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Compact linear operator in simple terms?

  1. Mar 11, 2013 #1

    I'm struggling to understand this concept. I think the term probably comes from functional analysis and I don't know any of the terms in that field so I'm having trouble understanding the meaning of what a compact linear operator is.

    I posted this in linear algebra because I'm reading about a matrix which acts as a "compact linear operator" on a vector space. Can anyone explain what this is using simple terms, and without referring to Banach spaces? Deciphering this concept from wikipedia is like going down a rabbit hole, first to learn what "relatively compact" means, then what "close is compact" means...

    Thanks for any help
  2. jcsd
  3. Mar 11, 2013 #2


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    I'll be using the language you said not to but don't worry it isn't esoteric. Let [itex]X,Y[/itex] be Banach spaces and let [itex]T:X\rightarrow Y[/itex] be a linear operator. We say [itex]T[/itex] is compact if for all bounded subsets [itex]B\subseteq X[/itex], [itex]T(B)\subseteq Y[/itex] is pre - compact (also called relatively compact) which means that [itex]\overline{T(B)}[/itex] is compact where [itex]\overline{T(B)}[/itex] is the closure of [itex]T(B)[/itex]. A compact operator is obviously necessarily bounded. Now at this point you can certainly understand all this given that you know what compactness, closure, and boundedness of sets are. If you don't then feel free to ask and it can be explained.
  4. Mar 11, 2013 #3


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    Unfortunately one cannot leave out the topological aspects of this notion, one can merely try to dillute them: a compact linear operator is the natural generalization from finite dimensional operators acting on vector spaces to the infinite dimensional case. And wbn's formulation which is severely abstract can be recast in terms of sequences of elements of a Banach space. Look this definition up on wikipedia.
    Last edited: Mar 11, 2013
  5. Mar 11, 2013 #4
    Compact operators are operators between Banach spaces that are "close to being finite-dimensional". This is the intuition you should have.

    If you have a linear map between finite-dimensional normed spaces [itex]T:\mathbb{R}^n\rightarrow \mathbb{R}^m[/itex], then there are a lot of nice things you can do with this. One of the nicest results is probably the spectral theorem which states that any hermitian linear map can be diagonalized. Of course, we want to generalize this property to linear maps between infinite-dimensional cases, but it fails there. One of the reasons is that a linear operator doesn't need to have any eigenvalues to begin with!! This is in sharp contrast with the finite-dimensional case. Compact linear operators are a class of operators for which the spectral theorem does generalize. So this is one reason why they are important. It should be no surprise that any operator [itex]T:\mathbb{R}^n\rightarrow \mathbb{R}^m[/itex] between finite dimensional spaces is compact. And even more generally: if the range is finite dimensional, then the operator is compact.

    There are many other reasons why compact operators are important. For example, the set of all compact operators in a Hilbert space forms a simple C*-algebra and that is one of the main examples.
    From the point-of-view of Noncommutative Geometry, a compact operator is the analogue of an "infinitesimal".

    There are also many examples of compact operators. Famous examples include the Fredholm operators (= integral operators). These are important tools in solving integral equations. That integral operators are compact is very nice, because it allows us to solve integral equations using techniques that are familiar from finite dimensions.
  6. Mar 11, 2013 #5
    I'm sorry, I don't know what compactness or closure mean. I don't know what a Banach space means really either. From some reading I get the sense that a compact space doesn't have any gaps in it? I'm not a great learner from rows of dense notation.

    Does it (very loosely) mean that if you take some close points in the domain and apply the operator, you end up with some similarly close points in the range?

    Is compactness only a characteristic of infinite-dimensional operators? I'm wondering if I misread somewhere. Can someone give an example of a non compact operator?

    One other thing that interests me, if an operator is compact, what does this mean about the inversion operator?
  7. Mar 11, 2013 #6


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    Any unbounded operator is an example of non-compactness. Take,e.g., a differential operator between normed spaces.

    Compactness is a sort-of second best to being finite, like I think someone here commented. The most general definition is that of every cover by sets having a finite subcollection that covers the set. I'm afraid you may have to spend some time reading up on , and doing exercises on closure, compactness, etc. to understand better. Feel free to ask any followups
  8. Mar 11, 2013 #7


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    Just to clarify, a topological space [itex]X[/itex] is compact if every open cover of [itex]X[/itex] has a finite subcover. Closure is very intuitive: given [itex]A\subseteq X[/itex], the closure [itex]\overline{A}[/itex] of [itex]A[/itex] is essentially the smallest closed subset that contains [itex]A[/itex]. For metric spaces, of which a Banach space is one, you can use the equivalent condition of sequential compactness if it is more intuitive to you: [itex]X[/itex] is sequentially compact if every sequence of points in [itex]X[/itex] has a subsequence that converges to a point in [itex]X[/itex].
    Last edited: Mar 11, 2013
  9. Mar 11, 2013 #8


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    For subsets of ℝ, "compact" is equivalent to "closed and bounded" and "closed" means that for any convergent sequence such that all the terms are in the set, the limit is also in the set. This can perhaps give you some intuition that will help you when you're dealing with other spaces as well, but if you really want to understand this, there's no way around it, you're going to have to study some topology.

    The "closure" is as WannabeNewton said, the smallest closed set that contains the set. For example, the open interval (0,1) isn't closed, because 1/n is a convergent sequence in that set that converges to 0, which isn't in the set. The closure of that set is [0,1], because no convergent sequences in (0,1) have limits that are less than 0 or greater than 1.

    The "no gaps" thing you seem to have heard of is probably "completeness".
  10. Mar 11, 2013 #9


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    Notice too, the standard argument used to show that linear operators are bounded (and therefore continuous) in finite-dimensional spaces, and see how this argument can fail in infinite-dimensional ones:

    Let L: V-->W be a linear operator between finite-dimensional normed spaces. Let v be

    in V, and let {v1,..,vn} be a basis for V. Then v=c1*v1+....+cn*vn

    Then|| L(v)||_W= ||L(c1*v1+...+cn*vn)||_W=|| c1*L(v1)+.....+cn*L(vn)||_W , and

    then use the triangle inequality to show the last expression is bounded. Now, assume

    you have an infinite basis, which you

    can normalize/rescale, so that each basis element has length 1 (notice that 0 is not

    an element of any basis) . Then define a map like T(ei)=i*ei , and extend it linearly

    to the whole space.

    Re Banach space, a Banach space is a normed (and with a metric derived from this norm) which is complete in the given norm/metric. There is a special

    subcategory of Banach spaces that is particularly-nice called Hilbert Spaces. For every finite n, R^n is a Hilbert space.
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