)$ containing $T$ must be contained in $\Cent_{\SU(n)}(T) = T$, hence equal $T$, so that some (hence every, because they are all conjugate) maximal torus is ($n - 1$)-dimensional.
Now to show that $\Cent_{\SU(n)}(T) = T$.  I'll show more generally (since $T$ is the group of diagonal matrices in $\SU(n)$) that $\Cent_{\GL(n)}(T)$ consists of diagonal matrices.  In fact, even $\Cent_{\GL(n)}(t)$ consists of diagonal matrices for any single diagonal matrix $t = \diag(z_1, \dotsc, z_n)$ where the $z_i$ are all different and all have norm $1$.  (By working with the full torus rather than just a single element, this argument can be made to work even over fields that don't have enough scalars to produce such a diagonal matrix.)  If $g$ commutes with $t$ and $i \ne j$, then the $(i, j)$ entry of $g t$ is $g_{i j}z_j$, but the $(i, j)$ entry of $t g$ is $z_i g_{i j}$, and the fact that these are equal (since $g t = t g$) but that $z_i \ne z_j$ implies that $g_{i j}$ is $0$.
