Orthogonal complement of kernel of linear functional has dimension n iff T surjective

Question

I have successfully proven that for every linear bounded map $T:H \to \mathbb{K}^n$ it holds that $\dim(\ker(T)^\bot)\leq n$ where $H$ is a hilbert space. Now, I need to prove that equality holds iff $T$ is surjective.

My attempt: I tried finding $n$ linearly independent vectors $x_1,\dots, x_n \in\ker(T)^\bot$ such that every component of $T$ is $x\mapsto (x,x_i)$. Then, I do not know how to proceed.

Answer 1

The operator $T$ restricted to $(\ker T)^\perp$ is injective. If it is surjective at the same time the spaces $(\ker T)^\perp$ and $K^n$ are isomorphic. Therefore they have the same dimension.