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Dual Spaces and Complementary Subspaces

Date: 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 

   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

      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

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 

   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

- Doctor Warren, The Math Forum 
Associated Topics:
College Linear Algebra

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