Closed cone in the euclidean space $\mathbb{R}^n$

Question

Let $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ be a linear map and let $\mathcal{C}$ be a closed cone in $\mathbb{R}^n.$ Prove that $T(\mathcal{C})$ is a closed cone in $\mathbb{R}^m$ provided $\ker(T)\cap \mathcal{C}=\{0\}$.

I have no problem to justify that $T(\mathcal{C})$ is a cone in $\mathbb{R}^m$. The issue is to verify that $T(\mathcal{C})$ is closed in $\mathbb{R}^m$. My idea is trying to use a sequence characterization for closed sets, by which I mean that for any given sequence in $T(\mathcal{C})$ that converges to some $x\in\mathbb{R}^m$, then $x\in T(\mathcal{C}).$ But it was not successful, since I do not know how to apply the given hypothesis $\ker(T)\cap \mathcal{C}=\{0\}$. Does anyone have a useful option/thought or recommendation on this problem? Or if you have a better idea, I would be happy to listen to it.

Answer 1

Let $y_n:=Tx_n \to y$ be a sequence in $T(C)$ with $x_n\in C$. If $(x_n)$ contains a bounded subsequence, then $x_{n_k}\to x$ and $y=Tx$. If $(x_n)$ contains no bounded subsequence, then $\|x_n\|\to\infty$. We can consider the sequence $v_n:= \frac{x_n}{\|x_n\|}\in C$, which is well-defined for all $n$ large enough. It has a converging subsequence $v_{n_k}\to v$ with $\|v\|=1$. Now $$ Tv_n = \frac{y_n}{\|x_n\|} \to 0 $$ as $(y_n)$ is a bounded sequence. This shows $v\in \ker T\cap C$ implying $v=0$. A contradiction. Hence $(x_n)$ cannot be unbounded.

Answer 2

Here is an example that shows what happens if the condition is violated: https://math.stackexchange.com/a/1457777/27978

Note that if $\ker T \cap C = \{ 0 \}$, then $\ker T \cap \operatorname{sp}C = \{ 0 \}$.

Define $A: {\cal R} T \to \operatorname{sp}C$ by $AT x = x$.

Now suppose $t_k =Tc_k$, with $c_k \in C$ and $t_k \to t$. Since $At_k = c_k$ and $A t_k \to At$, we see that $c_k \to At$, and since $C$ is closed, $At \in C$ and so $t \in TC$.