Dual Spaces and Complementary SubspacesDate: 08/13/2003 at 23:19:26 From: Esha Subject: Dual spaces and dual maps What exactly is a dual space of a vector space? And also can you explain why if f exists in V* then f exists in (Im L)(transposed) then why is it that (Im L)(transposed) = Ker L*? I know the definition, but there is no picture of a dual map. Is there such a thing as a dual transformation? I thought that a dual space was when you have a vector space V and you have a transformation L that takes V to another vector space U. And a dual map is a function that assigns U* to a field in V*. Date: 08/15/2003 at 03:25:13 From: Doctor Warren Subject: Re: Dual spaces and dual maps Hi Esha, Thanks for writing to Dr. Math. You've asked a lot of questions here, and most of the answers require quite a bit of explanation. I'm going to try to target my explanations to an audience that is rather far along in the study of linear algebra. If I confuse you at any time, please ask me for clarification. Also, I apologize for the lack of symbology here in plain ASCII text - it might make it hard to discern a vector from a scalar, for example, because I can't use boldface. I will therefore always explicitly list the "part of math" of each symbol I use. I will use greek letters spelled out (like "phi") to represent maps, the letters (a, b, c...) to represent scalars, and the letters (u, v, w...) to represent vectors. I will spell out operations, such as "the complementary subspace of" and "the transpose of." Let's start at the beginning. A "functional" is a black box that takes vector(s) as input and spits out real numbers as output. In other words, a linear functional is a map from vectors to real numbers: phi: V -> R where V is a real vector space and R is the real number line. Furthermore, a "linear functional" is a functional that satisfies the conditions: the additive property: phi(v + w) = phi(v) + phi(w) the homogeneity property: phi(av) = a phi(v) or both together: phi(av + bw) = a phi(v) + b phi(w) where a and b are scalars, and v and w are vectors in some vector space V. Every vector space V implies, by definition, a dual vector space. The dual space of V, called V*, is the space of all linear functionals on V. (Say this to yourself a couple of times, until it makes sense... or seems to make sense.) In other words, the dual space V* is the space of all possible maps from V -> R. The most common example of a dual vector space is as follows: If V is the space of row vectors, V* is the space of column vectors, because member V . member V* = real number / d \ (a b c) . ( e ) = ad + be + cf \ f / where . denotes matrix multiplication. Now, on to your questions. ---------------------------------------------------------------------- --------------- FIRST QUESTION is: "explain why if f exists V* then f exists in (Im L)(transposed)." ---------------------------------------------------------------------- --------------- We need to work on your notational precision a bit first! We reserve the following words and symbols to mean the following things: *: the dual of some vector space. Example: V*, the dual space of V. transpose (^T): the act of exchanging rows for columns in some matrix A. Example: / a b \ T / a c \ \ c d / = \ b d / complementary: the set of all vectors that are orthogonal to subspace some subspace of some vector space V. On the blackboard, this operator looks like a superscript upside down "T." The reason I'm being picky is that it seems you're mixing up some terminology. You've applied the word "transpose" to "Im L," which is the image of some linear operator L. The image is a subspace, and it makes no sense to transpose a subspace - only a matrix. I have the feeling you meant "the complementary subspace of Im(L)" when you wrote "(Im L)(transposed)." Unfortunately, I'm unable to do much to answer this question, because I do not know how V* and L are related. If you can provide me more information about this relation, I can probably help you with your question. Perhaps your teacher has tried to explain some things, but has not been very precise in the way he presented the various mathematical devices: V, V*, L, and so on? ----------------------------------------------------------------- SECOND QUESTION is: "why is it that (Im L)(transposed) = Ker L*." ----------------------------------------------------------------- Once again, your notation is fouled a little bit. I think what you mean is this: "Why is the complementary subspace of the image of L the same as the kernel of the transpose of L?" This is a very deep relation. To start with, let's consider a symmetric linear transformation from R^3 to R^3 represented by matrix L: T define L = / 1 0 0 \ and L = L ( 0 1 0 ) \ 0 0 0 / What does this linear transformation do when applied to 3-vectors in space V? It maps vectors of the form (x, y, z) to vectors of the form (x, y, 0). In other words, it projects the volume of a 3D space onto the 2D x-y plane. It "squashes" the volume of 3D space onto a plane. What are the kernel and image of this linear transformation L? The kernel is the space of all vectors that are mapped by L to the zero vector (0,0,0). Which vectors does L map to (0,0,0)? Any vector parallel or anti-parallel to the z-axis has x=0 and y=0, and thus L maps any vector that points straight up in the +z direction, or straight down in the -z direction, to the zero vector. The kernel of L is all vectors that point straight up in the +z direction, or straight down in the -z direction. The kernel of L is the set of vectors of the form (0,0,z). What is the image of L? If you think about L as a machine, the image of L is the set of all outputs that L can spit out. It's the range of the transformation. Since L maps everything onto the x-y plane, it is obvious that the image of L is the x-y plane. So what is a complementary subspace? A complementary subspace of a subspace U is the set of vectors which are orthogonal to all vectors in U. It should be obvious now how to construct the complementary subspace of the image of our little linear transformation L. The image of L is the x-y plane. The set of vectors which are orthogonal to all vectors in the x-y plane are all vectors which point straight up in the +z direction or straight down in the -z direction. You can describe such a vector as being of the form (0,0,z). It should also be obvious how to construct the kernel of the transpose of L. Since L equals its own transpose (it's symmetric), the kernel of L^T is also the set of all vectors of the form (0,0,z). The complementary subspace of the image of L therefore equals the kernel of the transpose of L, as we sought out to demonstrate. Let's see if we get a similar result from a non-symmetric linear transformation. Let's define a linear transformation, a map from R^2 to R^3, as being represented by the matrix J: / 2 1 \ T define J = ( 3 1 ) and J = / 2 3 1 \ \ 1 -1 / \ 1 1 -1 / What's the kernel of the tranpose of J? To find it, simply solve the linear equation: T / x \ / 0 \ J ( y ) = \ 0 / \ z / The result is x = 4z, 3 = -z, or, equivalently, the kernel of the transpose of J is the one-dimensional subspace of R^3 (a line) spanned by the single basis vector (4,-3,1). Any vector parallel to the line (x,y,z) = t(4,-3,1), where t is a scalar, is mapped to (0,0,0) by J^T. What's the image of J? J maps any vector in R^2, (x,y), into: J / x \ = / 2x + y \ \ y / ( 3x + y ) \ x - y / This is the set of all vectors which are orthogonal to (4,-3,1), as you can easily demonstrate by taking the dot product between them: / 2x + y \ / 4 \ ( 3x + y ) . ( -3 ) = 0 \ x - y / \ 1 / The complementary space of the image of J is, of course, the set of vectors orthogonal to it. These are the vectors parallel to the line t(4,-3,1), which we already identified as the kernel of J^T. I hope this helps you understand some of these relations. Let me know if you need any clarification, or if you have any additional questions! - Doctor Warren, The Math Forum http://mathforum.org/dr.math/ |
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