User:KYN/WhyDualSpace

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[edit] Why do we need dual spaces?

The concept of dual spaces is used frequently in abstact mathematics, but also has some practical applications. Consider a 2D vector space V = R2 on which a differentiable function f is defined. As an example, V can be the Cartesian coordinates of points in a topographic map and f = f(c1,c2) can be the ground altitude which varies with the coordinate x = (c1,c2) . According to theory, the infinitesimal change df of f at the point x = (c1,c2) as a consequenece of changing the position an infintesimal amount dx = (dc_{1}, dc_{2}) \in V is given by

df = dc_{1} \frac{df(x)}{dc_{1}} + dc_{2} \frac{df(x)}{dc_{2}}

the scalar product between the vector dx and the gradient of f. Clearly, df is a scalar and since it is constructed as a linear mapping on dx, by computing its scalar product with \nabla f(x), it follows from the above defintion that \nabla f(x) is an element of V^{\star}.

From the outset, both vectors dx and \nabla f(x) can be seen as elements of R3. Why is a dual space needed? What is the difference between V and V^{\star} in this case?

To see the difference between V and V^{\star}, remember that in practice both vectors dx and \nabla f(x) must be expressed as a set of three real number which are their coordinates relative to some basis of R3. Intuitively we may choose to use an orthogonal basis, with normalized basis vectors which are mutually perpendicular. Let E = {e1,e2,e3} be a such a basis for R3. This means that dx can be written as

dx = dx1e1 + dx2e2 + dx3e3

where dx1,dx2,dx3 are the (infinitesimal) coordinates of dx in the basis E. Similiarly, \nabla f(x) can be written as

\nabla f(x) = \nabla_{1} f(x) e_{1} + \nabla_{2} f(x) e_{2} + \nabla_{3} f(x) e_{3}

where \nabla_{1} f(x), \nabla_{2} f(x), \nabla_{3} f(x) are the coordinates of \nabla f(x) in the basis E. Given that the coordinates of both