linear algebra - Matrix decomposition problem

As Sergei Ivanov pointed out in his first comment, it is necessary and sufficient, to solve your (ii) and (iii), to have

$$sum_{i = 1}^{n} x_i = sum_{i=1}^{n} y_i ; ; .$$ If this is true then take $M = sum_{i = 1}^{n} x_i = sum_{i=1}^{n} y_i ; ; .$ The most natural solution to (ii) and (iii) is the rank-one matrix $C^0$ given by $$c_{ij}^{0} = frac{x_i y_j}{M}$$

Now, there is a kernel involved next of dimension $(n-1)^2,$ these being matrices $F$ satisfying $F 1 = 0$ and $F^t 1 = 0.$ One may specify any entries desired in the upper left square $n-1$ by $n-1$ block of $F$, then fill in the final column and row. Any solution of (ii) and (iii) must be of the form

$$C^0 + F ; ; .$$

Progress: for your purpose it is better to specify the matrix $F$ as shown below for $n=4,$ the other entries of $F$ are forced by the condition that all row sums and all column sums are zero.

$$F = left( begin{array}{cccc} & r & s & t \ & & u & v \ a & & & w \ b & c & & end{array} right).$$
As a result, $C^0 + F$ can be arranged to have all zeroes above the diagonal, then zeroes below a single layer alongside the main diagonal. The result is slightly better than what is called tridiagonal in that the entries above the diagonal are also 0.

http://en.wikipedia.org/wiki/Tridiagonal_matrix

We have arranged

$$C^0 + F = left( begin{array}{cccc} a_1 & & & \ r_1 & b_1 & & \ & s_1 & c_1 & \ & & t_1 & d_1 end{array} right) .$$

Now that we know that we can insist on this shape, we can just start out with this and a simple scheme involving your (ii) and (iii) defines the values for all the nonzero positions. Furthermore,
if in addition $x = y,$ then it follows from (ii) and (iii) that
$C^0 + F$ is actually diagonal. Done.

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