# Orthogonal Complement to the Kernel of a Linear Transformation

1. Jul 29, 2010

### Kreizhn

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*} 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*}

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

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

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
$$A \cong B^\perp \not\Rightarrow A^\perp \cong B$$?

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

Anyway, I tried this on the toy example

$$A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{pmatrix}, y = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

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
$$x = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$$
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.

Can anyone see where I went wrong?

2. Jul 29, 2010

### Kreizhn

I think I may know what the problem is. Is it perhaps that $\mathbb R^n/\ker A$ is the set of all equivalence classes under the kernel of A, but we want only equivalence under y?

3. Jul 29, 2010

### Office_Shredder

Staff Emeritus
Yeah your description for S is bunk. Obviously it needs to depend on y. S is the set y+ker(A)

4. Aug 2, 2010

### vigvig

You are assuming that y is fixed aren't you? So perhaps, define the map
T:$$R^{n}$$ $$\rightarrow$$ $$R^{m}$$, such that T(x)= A(x-y). Thus, your set S = kerT.
Vignon S. Oussa