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Homography – Part II: useful facts

04/04/2011

In order to prove the second largest singular value of Homography matrix is 1, or decompose the Homography matrix, the following facts are very important.

Fact 1: Let a\in R^3, b\in R^3 and A=ab^T+ba^T+bb^T. A vector x\in R^3 makes x^TAx=0 if and only if x\perp b or x\perp 2a+b.

Proof:

x^TAx=(x^Ta)(b^Tx)+(x^Tb)(a^Tx)+(x^Tb)(b^Tx)=(x^Tb)(2x^Ta+x^Tb)

That is

x^TAx=(x^Tb)[x^T(2a+b)]

Hence x^TAx=0 implies (x^Tb)=0 or x^T(2a+b)=0. \blacksquare

People usually confuse Fact 1 and Fact 2. Just note for a symmetric matrix, we can not have x^Ax=0 implies Ax=0. Compare the Fact 1 and Fact 2.

Fact 2: Let a\in R^3, b\in R^3 and A=ab^T+ba^T+bb^T. A vector x\in R^3 makes Ax=0 if and only if x\perp a and x\perp b.

Proof: Sufficiency is apparent. Next prove necessity:

case 1: a and b are collinear. Then a=kb, hence A=(2k+1)bb^T. In general, ak+1\neq 0. So it is easy to show A has two zeros eigenvalues , and associating eivenvectors are perpendicular to b.

case 2: a and b are not collinear.

A=ab^T+ba^T+bb^T=(a+b)(a+b)^T-aa^T

a) if x\perp a, x\perp b, then x=a\times b, Ax=0. Easy

b) There exists x\perp a but no perpendicular to a+b. Then x^TAx=x^T(a+b)(a+b)^Tx>0

c) There exists x\perp a+b but no perpendicular to a. Then x^TAx=-x^Taa^Tx<0

So if \lambda (A)={\lambda_1\ge \lambda_2 \ge \lambda_3}, then \lambda_1>0, \lambda_2=0, \lambda_3<0. Hence the eigenspace associating zero eigenvalue is one dimensional. So Ax=0 implies x is collinear with a \times b.

In summary, Ax=0 if and only if x\perp a and x\perp b. \blacksquare

Fact 3: The homography matrix is H=R+TN^T. Let x\in R^3 is a vector. Then \|Hx\|=\|x\| if and only if x\perp N or x\perp 2R^TT/\|T\|+\|T\|N.

Proof: \|Hx\|=\|x\| is equivalent to x^TH^THx=x^Tx. That is

x^T(ab^T+b^Ta+bb^T)x=0

where a=\frac{R^TT}{\|T\|}, b=\|T\|N. According to the Fact 1, we have x\perp N or x\perp 2R^TT/\|T\|+\|T\|N. \blacksquare

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