2

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.

balddraz
  • 7,800
Nothingone
  • 299

2 Answers2

1

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 $t \in TC$.

copper.hat
  • 178,207
  • Well, what do $sp \mathcal{C}$ and $\mathcal{RT}$ stand for? I have no background in functional analysis (FA). My intention was try to solving this problem without touching tools from FA. Btw, thanks for your solutions. – Nothingone Jul 04 '19 at 06:26
  • @Nothingone: There is no functional analysis here, just linear algebra and some basic analysis. $\operatorname{sp} C$ is the span of $C$, or the smallest linear subspace containing $C$ and ${\cal R T}$ is the range space of $T$. Basically it shows that $T$ has an inverse on the smallest linear space containing $C$. – copper.hat Jul 04 '19 at 22:39
0

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.

daw
  • 54,637
  • 2
  • 44
  • 85
  • Thank you for sharing your idea. Anyway, I have a few questions. What arguments do we need to say that $v_n$ has a convergent subsequence here. I probably cannot catch up why the sequence ${y_n}$ is bounded and then $Tv_n$ approaching to $0$. – Nothingone Jul 04 '19 at 06:23
  • $(v_n)$ is bounded in $\mathbb R^n$, so has a convergent subsequence; $(y_n)$ is convergent, hence bounded. By assumption $(x_n)$ is unbounded, so $Tv_n = y_n / |x_n|\to0$. – daw Jul 04 '19 at 10:13
  • I actually just noticed that ${x_n}$ is unbounded, it seems not clear to me why ${v_n}$ is bounded in $\mathbb{R}^n.$ – Nothingone Jul 04 '19 at 15:29
  • Because $|v_n|=1$ – daw Jul 05 '19 at 06:43
  • @daw A few details are missing.
    1. Some of the $x_n$'s may be zero. So the sequence $v_n = \frac{x_n}{||x_n||}$ may not make sense. It is better to split the cases into $y = 0$ (in which case $y \in T(C)$ trivially) and $y \neq 0$ in which case WLOG $y_n \neq 0$ (as a tail of non-zero elements must exist as $y_n \to y$) which forces $x_n \neq 0$ as $\ker T \cap C$ is trivial.
     – balddraz Jul 15 '19 at 08:56
  • Even if $x_n$ is unbounded and $y_n$ is convergent, it does not imply that $\frac{y_n}{||x_n||} \to 0$. As an example, pretend we are using real sequences $y_n: 1, 1, 1, \ldots$ and $x_n: 1, 10, 1, 100, 1, 1000, 1, \ldots$. Then $x_n$ is unbounded and $y_n$ is convergent but $\frac{y_n}{||x_n||}$ is not even convergent because there is at least a subsequence going to $1$. What is true is that if $x_n$ is unbounded, then it has a subsequence $x_{n_k}$ whose sequence of sizes $||x_{n_k}|| \to \infty$ and now $\frac{y_{n_k}}{||x_{n_k}||} \to 0$ indeed.
    • – balddraz Jul 15 '19 at 09:06
  • @ZeroXLR thanks for the comment. Edited accordingly. – daw Jul 15 '19 at 09:12