I came across the statement that for $\mathfrak{g}$ a Lie subalgebra of $\mathfrak{u}(n)$, $\text{dim}(\mathfrak{g})=n^2-1$ implies $\mathfrak{g}=\mathfrak{su}(n)$.
I've tried the following. Does it seem right? For some reason it feels not quite airtight to me.
Let $X_i$ be an orthonormal basis for $\mathfrak{su}(n)$ with $X_1,X_2,X_3$ being $n\times n$ matrices that are zero except for the upper left $2\times 2$ block, where they are equal to the standard $X$, $Y$, $Z$ generators of $\mathfrak{su}(2)$. Then since $\left[X_1,X_2\right]\propto X_3$, there are no more than $n^2-2$ generators (in the algebra sense) of $\mathfrak{su}(n)$.
Suppose that there is a subalgebra $\mathfrak{g}$ of with $\dim\left(\mathfrak{g}\right)\geq n^2-1$ of $\mathfrak{u}(n)$. Such a subalgebra is in particular a linear subspace of $\mathfrak{u}(n)$ with codimension 1. Then the intersection of $\mathfrak{g}$ with the codimension 1 subspace $\mathfrak{su}(n)$ is a linear subspace of $\mathfrak{su}(n)$ with codimension 0 or 1. If the codimension is 0, then $\mathfrak{g}=\mathfrak{su}(n)$. If the codimension is 1, let $X$ be an normalized element of $\mathfrak{u}(n)$ orthogonal to this subspace of $\mathfrak{su}(n)$. By the transitivity of the unitary group acting on the special unitary group, there is a unitary $U$ such that $UX_3U^\dagger=X$. Then $\left[UX_1U^\dagger,UX_2U^\dagger\right]\propto X$. Since unitary operators preserve the inner product, $UX_1U^\dagger$ and $UX_2U^\dagger$ are orthogonal to $X$ and traceless, so still in $\mathfrak{su}(n)$, and therefore are in the codimension 1 subspace of $\mathfrak{su}(n)$. Then $\mathfrak{g}=\mathfrak{u}(n)$. Therefore, the only Lie subalgebra of $\mathfrak{u}(n)$ of dimension $n^2-1$ is $\mathfrak{su}(n)$.