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When calculating a homography with line correspondences instead of point correspondences, what is the derivation of the formula:

$$ l_i = H^T\cdot l^{'}_i $$

I know that:

$$ l^T\cdot x = 0 \quad\text{and}\quad l^{'T}\cdot x^{'} = 0\\ \text{with}\quad x^{'} = H\cdot x\\ \text{then}\quad l^{'T}\cdot H\cdot x = 0 $$

But I don't know how to continue.

Matthias Preu
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  • Could you provide some context? What is $H$, $\ell_i$, $(\cdot)'$, $x$? – martini Feb 01 '16 at 11:37
  • I'm referring to section 4.1.4 (derivation of the Direct Linear Transformation algorithm) in Hartley, Zisserman. The formula above is shown in the first section.

    Some context: There are two images $I$ and $I^{'}$. A line $l^{'}_i$ in the second image is projected into the first image as line $l_i$ with the homography matrix $H$. The homography matrix can also relate a corresponding point pair $x$ and $x^{'}$ in the images $I$ and $I^{'}$.

     – Matthias Preu Feb 01 '16 at 12:07

2 Answers2

3

You are almost there. Now, just note that your result $$0=l'^THx=(H^Tl')^Tx$$ should hold for all points on the line $l$ in the first image. As a consequence, it follows that $l\sim H^Tl'$, as desired.

Mårten W
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I will provide a complete proof.

Assumptions

  1. $\mathbf{l}^T \mathbf{x} = 0$, for all 2d points $\mathbf{x} \in \mathbb{R}^3$ represented in homogenous coordinates that belong to $\mathbf{l}^T \in \mathbb{R}^3$ (i.e. a homogenous representation of a line, in a plane).
  2. Similarly, $\mathbf{l}'^T \mathbf{x}' = 0$, for all points $\mathbf{x}' \in \mathbb{R}^3$ that lie on the other line $\mathbf{l}' \in \mathbb{R}^3$ (i.e. a homogenous representation of a line, in another plane).
  3. $\mathbf{x}' = \mathbf{H} \mathbf{x}$, for some homography $\mathbf{H} \in \mathbb{R}^{3 \times 3}$.

Theorem

We want to prove that $\mathbf{l} = \mathbf{H}^T \mathbf{l}'.$

Proof

We can combine 3 with 2 to obtain

\begin{align} \mathbf{l}'^T \mathbf{x}' &= 0 \iff \\ \mathbf{l}'^T \left(\mathbf{H} \mathbf{x} \right)&= 0 \end{align} Given that matrix multiplication is associative, we don't need the parentheses \begin{align} \mathbf{l}'^T \mathbf{H} \mathbf{x} = 0 \label{1}\tag{1} \end{align} Now, note that $\mathbf{l}'^T \mathbf{H} = (\mathbf{H}^T \mathbf{l}')^T$, then \ref{1} becomes \begin{align} (\mathbf{H}^T \mathbf{l}')^T \mathbf{x} = 0 \end{align} Now, we know that $\mathbf{l}^T \mathbf{x} = 0$, so \begin{align} (\mathbf{H}^T \mathbf{l}')^T \mathbf{x} &= \mathbf{l}^T \mathbf{x} \iff \\ (\mathbf{H}^T \mathbf{l}')^T &= \mathbf{l}^T \iff \\ \mathbf{H}^T \mathbf{l}' &= \mathbf{l} \end{align}

Note that this equality holds up to scale (because we are working in projective space).