# Proof that the image of a closed cone under a linear transformation is a closed cone [duplicate]

The question is:

Let $$T : \mathbb{R}^n \rightarrow \mathbb{R}^m$$ be a linear map. Let $$C$$ be a closed subset of $$\mathbb{R}^n$$ which is also a cone. Prove that if $$\ker(T) \cap C = \{0\}$$ then $$T(C)$$ is a closed cone in $$\mathbb{R}^m$$.

I can show that $$T(C)$$ is a cone but I can't figure out how to show it is closed.

This question was on my PhD qualifying exam I took recently and I'm going back over it to study.

• What is a cone for you? Jul 15, 2019 at 4:47
• Possible duplicate of Closed cone in the euclidean space $\mathbb{R}^n$ The accepted answer has details missing and does not consider all cases but its general steps are correct. Jul 15, 2019 at 6:50
• @FedericoFallucca A cone is a set where if $x \in C$ then $\alpha x \in C \; \forall \alpha \geq 0$ Jul 16, 2019 at 2:00
• For those looking for a reference: this can be found in Rockafellar's "Convex Analysis" (1970) under Theorem 9.1 Jan 14, 2023 at 10:22

If the map is surjective it is obvious because in that case

$$T^\sim : \mathbb{R}^n/ker(T)\to \mathbb{R}^m$$ is an omeomorphism (by open map theorem because $$T$$ is linear) and so $$T^\sim(\pi(C))$$ is closed if and only if $$(T^\sim)^{-1}(T^\sim(\pi(C))=\pi(C)$$ is closed in the quotient space that it is closed because $$C$$ is closed (infact $$\pi^{-1}(\pi(C)=C$$ if you have that $$T$$ satisfies the following condition:

If $$T(x)\in T(C)$$ then $$x\in C$$ )

So $$T^\sim (\pi(C))$$ is closed in $$\mathbb{R}^m$$ but you can observe that

$$T^\sim(\pi(C))=T(C)$$ because

if $$y\in T^\sim (\pi(C))$$ then there exist $$\pi(x)\in \pi(C)$$ such that

$$y=T^\sim(\pi(x))$$

You have that $$\pi(x)\in \pi(C)$$ so there exist $$a\in C$$ such that $$\pi(x)=\pi(a)$$ so $$x-a\in ker(T)$$ and you have that

$$T(x)=T(a)$$

Then $$y=T(a)\in T(C)$$

If $$T(x)\in T(C)$$ with $$x\in C$$ then it is clear that

$$T(x)=T^\sim(\pi(x))$$ so

$$T^\sim(\pi(C))=T(C)$$