The proof on Ma Yi’s book is not so clear. Here I want to give my proof.

Property: The homography matrix is $H=R+\frac{1}{d}TN^T$. Its singular values are $\sigma_1\le \sigma_2 \le \sigma_3$. Then $\sigma_2=1$.

Proof: For the sake of simplicity, just use $H=R+TN^T$. The original problem is equivalent to prove the second largest eigenvalue of $H^TH$ is 1.

$H^TH=(R+TN^T)^T(R+TN^T)=I+R^TTN^T+NT^TR+NT^TTN^T$

Let $A=R^TTN^T+NT^TR+NT^TTN^T$. Then our goal becomes to prove the second largest eigenvalue of $A$ is zero. Let $u=R^TT$, then

$A=R^TTN^T+NT^TR+NT^TTN^T=uN^T+Nu^T+\|u\|^2NN^T$

Let $a=\frac{u}{\|u\|}, b=\|u\|N$, then

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

Hence $A$ is the difference of two positive semi-definite matrices.

step 1: prove $A$ has a zero eigenvalue

Consider $x=a\times (a+b)=a\times b$, it is obvious $Ax=0$. So $A$ has a zero eigenvalue and its associating eigenvector is  $x=a\times b=(R^TT)\times N$.

step 2: prove the zero eigenvalue is the second largest eigenvalue of $A$

Note $A=(a+b)(a+b)^T-aa^T$. There exists a vector $x_1$ which is orthogonal to $a$ but not $a+b$, then $x^TAx=x_1^T(a+b)(a+b)^Tx_1>0$.  There exists a vector $x_2$ which is orthogonal to $a+b$ but not $a$, then $x_2^TAx_2=-x_2^Taa^Tx_2<0$. Therefore, $A$ 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 $A$ is $x=a\times b=(R^TT)\times N$. Remember what $-R^TT$ mean? If $-R^T$ and $N$ is collinear, then the eigenspace associating zero eigenvalue is a plane, which is the plane orthogonal to the vector $N$.

b) We can also use $HH^T$ to prove. The result is the same.

Here is the brief proof. Note $N^TN=1$, so we have

$HH^T=(R+TN^T)(R+TN^T)^T=I+TN^TR^T+RNT^T+TT^T$

Hence

$HH^T=I+(T+RN)(T+RN)^T-RNN^TR=I+(a+b)(a+b)^T-bb^T$

Next it is easy to show $HH^T$ has a eigenvalue=1 and it’s the second largest one.