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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!

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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'.

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  • $\begingroup$ So I am supposed to prove that it is always possible to find a $\lambda$ so that $K = \lambda L$? Or should I find a specific $\lambda$ that will make this statement true? I already did some rather "dodgy" maths with said theorem to find that $\ker(L) = \ker(K)$, and thus as $K(v)$ and $L(v)$ are both elements of $\mathbb{R}$, let's say not equal to $0$, then I can find a $\lambda$ for every $v$. But that seems rather trivial, no? Shouldn't I do something else? $\endgroup$ – Linde Jan 21 '18 at 23:58
  • $\begingroup$ Given some fixed $K$ and $L$, there should be an element $\lambda$ such that, for every $v\in V$, we have $K(v) = \lambda L(v)$. That is, $\lambda$ is allowed to depend on $K$ and $L$, but not on $v$. $\endgroup$ – Billy Jan 22 '18 at 0:05
  • $\begingroup$ But the rest of what you've said looks fine to me. You should indeed find that $\ker(L) = \ker(K)$ (and you should be able to write down exactly what the dimension of this kernel is too). After that, well, perhaps pick a nice basis for $V$? $\endgroup$ – Billy Jan 22 '18 at 0:07
  • $\begingroup$ Wow, thanks, I hadn't thought of choosing a basis! $\endgroup$ – Linde Jan 22 '18 at 0:37
  • $\begingroup$ If I'm not mistaken, it just gives me that $\dim(\ker(K))=n-1$, which means I can choose a basis for V that consists of $n-1$ vectors that are elements of $\ker(L)$, and then one extra vector, say $v$. Since every $\lambda$ would do for the first $n-1$ vectors, I can say “let $\lambda$ be $K(v)/L(v)$, in which $L(v)\neq 0$ since $ v \notin ker(L)$“, and that should work, since a linear transf. is determined entirely by the images of the basis of V? If what I just said is correct, then thank you a huge lot for helping me out! $\endgroup$ – Linde Jan 22 '18 at 0:47
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Yes, that is what is meant. Asserting that $K=0$ means that $(\forall v\in V):K(v)=0$.

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