In mathematicsany vector space V has a corresponding dual vector space or just dual space for short consisting of all linear functionals on Vtogether with the vector space structure of pointwise addition and scalar multiplication by constants. The dual space as defined above is defined for all vector spaces, and to avoid ambiguity may also be called the algebraic dual space.
When defined for a topological vector spacethere is a subspace of the dual space, corresponding to continuous linear functionals, called the continuous dual space.
Dual vector spaces find application in many branches of mathematics that use vector spaces, such as in tensor analysis with finite-dimensional vector spaces. When applied to vector spaces of functions which are typically infinite-dimensionaldual spaces are used to describe measuresdistributionsand Hilbert spaces. Consequently, the dual space is an important concept in functional analysis.
The term dual is due to Bourbaki Since linear maps are vector space homomorphismsthe dual space is also sometimes denoted by Hom VF. In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations. This property is referred to as biorthogonality property. Note that the basis vectors are not orthogonal to each other. Note: The superscript here is the index, not an exponent. We can express this system of equations using matrix notation as.
The biorthogonality property of these two basis sets allows us to represent any point x in V as. In particular, if we interpret R n as the space of columns of n real numbersits dual space is typically written as the space of rows of n real numbers. Such a row acts on R n as a linear functional by ordinary matrix multiplication.
One way to see this is that a functional maps every n -vector x into a real number y. To compute the value of a functional on a given vector, one needs only to determine which of the lines the vector lies on. Or, informally, one "counts" how many lines the vector crosses. Note that F A 0 may be identified essentially by definition with the direct sum of infinitely many copies of F viewed as a 1-dimensional vector space over itself indexed by Ai.
On the other hand, F A is again by definitionthe direct product of infinitely many copies of F indexed by Aand so the identification. Thus if the basis is infinite, then the algebraic dual space is always of larger dimension as a cardinal number than the original vector space.
This is in contrast to the case of the continuous dual space, discussed below, which may be isomorphic to the original vector space even if the latter is infinite-dimensional. But there is in general no natural isomorphism between these two spaces. In other words, the bilinear form determines a linear mapping.
Thus there is a one-to-one correspondence between isomorphisms of V to a subspace of resp. If the vector space V is over the complex field, then sometimes it is more natural to consider sesquilinear forms instead of bilinear forms.
Note that infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals. This identity characterizes the transpose,  and is formally similar to the definition of the adjoint. In the language of category theorytaking the dual of vector spaces and the transpose of linear maps is therefore a contravariant functor from the category of vector spaces over F to itself.
These points of view are related by the canonical inner product on R nwhich identifies the space of column vectors with the dual space of row vectors. Let S be a subset of V.
Within finite dimensional vector spaces, the annihilator is dual to isomorphic to the orthogonal complement. The annihilator of a subset is itself a vector space. If A i is any family of subsets of V indexed by i belonging to some index set Ithen. If V is finite-dimensional, and W is a vector subspacethen.The dual vector space to a real vector space is the vector space of linear functionsdenoted.
In the dual of a complex vector spacethe linear functions take complex values. In either case, the dual vector space has the same dimension as. Given a vector basisAnother way to realize an isomorphism with is through an inner product. A real vector space can have a symmetric inner product in which case a vector corresponds to a dual element by. Then a basis corresponds to its dual basis only if it is an orthonormal basisin which case. A complex vector space can have a Hermitian inner productin which case is a conjugate-linear isomorphism of withi.
Dual vector spaces can describe many objects in linear algebra. When and are finite dimensional vector spaces, an element of the tensor productsaycorresponds to the linear transformation. That is. For example, the identity transformation is. A bilinear form onsuch as an inner product, is an element of. This entry contributed by Todd Rowland. Rowland, Todd. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more.
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Do the two sets have the same cardinality? They are not identical sets are they? Homework Helper. They are of course not identical as sets, but have exactly the same vector space structure. By "vector space structure" do you just mean they have the same number of bases and the same cardinality or do these two vector spaces share other characteristics that other vector spaces might not necessarily share?
Do they have any other internal orderings in common or is the above description sufficient? Gold Member. Last edited: Jan 17, You are familiar with functions evaluated on points, such as f p. In this setting p is like the vector and f is like the covector or dual vector, i. Obviously f and p are not the same. Each one can be viewed as a function on the other.
Thus f and p are "dual" to each other. Let's denote points by letters like p, and we may think of them as vectors, namely the arrow from 0,0 to p. If we want to describe a vector p by its coordinates, then the x and y coordinates are numbers, x p and y p. In these expressions, x and y are functions and p is the argument. Thus the coordinate functions x and y, on the plane, are covectors. So essentially a dot product can be viewed as a linear map from vectors to covectors.
We usually make some assumptions about this map as well, e.In functional analysisthe dual norm is a measure of the "size" of each continuous linear functional defined on a normed vector space. See Theorems 1 and 2 below. The dual norm is a special case of the operator norm defined for each bounded linear map between normed vector spaces. This can be shown to be a norm. This need not hold in infinite-dimensional vector spaces.
The dual of the Euclidean norm is the Euclidean norm, since. The associated dual norm is. The Frobenius norm defined by. It follows.
This establishes the completeness of L XY. The proof for c  now follows directly.
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From Wikipedia, the free encyclopedia. Similarly, for any z there is an x that gives equality. Hahn—Banach theoremclosed graph theoremuniform boundedness principleKakutani fixed-point theoremKrein—Milman theoremmin-max theoremGelfand—Naimark theoremBanach—Alaoglu theorem. Besov spaceHardy spacespectral theory of ordinary differential equationsheat kernelindex theoremcalculus of variationfunctional calculusintegral operatorJones polynomialtopological quantum field theorynoncommutative geometryRiemann hypothesis.
It only takes a minute to sign up. When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are, but I don't really understand why we want to study them within linear algebra.
I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they "show up" as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings? Hopefully, this clarifies my question. Recall that solving such equations or simultaneous sets of such equations is one of the basic motivations for developing linear algebra.
Or better, we can, but only after choosing coordinates, and this is not canonical. For this we need a conceptual interpretation of the above equation.
And here linear functionals come to the rescue. More precisely, the map. Now you might ask: why do we make them a vector space themselves? There are lots of reasons for this; here is one: Remember how important it is, when you solve systems of linear equations, to add equations together, or to multiply them by scalars here I am referring to all the steps you typically make when performing Gaussian elimination on a collection of simultaneous linear equations?
Well, under the dictionary above between linear equations and linear functionals, these processes correspond precisely to adding together linear functionals, or multiplying them by scalars.
If you ponder this for a bit, you can hopefully convince yourself that making the set of linear functionals a vector space is a pretty natural thing to do. Since there is no answer giving the following point of view, I'll allow myself to resuscitate the post. The dual is intuitively the space of "rulers" or measurement-instruments of our vector space. Its elements measure vectors. This is what makes the dual space and its relatives so important in Differential Geometry, for instance.
This immediately motivates the study of the dual space. For motivations in other areas, the other answers are quite well-versed. This also happens to explain intuitively some facts. For instance, the fact that there is no canonical isomorphism between a vector space and its dual can then be seen as a consequence of the fact that rulers need scaling, and there is no canonical way to provide one scaling for space.
However, if we were to measure the measure-instrumentshow could we proceed? Is there a canonical way to do so? Well, if we want to measure our measures, why not measure them by how they act on what they are supposed to measure?
We need no bases for that. This justifies intuitively why there is a natural embedding of the space on its bidual. Note, however, that this fails to justify why it is an isomorphism in the finite-dimensional case.