Homography – Part I: second largest singular value=1
The proof on Ma Yi’s book is not so clear. Here I want to give my proof.
Property: The homography matrix is . Its singular values are . Then .
Proof: For the sake of simplicity, just use . The original problem is equivalent to prove the second largest eigenvalue of is 1.
Let . Then our goal becomes to prove the second largest eigenvalue of is zero. Let , then
Let , then
Hence is the difference of two positive semi-definite matrices.
step 1: prove has a zero eigenvalue
Consider , it is obvious . So has a zero eigenvalue and its associating eigenvector is .
step 2: prove the zero eigenvalue is the second largest eigenvalue of
Note . There exists a vector which is orthogonal to but not , then . There exists a vector $x_2$ which is orthogonal to but not , then . Therefore, has at least one positive eigenvalue and at least one negative eigenvalue. QED
a) This is very important for the decomposition! The eigenvector associating zero eigenvalue of is . Remember what mean? If and is collinear, then the eigenspace associating zero eigenvalue is a plane, which is the plane orthogonal to the vector .
b) We can also use to prove. The result is the same.
Here is the brief proof. Note , so we have
Next it is easy to show has a eigenvalue=1 and it’s the second largest one.