- #1
Kreizhn
- 714
- 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 A \in M_{m\times n}(\mathbb R^n), y \in \mathbb R^n and define the space S = \left\{ x \in \mathbb R^n : A(x-y) = 0 \right\}. My goal is to characterize S^\perp.
I performed the following calculation
\begin{align*}<br /> S &= \left\{ x \in \mathbb R^n : A(x-y) = 0 \right\} \\<br /> &= \left\{ x \in \mathbb R^n : x-y \in \ker A \right\} \\<br /> &= \mathbb R^n/ \ker A<br /> \end{align*}<br />
where I've used the fact that x-y \in \ker A defines an equivalence relation to turn \mathbb R^n/\ker A into a quotient space. In particular, since \ker A is a closed linear subspace and \mathbb R^n is a Hilbert space in the standard inner product, we must have that
\mathbb R^n /\ker A \cong (\ker A )^\perp [/itex]<br /> and since the orthogonal complement is "reflexive" in finite dimensions, we conclude that<br /> \left( \mathbb R^n/\ker A \right)^\perp \cong \ker A<br /> <br /> 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<br /> A \cong B^\perp \not\Rightarrow A^\perp \cong B?<br /> <br /> Alternatively, I've also calculated that<br /> S^\perp = \text{Row}(A) \cap \text{span}\{y\}<br /> but this is far less useful than a simple result. <br /> <br /> Anyway, I tried this on the toy example<br /> <br /> A = \begin{pmatrix} 1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 \\ 1 &amp; -1 &amp; 0 \end{pmatrix}, y = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}<br /> <br /> Now \ker A = \left\{ (-t,t,t) : t \in \mathbb R \right\}. Choosing an arbitrary representation with t=1, we would get a point x \in S given by<br /> x = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}<br /> which is certainly in S since x-y \in \ker A. However, there is no nonzero t \in \mathbb R^n such that (0, 1, 4) \cdot(-t,-t,t) = 0 and so my result cannot be correct.<br /> <br /> Can anyone see where I went wrong?
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 A \in M_{m\times n}(\mathbb R^n), y \in \mathbb R^n and define the space S = \left\{ x \in \mathbb R^n : A(x-y) = 0 \right\}. My goal is to characterize S^\perp.
I performed the following calculation
\begin{align*}<br /> S &= \left\{ x \in \mathbb R^n : A(x-y) = 0 \right\} \\<br /> &= \left\{ x \in \mathbb R^n : x-y \in \ker A \right\} \\<br /> &= \mathbb R^n/ \ker A<br /> \end{align*}<br />
where I've used the fact that x-y \in \ker A defines an equivalence relation to turn \mathbb R^n/\ker A into a quotient space. In particular, since \ker A is a closed linear subspace and \mathbb R^n is a Hilbert space in the standard inner product, we must have that
\mathbb R^n /\ker A \cong (\ker A )^\perp [/itex]<br /> and since the orthogonal complement is "reflexive" in finite dimensions, we conclude that<br /> \left( \mathbb R^n/\ker A \right)^\perp \cong \ker A<br /> <br /> 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<br /> A \cong B^\perp \not\Rightarrow A^\perp \cong B?<br /> <br /> Alternatively, I've also calculated that<br /> S^\perp = \text{Row}(A) \cap \text{span}\{y\}<br /> but this is far less useful than a simple result. <br /> <br /> Anyway, I tried this on the toy example<br /> <br /> A = \begin{pmatrix} 1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 \\ 1 &amp; -1 &amp; 0 \end{pmatrix}, y = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}<br /> <br /> Now \ker A = \left\{ (-t,t,t) : t \in \mathbb R \right\}. Choosing an arbitrary representation with t=1, we would get a point x \in S given by<br /> x = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}<br /> which is certainly in S since x-y \in \ker A. However, there is no nonzero t \in \mathbb R^n such that (0, 1, 4) \cdot(-t,-t,t) = 0 and so my result cannot be correct.<br /> <br /> Can anyone see where I went wrong?