Linear Algebra, Hoffman and Kunze, Chapter 8.3, page 292

Question

A small problem that I somehow cannot figure out.

Let $V$ be a finite-dimensional inner product space, $f$ a linear functional on $V$, and $W$ be the null space of $f$.
The author tries to prove that $V = W + W^{\perp}$.

On page 292, suppose $f\neq 0$. Then $f$ is of rank $1$ and $\dim(W^{\perp}) = 1$.
(page 292, see upper right)

Why $\dim(W^{\perp}) = 1$?

[Reference]
Theorem 4
Theorem 5

Answer 1

Because $\dim W=\dim\ker f=n-1$ (unless $f=0$, in which case $\ker f=V$). So,\begin{align}\dim W^\perp&=\dim V-\dim W\\&=n-(n-1)\\&=1.\end{align}