A real symmetric matrix decomposed by diagonal entries Suppose you have a real symmetric $n\times n$ matrix $A$, with entries $a_{ij}$. Using diagonal terms $a_{ii}$ let each $a_{ij}$ terms be decomposed into $a_{ij} = b_{ij,i} + b_{ij,j}$, with ratio constraint $b_{ij,i}/b_{ij,j} = a_{ii}/a_{jj}$.
Let $b_i = \sum_j b_{ij,i} + \sum_j b_{ji,i} - b_{ii,i}$. Would $b_i/b_j = a_{ii}/a_{jj}$?
 A: I'm assuming throughout that the diagonal terms are non-zero. We have $$\frac{b_{ji,j}}{b_{ji,i}} = \frac{a_{jj}}{a_{ii}} \implies b_{ji,j} = \frac{b_{ji,i}a_{jj}}{a_{ii}}$$ Also,
$$a_{ij} = a_{ji} = b_{ji,j} + b_{ji,i}$$
Substituting $b_{ji,j}$ in the above equation,
$$b_{ji,i}\bigg(1+\frac{a_{jj}}{a_{ii}}\bigg) = a_{ij} \implies b_{ji,i} = \frac{a_{ij}a_{ii}}{a_{ii}+a_{jj}} = \frac{b_{ij,i}a_{ii}+b_{ij,j}a_{ii}}{a_{ii}+a_{jj}}$$
Subtituting this in the expression for $b_i$,
$$b_i = \sum_j\bigg[\frac{b_{ij,i}a_{ii}+b_{ij,i}a_{jj}+b_{ij,i}a_{ii}+b_{ij,j}a_{ii}}{a_{ii}+a_{jj}}\bigg] - b_{ii,i} = \sum_j\bigg[\frac{2b_{ij,i}a_{ii}+2b_{ij,i}a_{jj}}{a_{ii}+a_{jj}}\bigg] - b_{ii,i}$$
where we used the fact that $b_{ij,i}a_{jj}=b_{ij,j}a_{ii}$, and we get
$$b_i = 2\sum_jb_{ij,i} - a_{ii}/2$$
Similarly $b_j = 2\sum_ib_{ji,j} - a_{jj}/2$. So effectively we must prove that $b_ja_{ii} = b_ia_{jj}$, or
$$2a_{ii}\sum_kb_{jk,j} - a_{ii}a_{jj}/2 = 2a_{jj}\sum_kb_{ik,i} - a_{jj}a_{ii}/2 \implies a_{ii}\sum_kb_{jk,j} = a_{jj}\sum_kb_{ik,i}$$
You can also get the following expression for $b_{ik,i}$:
$$b_{ik,i} = \frac{a_{ik}a_{ii}}{a_{ii}+a_{kk}}$$
and similarly for $b_{jk,j}$. Assuming all diagonal terms equal $1$, we would have to prove that $\sum_k(a_{ik}-a_{jk}) = 0$, which need not be true.
