When is the kernel of a linear operator closed?

In summary, If T:V\rightarrow W is a continuous bounded linear operator between two Hausdorff topological vector spaces, then T^{-1}(\{0\}) is always closed. The requirement for W to be finite-dimensional in the theorem is necessary for the statement to be an iff statement.
  • #1
AxiomOfChoice
533
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 )) 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:
Physics news on Phys.org
  • #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!
 

1. What is the kernel of a linear operator?

The kernel of a linear operator is the set of all inputs that produce an output of zero when operated on by the operator. In other words, it is the set of all vectors that are mapped to the zero vector by the operator.

2. Why is the kernel of a linear operator important?

The kernel of a linear operator is important because it allows us to understand the behavior of the operator and its effect on different vectors. It helps us to identify which vectors will be mapped to the zero vector and which ones will not be affected by the operator.

3. What does it mean for the kernel of a linear operator to be closed?

A closed kernel means that the set of all inputs that produce an output of zero is also a closed set. In other words, any limit points of the set are also contained within the set. This is an important property because it ensures that the operator is well-defined and behaves predictably.

4. How can we determine if the kernel of a linear operator is closed?

To determine if the kernel of a linear operator is closed, we need to check if the set of all inputs that produce an output of zero satisfies the definition of a closed set. This can be done by checking if any limit points of the set are also contained within the set.

5. What are the implications of a closed kernel for a linear operator?

A closed kernel has several implications for a linear operator. It ensures that the operator is well-defined and behaves predictably. It also allows us to make important conclusions about the operator, such as the existence of a bounded inverse. In addition, a closed kernel can help us to simplify calculations and solve equations involving the operator.

Similar threads

I If T is diagonalizable then is restriction operator diagonalizable?
    • Linear and Abstract Algebra
Replies
3
Views
1K
I Orthogonality of Eigenvectors of Linear Operator and its Adjoint
    • Linear and Abstract Algebra
Replies
3
Views
3K
I Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
    • Linear and Abstract Algebra
Replies
3
Views
1K
POTW A Linear Operator with Trace Condition
    • Math POTW for University Students
Replies
1
Views
635
Irreducible linear operator is cyclic
    • Linear and Abstract Algebra
Replies
4
Views
1K
I Question about an eigenvalue problem: range space
    • Linear and Abstract Algebra
Replies
1
Views
1K
MHB Uniquely Determined Linear Operator
    • Linear and Abstract Algebra
Replies
7
Views
2K
A Karhunen–Loève theorem expansion random variables
    • Set Theory, Logic, Probability, Statistics
Replies
1
Views
96
I Proof that if T is Hermitian, eigenvectors form an orthonormal basis
    • Linear and Abstract Algebra
Replies
3
Views
2K
Linear operators & mappings between vector spaces
    • Linear and Abstract Algebra
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top