Given a matrix $A$:
$$ A = \begin{pmatrix} a_{11} & 0 & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & 1 \\ \end{pmatrix} $$
All the parameters different from $0$ are strictly positive. I was wondering if in this specific case there's like a simple solution for the SVD decomposition.
My attempt was to find the eigenvalues and then the eigenvectors, but I've realized that I don't know a specific algorithm for this.
I think I can use somehow the Gram Schmidt orthonormalization procedure, but I might be wrong here.
Any suggestion is welcome.
Let us use the following simplified notations :
$$M:= \begin{pmatrix} a & 0 & b \\ 0 & c & d\\ 0 & 0 & 1 \\ \end{pmatrix}$$
We are looking for the SVD decomposition $M=USV^T$ of $M$.
$$N:=M^TM= \begin{pmatrix} a^2 & 0 & ab \\ 0 & c^2 & cd\\ ab & cd & (b^2+d^2+1) \\ \end{pmatrix}=(USV^T)^T(USV^T)=VS^2V^T.$$
The characteristic polynomial of $N$ (whose roots are the $\sigma_k^2$s) depends only on $A=a^2, B=b^2, C=c^2, D=d^2$ ; it is :
$$- x^3 + \underbrace{(A+B+C+D+1)}_{\text{trace}(N)}x^2 - (AC+AD+BC+A+C)x +\underbrace{AC}_{\det(N)} \tag{1}$$
(see how simple is the determinant !)
Now, either one takes a "numerical attitude", and it's rather easy, or, for the fun, one can try a Computer Algebra System...
providing explicit expressions for the 3 roots of polynomial above, but these are one kilometer long...(I don't try to reproduce them here).
Having the $\sigma_k$s, one can easily obtain $V$ as the matrix of eigenvectors associated with the $\sigma_k^2$s, either formally (desperate !) or numerically.
Now a similar treatment can be done on $MM^T$ for obtaining $U$ (but one can short-circuit the computation in different ways...).