Prove for $L,K\colon V \rightarrow \mathbb{R} $ if $\ker(L) \subset \ker(K)$ then $K=\Lambda L$?

Question

I saw the following question in my linear algbra book, and found it rather strange:

"Let $V$ be a real vector space of dimension $n$. Let $L,K \colon V \rightarrow \mathbb{R}$ be linear transformations, so that $\ker(L) \subset \ker(K)$. Prove that $K=\lambda L$ for a $\lambda \in \mathbb{R}$, using the following steps:

1. Prove this for $K=0$.
2. Assume $K \neq 0$. Prove that $\dim(\ker(L)) = \dim(\ker(K))$.
3. Now prove that $K = \lambda L$ for $K \neq 0$."

What is meant by "$K=0$"? How is a linear transformation ever equal to a number? I have a feeling that they might mean something else than just "$K$", but I'm not sure.

I tried something for "K maps every vector to $0$", but that doesn't really seem to be what is meant, because then I obviously get that $K(v) = \lambda L(v)$ when $\lambda = 0$, which would mean that this question allows me to just choose whatever $\lambda$ to make the statement true and I'm not sure that is the case here. Also, how would I continue for $K \neq 0$?

Can anyone give me a hint?

Thanks!

Answer 1

How is a linear transformation ever equal to a number?

Here, you're supposed to understand (from context) that $0$ doesn't mean the number, it means the function $0: V\to \mathbb{R}$ which sends every element to (the number) $0$. It's confusing notation.

then I obviously get that $K(v)=\lambda L(v)$ when $\lambda = 0$

That's correct.

Also, how would I continue for $K\neq 0$?

Have you tried using the hint in part 2 of the question? You're going to need to use a special fact about the target vector space $\mathbb{R}$. Think 'rank-nullity theorem'.

Answer 2

Yes, that is what is meant. Asserting that $K=0$ means that $(\forall v\in V):K(v)=0$.