# Proof that the Dual Transformation of a Dual Transformation is Itself

Let $$A$$ be an invertible linear transformation in $$\mathbb{R}^{n\times n}$$. Then, we know that $$A$$ takes $$\mathbb{R}^{n-1}$$ hyperplanes in its domain to $$\mathbb{R}^{n-1}$$ hyperplanes in its image. For example, if $$n=2$$, for some line $$L$$ in the 2D domain of $$A$$, the image of every vector along $$L$$ under transformation by $$A$$ would be another line in the 2D codomain of $$A$$.

For every $$\mathbb{R}^{n-1}$$ hyperplane in $$A$$'s domain, let the dual vector $$\mathbf{v}_L$$ of this hyperplane be the vector perpendicular to this hyperplane, with length equal to $$1/d_L$$, where $$d_L$$ is the distance from the origin to the hyperplane. In this manner, $$\mathbf{v}_L \cdot \mathbf{l} = 1$$ for any $$\mathbf{l} \in L$$.

I'd like to prove two things:

1. There is a linear transformation $$B$$ (we'll call it the dual transformation) that takes dual vectors in the domain of $$A$$ to dual vectors in the image of $$A$$. That is, when $$A$$ takes $$L$$ to $$L'$$, $$B$$ takes $$\mathbf{v}_L$$ to $$\mathbf{v}_{L'}$$.
2. The dual transformation of $$B$$ is $$A$$.

~ ~ ~

I came across this problem while trying to visualize the geometric meaning of the transpose without using the matrix definition (i.e. that $$A^T$$ is $$A$$ in matrix form reflected over the diagonal). So it turns out that $$B$$ is simply equal to $${A^T}^{-1}$$ which we can prove by a sequence of dot product manipulations, but this doesn't feel very satisfactory as a proof to me.

I feel like this property is quite basic: that if $$A$$ takes hyperplanes to hyperplanes, and $$B$$ takes dual vectors to dual vectors, then if we invert the relationship and let $$B$$ takes hyperplanes to hyperplanes, then $$A$$ would take $$B$$'s dual vectors to dual vectors. I'd appreciate any insight or intuition. Thank you!

Let $$X$$ be a finite-dimensional vector space. For each $$x \in X$$ and $$\xi \in X^*$$, denote $$\langle \xi,x\rangle = \langle x,\xi\rangle = \xi(x).$$ Recall that $$X^{**}=X$$.

If $$L_1, L_2: X \rightarrow Y$$ are linear maps, then $$L_1 = L_2$$ if and only if for any $$x \in X$$ and $$\eta \in Y^*$$, $$\langle \eta, L_1 x\rangle = \langle\eta,L_2x\rangle.$$

Since for any $$x \in X$$ and $$\eta \in Y$$, $$\langle Lx,\eta\rangle = \langle x,L^*\eta\rangle,$$ it follows that for any $$x \in X$$ and $$\eta \in Y^*$$, $$\langle \eta, L^{**}x\rangle = \langle \eta, (L^*)^* x\rangle = \langle L^* \eta, x\rangle = \langle x,L^*\eta\rangle = \langle Lx, \eta\rangle = \langle \eta, Lx\rangle.$$ Therefore, $$L^{**} = L$$.

$$\newcommand\R{\mathbb R}$$Let $$V = \R^n$$. A covector $$\omega\in V^*$$ is a weighted, oriented hyperplane in the sense that $$\ker\omega$$ is a hyperplane, $$\alpha\omega$$ for any positive scalar $$\alpha$$ represents the same hyperplane, and $$-\omega$$ also represents the same hyperplane.

The map on hyperplanes induced by $$A$$ that you describe is precisely $$\omega \mapsto \omega\circ A^{-1}$$. This is easy to see, since if $$\omega(v) = 0$$ then $$A(v)$$ is mapped to $$(\omega\circ A^{-1})(A(v)) = \omega(v) = 0$$. Another way to denote this map is $$(A^{-1})^*$$, reminiscent of your inverse transpose.

Now, the inner product gives us an isomorphism $$\flat : V \cong V^*$$, $$v \mapsto v^\flat$$ where $$v^\flat(w) = v\cdot w$$. This is precisely the map from a weighted, oriented line $$v$$ to it's weighted, oriented orthogonal complement hyperplane $$v^\flat$$. The inverse is denoved $$\omega \mapsto \omega^\sharp$$.

To get your dual map all we need to do is apply this orthogonalization: $$B(v) = [(A^{-1})^*(v^\flat)]^\sharp = [v^\flat\circ A^{-1}]^\sharp.$$ Every step of this is geometric:

1. Give a vector $$v$$, we apply $$\flat$$ to get its orthogonal hyperplane $$v^\flat$$.
2. Then, we take the image of this hyperplane under $$A$$ by applying $$(A^{-1})^*$$, yielding $$v^\flat\circ A^{-1}$$.
3. Finally, we take the orthogonal complement of this hyperplane using $$\sharp$$, yielding the above formula for $$B(v)$$.

When we dot with arbitrary $$w$$ we get $$B(v)\cdot w = v^\flat(A^{-1}(w)) = v\cdot A^{-1}(w).$$ Thus $$B$$ is the adjoint of $$A^{-1}$$ under the standard inner product; such an adjoint is precisely the transpose: $$w^TBv = B(v)\cdot w = [A^{-1}w]^Tv = w^T[A^{-1}]^Tv.$$