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When is the kernel of a linear operator closed?

  1. Jan 26, 2013 #1
    If you consider a bounded linear operator between two Hausdorff topological vector spaces, isn't the kernel *always* closed? I mean, if you assume singleton sets are closed, then the set [itex]\{0\}[/itex] in the image is closed, so that means [itex]T^{-1}(\{0\})[/itex] is closed, right (since T is assumed continuous)? I keep finding these weird provisos all over the Internet (see, for instance, http://en.wikipedia.org/wiki/Kernel_(linear_algebra [Broken])) that require the target space to be finite-dimensional, and I don't see why this is a necessary hypothesis.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 27, 2013 #2
    You are right, if [itex]T:V\rightarrow W[/itex] is continuous and if W is Hausdorff, then [itex]T^{-1}({0})[/itex] is always closed.

    Why the weird proviso's like finite-dimensional? Well, because the result in the wiki is an iff statement. So the statement in the theorem also says that if [itex]T^{-1}(\{0\})[/itex] is closed, then T is continuous. This is not true in general but only if W is finite dimensional!
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