How to diagonalize this matrix? Consider the $n\times m$ matrix $M=[M_1, \ldots, M_m]$ where the $i$-th column reads 
$$
M_i= \,^t(\underbrace{1,\ldots,1}_{a_i},0,\ldots,0)
$$
where the $a_i$'s are given positive natural numbers. 
Is it possible to compute the  singular values of $M$ in terms of the $a_i$'s ?
Maybe it is simpler if one choose the $a_i$'s non-decreasing.
I am interested  in this problem since it somehow generalize the case where  $n=m$ and $a_i=n$ for all $i=1,\ldots,n$, and for which the singular values are $0$ with multiplicity $n-1$ and $n^2$ with multiplicity $1$. This comes from an easy kernel size plus trace argument. And I was wondering if it could be applied to this general case, but I didn't succeed ...
 A: Since we are now asking about singular values, arranging the columns for non-decreasing $a_i$ is without loss of generality.  A permutation of columns does not affect a matrix's singular values.
Consider $A = M'M$, a symmetric positive semi-definite matrix, whose eigenvalues are squares of the singular values of $M$.  $A$ is a structured matrix with $A_{ij} = \min \{a_i,a_j\} = a_{\min \{i,j\} }$.
The nonsingular case ($n \times n$ matrix, strictly increasing $a_i = i$) is included in a "test case gallery" as minij in MatLab and other suites.  According to comments in this test C code, the eigenvalues of $A$ have an explicit formula:
$$ \frac{0.5}{ 1 - \cos \left( \frac{( 2k - 1 )\pi}{ 2n + 1 } \right) } \;\; \forall \; 1 \le k \le n $$
Accordingly the singular values of $M$ should be the respective square roots of those.  For $n=2$ we get singular values $\varphi = 1.61803\ldots$ and $\varphi - 1 = 0.61803\ldots$ as a perhaps lucky connection to the golden ratio.
A similarly simple expression for singular values of the singular cases seems a daunting challenge, if only because of the large number of possibilities.  However the rank of $M$ (resp. of $A$) can be found by counting the number of columns containing a "leading one", since removing duplicate rows puts $M$ in row-echelon form (due to column sorting).  Therefore the nullity (dimension of nullspace) of $M$ is easily found, or in other words the count of singular values which are zero.
