Paper: R. Hartly, "In Defence of the 8-point Algorithm," GE-Corpoate Research Dev.
Summary
- The 8-point algorithm is the algorithm that solve Eipolar matrix estimation problem.
- But it is susceptible to noise.
- The paper refer to basic 8-point algorithm and it's numerical constraint called The singularity constraint related to Singular Value Decomposition.(Reduce Rank of least singular value)
- The important sections of this paper are after Section 3. These are the main ideas.
- Section 4. Condition Number
- From Interlacing Property, Symmetric matrix has some relation between eigenvalues of Submatrix.
- $trace(A)= \sum_{i=1}^N\lambda_i$
- Suppose $X=A^TA$, $X_r$ is the trailing $r\times r$ principal submatrix $A^TA$ and by $\lambda_i(X_r)$ its $i-th$ largest eigenvalue. Thus, $X_9=A^TA$ and $\kappa=\lambda_i(X_9)/\lambda_8(X_9)$.
- Suppose we have image point form $u=(100,100,1)$ if $u'=u$. we got some row of equation matrix $A$ like $\bold a_i=u\otimes u'=(10^4, 10^4, 10^2, 10^4, 10^4, 10^2, 10^2, 10^2, 1)$.
- Summing over all point correspondences will result in a matrix $A^TA$ for which the diagonal entries are approximately in this portion.
- The contribution to the matrix $A^TA$ is of the form $\bold {rr}^T$, which will contain entries ranging between $10^8$ and 1. For instance, the diagonal entries of $A^TA$ will be $(10^8, 10^8, 10^4, 10^8, 10^8, 10^4, 10^4, 10^4, 1)$.
- First we consider the eigenvalues of $trace(X_2)=10^4+1$.
- Since the matrix is positive semi-definite, both eigenvalues are non-negative, so we may deduce that $\lambda_1(X_2)≤10^4+1$.
- From the interlacing Property, we deduce that $$\lambda_8(X_9)≤
- $Ah=0$이 안정적인 해를 가지려면, $A^TA$가 보다 좋은 condition number를 가져야 한다.
- $||A||F=\Sigma{i=1}^9\sqrt{diag(A^TA)i}=\begin{pmatrix}\Sigma{i=1}^9d_i^2\end{pmatrix}^{1/2}$이기 때문에, $\kappa$가 매우 크면, $d_8$은 Frobenius norm의 매우 작은 영향을 끼친다.
- $d_8$이 노이즈에 의하여 변화가 생겨도 $A^TA$에는 상대적으로 작은 변화를 가져온다.
- $A^TA=d_1 x_1x_1^T+...+d_8x_8x_8^T+...$에서 $d_8$이 매우 작으면 $x_8x_8^T$의 영향이 작아짐
- Condition number가 커지면 $Ah=0$의 해인 $h$의 변화도 커지기 때문에 Least eigenvector에 큰 변화를 가져온다.
Supplimentary
Suppose $A\in{\rm I\!R}^{n\times n}$ is symmetric, Let $B\in{\rm I\!R}^{m\times m}$ with $m<n$ be a principal matrix.
Suppose A has eigen values $\lambda_1 ≤... ≤ \lambda_n$ and $\beta_1 ≤... ≤ \beta_m$. Then,
$$
\lambda_k ≤\beta_k ≤\lambda_{k+n-m} \text{ for }k=1,...,m.
$$
And if $m=n-1$,
$$
\lambda_1 ≤\beta_1 ≤\lambda_2≤\beta_2≤...≤\beta_{n-1}≤\lambda_n.
$$
Proof: WLOG, assume $A=\begin{bmatrix} B & X^T \\X & Z\end{bmatrix}$. Let $\{ x_1, ..., x_n\}$ be eigenvectors of $A$, $\{y_1, ..., y_m\}$ be eigenvectors of $B$. We define the following vector spaces:
$$
V=span(\bold{x_k, ..., x_n}), \ \ W=span(\bold{y_1, ...,y_k}), \tilde{W}\begin{Bmatrix}\begin{pmatrix}\bold w \\ \bold 0\end{pmatrix}\in {\rm I\!R}^{n}, \bold w \in W\end{Bmatrix}
$$
Sine $dim(V)=n-k+1$ and $dim(\tilde{W})=dim(W)=k$, there exists $\tilde{w}\in V\cap \tilde W$ and $\tilde{w}=\begin{pmatrix} \bold w \\ \bold 0\end{pmatrix}$ for some $\bold w \in W$. Then
$$
\tilde{\bold w}^TA\tilde{\bold w}=\begin{bmatrix} \bold w^T & \bold 0^T\end{bmatrix} \begin{bmatrix} B & X^T \\ X & Z\end{bmatrix}\begin{bmatrix}\bold w \\ \bold 0\end{bmatrix}=\bold w^T B \bold w
$$