# 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**

**Remark:**

**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

Hence

Next it is easy to show has a eigenvalue=1 and it’s the second largest one.

I think A should be: A = a b^T + b a^T + b b^T

you wrote A = a b^T + b^T a + b b^T