Eigenbackground Revisited: Can We Model the Background with Eigenvectors?

Abstract

Using dominant eigenvectors for background modeling (usually known as Eigenbackground) is a common technique in the literature. However, its results suffer from noticeable artifacts. Thus, there have been many attempts to reduce the artifacts by making some improvements/enhancements in the Eigenbackground algorithm. In this paper, we show the main problem of the Eigenbackground is at its own core and in fact, it may not be a good idea to use the strongest eigenvectors for modeling the background. Instead, we propose an alternative solution by exploiting the weakest eigenvectors (which are usually thrown away and treated as garbage data) for background modeling. MATLAB codes are available at the GitHub of the paper.

Appendix

Theorem 3

Suppose that $A \in {\mathbb {C}}^{n\times n}$ is a Hermitian matrix, $A'=A+yy^*$ is rank one updated of A and $y \in {\mathbb {C}}^n$ is column vector. If v and $v'$ are the normalized eigenvectors of A and $A'$ corresponding to their largest eigenvalues, then:

$$\begin{aligned} \left\Vert v-v'\right\Vert \le \beta \left\Vert E\right\Vert \end{aligned}$$

(13)

where $E=yy^*$ is a small perturbation of A and $\beta$ is a value unrelated to E.

Proof

Suppose that $\lambda$ and $\lambda '$ are the largest eigenvalues of A and $A'$. By definition of eigenvector we have

$$\begin{aligned} Av&= \lambda v \\ A'v'&= \lambda ' v' \end{aligned}$$

According to Proposition 2, there is an $\epsilon >0$ such that $\lambda +\epsilon = \lambda '$. Hence:

$$\begin{aligned} (A+E)v' - Av&= \lambda ' v' - \lambda v \nonumber \\&= (\lambda + \epsilon )v' - \lambda v \end{aligned}$$

(14)

Thus:

$$\begin{aligned} A(v' -v) + Ev'= & {} \lambda v' + \epsilon v' - \lambda v \nonumber \\\Rightarrow & {} \nonumber \\ (A-\lambda I)(v'-v)= & {} 7 (\epsilon I - E)v' \nonumber \\\Rightarrow & {} \nonumber \\ v'-v= & {} {(A-\lambda I)^{-1} (\epsilon I - E)v' } \nonumber \\\Rightarrow & {} \nonumber \\ \left\Vert v'-v\right\Vert= & {} \left\Vert (A-\lambda I)^{-1} (\epsilon I - E)v' \right\Vert \nonumber \\&\le \underbrace{\left\Vert (A-\lambda I)^{-1}\right\Vert }_{\alpha }\left\Vert (\epsilon I - E)\right\Vert \underbrace{\left\Vert v'\right\Vert }_1 \nonumber \\&= \alpha \left\Vert (\epsilon I - E)\right\Vert \nonumber \\&\le \alpha (\left\Vert \epsilon I\right\Vert + \alpha \left\Vert -E\right\Vert ) \quad (\mathrm{Triangle Inequality})\nonumber \\&= \alpha ( \epsilon + \left\Vert E\right\Vert )\nonumber \\&\le \alpha (\left\Vert E\right\Vert + \left\Vert E\right\Vert ) (\mathrm{According to Proposition}\,\nonumber \\&\quad 2: \epsilon \le \left\Vert E\right\Vert )\nonumber \\&= \beta \left\Vert E\right\Vert \end{aligned}$$

(15)

where $\beta = 2\alpha$ $\square$

Lemma 3

For a given Hermitian matrix $A \in {\mathbb {C}}^{n\times n}$ and a column vector $y \in {\mathbb {C}}^n$, we have:

$$\begin{aligned}\lambda _\mathrm{max}(A) \le \lambda _\mathrm{max}(A+yy^*) \le \lambda _\mathrm{max}(A) + \left\Vert y\right\Vert ^2 \end{aligned}$$

Proof

See [36] . $\square$

Lemma 4

The norm of a symmetric matrix is maximum absolute value of its eigenvalue.

Proof

We have

$$\begin{aligned} \Vert A\Vert _2=\max _{\Vert x\Vert =1}\Vert Ax\Vert \end{aligned}$$

where $\Vert \cdot \Vert$ denotes the ordinary Euclidean norm. This is a constrained optimization problem with Lagrange function:

$$\begin{aligned} L(x,\lambda )=\Vert Ax\Vert ^2-\lambda (\Vert x\Vert ^2-1)=x^TA^2x-\lambda (x^Tx-1) \end{aligned}$$

Taking squares makes the following step easier. Taking derivative with respect to x and equating it to zero, we get

$$\begin{aligned} A^2x-\lambda x=0 \end{aligned}$$

the solution for this problem is the eigenvector of $A^2$. Since $A^2$ is symmetric, all its eigenvalues are real. So $x^TA^2x$ will achieve maximum on set $\Vert x\Vert ^2=1$ with maximal eigenvalue of $A^2$. Now since A is symmetric, it admits representation

$$\begin{aligned} A=Q\Lambda Q^T \end{aligned}$$

with Q the orthogonal matrix and $\Lambda$ diagonal with eigenvalues in diagonals. For $A^2$ we get

$$\begin{aligned} A^2=Q\Lambda ^2 Q^T \end{aligned}$$

so the eigenvalues of $A^2$ are squares of eigenvalues of A. The norm $\Vert A\Vert _2$ is the square root taken from maximum $x^TA^2x$ on $x^Tx=1$, which will be the square root of maximal eigenvalue of $A^2$ which is the maximal absolute eigenvalue of A [37]. $\square$

Lemma 5

Suppose $A=uv^T$ where u and v are nonzero column vectors in ${{\mathbb {R}}}^n$, $n\ge 3$. Then, $\lambda =0$ and $\lambda =v^Tu$ are the only eigenvalues of A.

Proof

$\lambda =0$ is an eigenvalue of A since A is not of full rank. $\lambda =v^Tu$ is also an eigenvalue of A since

$$\begin{aligned} Au = (uv^T)u=u(v^Tu)=(v^Tu)u. \end{aligned}$$

We assume $v\ne 0$. The orthogonal complement of the linear subspace generated by v (i.e., the set of all vectors orthogonal to v) is therefore $(n-1)$-dimensional. Let $\phi _1,\dots ,\phi _{n-1}$ be a basis for this space. Then, they are linearly independent and $uv^T \phi _i = (v\cdot \phi _i)u=0$. Thus, the eigenvalue 0 has multiplicity $n-1$, and there are no other eigenvalues besides it and $v\cdot u$. $\square$

Proposition 2

In the previous lemmas (3, 4 and 5), assume that $E=yy^*$, then there exists an $0 \le \epsilon \le \left\Vert E\right\Vert$ such that:

$$\begin{aligned} \lambda _\mathrm{max}(A) + \epsilon = \lambda _\mathrm{max}(A+yy^*)&\le \lambda _\mathrm{max}(A) + \left\Vert y\right\Vert ^2 \\&= \lambda _\mathrm{max}(A) + \left\Vert E\right\Vert \end{aligned}$$