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Orthogonal Complement to the Kernel of a Linear Transformation

  1. Jul 29, 2010 #1
    Hey all,

    I'm trying to find an orthogonal complement (under the standard inner product) to a space, and I think I've found the result mathematically. Unfortunately, when I apply the result to a toy example it seems to fail.

    Assume that [itex] A \in M_{m\times n}(\mathbb R^n), y \in \mathbb R^n[/itex] and define the space [itex] S = \left\{ x \in \mathbb R^n : A(x-y) = 0 \right\} [/itex]. My goal is to characterize [itex] S^\perp [/itex].

    I performed the following calculation

    [tex] \begin{align*}
    S &= \left\{ x \in \mathbb R^n : A(x-y) = 0 \right\} \\
    &= \left\{ x \in \mathbb R^n : x-y \in \ker A \right\} \\
    &= \mathbb R^n/ \ker A
    \end{align*}
    [/tex]

    where I've used the fact that [itex] x-y \in \ker A [/itex] defines an equivalence relation to turn [itex] \mathbb R^n/\ker A [/itex] into a quotient space. In particular, since [itex] \ker A [/itex] is a closed linear subspace and [itex] \mathbb R^n [/itex] is a Hilbert space in the standard inner product, we must have that

    [tex] \mathbb R^n /\ker A \cong (\ker A )^\perp [/itex]
    and since the orthogonal complement is "reflexive" in finite dimensions, we conclude that
    [tex] \left( \mathbb R^n/\ker A \right)^\perp \cong \ker A [/tex]

    However, this does not seem to produce a correct result. I've checked my work and the only place I can possibly see an error is that even when the orthogonal complement is reflexive, perhaps
    [tex] A \cong B^\perp \not\Rightarrow A^\perp \cong B [/tex]?

    Alternatively, I've also calculated that
    [tex] S^\perp = \text{Row}(A) \cap \text{span}\{y\} [/tex]
    but this is far less useful than a simple result.

    Anyway, I tried this on the toy example

    [tex] A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{pmatrix}, y = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} [/tex]

    Now [itex] \ker A = \left\{ (-t,t,t) : t \in \mathbb R \right\} [/itex]. Choosing an arbitrary representation with t=1, we would get a point [itex] x \in S [/itex] given by
    [tex] x = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix} [/tex]
    which is certainly in S since [itex] x-y \in \ker A [/itex]. However, there is no nonzero [itex] t \in \mathbb R^n [/itex] such that [itex] (0, 1, 4) \cdot(-t,-t,t) = 0 [/itex] and so my result cannot be correct.

    Can anyone see where I went wrong?
     
  2. jcsd
  3. Jul 29, 2010 #2
    I think I may know what the problem is. Is it perhaps that [itex] \mathbb R^n/\ker A [/itex] is the set of all equivalence classes under the kernel of A, but we want only equivalence under y?
     
  4. Jul 29, 2010 #3

    Office_Shredder

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    Gold Member

    Yeah your description for S is bunk. Obviously it needs to depend on y. S is the set y+ker(A)
     
  5. Aug 2, 2010 #4
    You are assuming that y is fixed aren't you? So perhaps, define the map
    T:[tex]R^{n}[/tex] [tex]\rightarrow[/tex] [tex]R^{m}[/tex], such that T(x)= A(x-y). Thus, your set S = kerT.
    Vignon S. Oussa
     
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