Summing rows and columns for a rank-1 matrix approximation Let $M$ be a matrix, and form a column vector $R$ by summing the rows of $M$, and a row vector $C$ by summing the columns of $M$. We think of the matrix $RC / \Sigma M$ as a rank-1 approximation to $M$ (here $\Sigma M$ denotes the sum of all elements of $M$. You can assume it's nonzero).
If $M$ is rank-1, $RC / \Sigma M = M$.
Is this approximation optimal among rank-1 matrices with respect to some matrix norm? What exactly is the matrix $RC / \Sigma M$?
Statistical motivation:
If $M$ represents a joint probability distribution $P(X, Y)$, then $R$ represents the marginal distribution $P(X)$, $C$ represents the marginal distribution $P(Y)$, and $RC$ represents a new probability distribution on the same space where now the coordinates are independent.
 A: It's not a norm, but this is the optimal rank-1 approximation using the Kullback-Leibler divergence.
Let $M = P(X, Y)$ represent a joint probability distribution. Let $RC$ be the product of marginals. Then it's one definition of mutual information that
$$D_{KL}(P(X, Y), RC) = I(X;Y)$$
And in fact this is optimal for rank-1 matrices: for any rank-1 matrix $r \otimes c$ where both $r$ and $c$ are probability distributions, 
\begin{align*}
D_{KL}(P(X, Y), r \otimes c) &= \displaystyle\sum_{x,y} P(X = x, Y = y) \log \left(P(X = x, Y = y) \over r_x c_y\right)\\
&= \displaystyle\sum_{x,y} P(X = x, Y = y) \log P(X = x, Y = y) - \displaystyle\sum_{x,y} P(X = x, Y = y) \log r_x - \displaystyle\sum_{x,y} P(X = x, Y = y) \log c_y\\
&= H(X, Y) - \displaystyle\sum_{x} P(X = x) \log r_x - \displaystyle\sum_{y} P(Y = y) \log c_y
\end{align*}
With the Gibb's Inequality (the inequality proving KL divergence is nonnegative) this is at least
$$H(X, Y) - H(X) - H(Y) = I(X;Y)$$
If you don't restrict yourself to probability distributions, you can send $D_{KL}$ to $-\infty$ by taking huge elements of $r$ and $c$.
If $M$ has all entries nonnegative (at least one positive), $M / \Sigma M$ will be a probability distribution, and the above will apply. 
