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 $x\perp N or x\perp 2R^TT/\|T\|+\|T\|N$. $\blacksquare$