Rank and Kernel of subspace where dot product = 0.

Question

So if V ∈ R^n and is the set of all vectors ∈ with the property that ⟨,⟩=0. let ∈ be a fixed non-zero vector

we have a function with domain by ()=⟨,⟩ ...(dot product)

Whats the rank and dimension kernel of T? By rank I mean dimension of the Image/range of the codomain, and kernel is The vectors that go to zero, so that would be W.

I see it as T: R^n -> R

I thought the rank would be 0 because w vectors dot product to zero but this was wrong. Any help in understanding would be great!

Answer 1

The rank is $1$ since $T(cv)=c\|v\|^{2}$ so the range includes all points of $\mathbb R$. The dimension of the kernel is $n-1$ by the Rank Nullity Theorem.

Answer 2

The set $W$ is by definition the kernel of $T$, and the range is the whole real line since for any $c \in \mathbb R$ we have $$c = T\Big( \frac{c}{\langle v,v \rangle}v\Big).$$