Doubt on singular value decomposition
I think this is way more fit to MathSE, but...


  First of all, I see that the last two columns of the singular value matrix are zeros, what does this mean?


The singular values matrix can always be seen as the actual singular values \$\Sigma\$ and some "padding" zeros \$\mathbf{0}\$, so that you could match the size of \$U\$ and \$V\$. You could have either:

\$ S = \begin{bmatrix} \Sigma &amp; \mathbf{0}\end{bmatrix}\$,
\$ S = \begin{bmatrix} \Sigma \\ \mathbf{0}\end{bmatrix}\$,
\$ S = \begin{bmatrix} \Sigma\end{bmatrix}\$.

in your case, the last columns of \$S\$ are the "padding" zeros. while the first 3 columns make up the diagonal matrix with the singular values \$\Sigma\$, it just happens that you have a zero singular value.  


  And also I can see that inputs directions are three, while the output directions are 4, and I have only two singular values, since one is zero, so I have at most three singular values.


Not sure what you mean by this, but that's what happens when you do the SVD of a matrix \$A \in \mathbb{R}^{4\times 3}\$, you get a \$U \in \mathbb{R}^{4\times 4}\$ and \$V \in \mathbb{R}^{3\times 3}\$. And since those matrices should be orthogonal (\$V^\top V = I\$) some of the rows/columns are associated with the left null space or the right null space. 

For \$A = USV^\top\$, some of the columns of \$U\$ would be associated with the left null space, and for \$V\$ some of its columns (rows of \$V^\top\$) would be associated with the (right) null space.