Orthogonal Complement to the Kernel of a Linear Transformation

[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

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

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?

 

[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?

 

[Office_Shredder]

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

 

[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

 

[blue_raver22]

Hi there,

It seems like your calculations are correct. However, there may be an issue with your toy example. In general, the orthogonal complement of a subspace S is not unique, so there may be multiple spaces that satisfy the conditions. It's possible that your calculated result is correct, but it just doesn't match the specific example you chose.

Additionally, it's worth considering that the orthogonal complement of a subspace S is defined as the set of all vectors that are orthogonal to every vector in S. So in your toy example, the vector (0,1,4) may not be orthogonal to every vector in S, even though it satisfies the condition of being orthogonal to the vectors in the kernel of A.

I would suggest trying a different example or checking your calculations with a different method to see if you get the same result. It's also helpful to keep in mind that the orthogonal complement is not unique, so there may be other spaces that satisfy the conditions. I hope this helps!