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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.

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  • $\begingroup$ What is a cone for you? $\endgroup$ Jul 15, 2019 at 4:47
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    $\begingroup$ 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. $\endgroup$
    – balddraz
    Jul 15, 2019 at 6:50
  • $\begingroup$ @FedericoFallucca A cone is a set where if $x \in C$ then $\alpha x \in C \; \forall \alpha \geq 0$ $\endgroup$
    – Aphyd
    Jul 16, 2019 at 2:00
  • $\begingroup$ For those looking for a reference: this can be found in Rockafellar's "Convex Analysis" (1970) under Theorem 9.1 $\endgroup$ Jan 14, 2023 at 10:22

1 Answer 1

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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

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

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