# Proof of a property of the inner rank of a matrix.

While reading an article about Sylvester matrix rank functions (generalization of the rank of a matrix for arbitrary rings) I needed to study the concept of inner rank, which is defined as follows:

Let $$R$$ be a ring and let $$A$$ be an $$n\times m$$ matrix over $$R$$. The inner rank of $$A$$, denoted as $$\rho(A)$$, is defined to be the least integer $$k$$ such that there exists two matrices $$B$$ and $$C$$, with dimensions $$n\times k$$ and $$k\times m$$ respectively, such that $$A=BC$$.

One of the properties the inner rank satisfies is the following:

$$\rho\left(\begin{matrix} A \\ B \end{matrix}\right)\geq\max\lbrace\rho(A),\rho(B)\rbrace$$ and $$\rho\left(\begin{matrix} A & C \end{matrix}\right)\geq\max\lbrace\rho(A),\rho(C)\rbrace$$ for any matrices $$A,B$$ and $$C$$ of proper size.

This was the only property I was not able to prove, since I don't know how to decompose a block matrix of this form into a product of two matrices, the other properties where easier since there was an obious decomposition of the matrix, but here either there is a more sophisticated argument or there is also an obvious decomposition that I can't see.

Any hint will be thanked.

For the first one: Let $$A\in R^{n_1\times m},B\in R^{n_2\times m},\rho=\rho\binom AB$$, which means there exist $$C\in R^{(n_1+n_2)\times \rho},D\in R^{\rho\times m}$$ s.t. $$\binom AB=CD$$. Now split $$C$$ into two matrizes $$C_1\in R^{n_1\times\rho},C_2\in R^{n_2\times\rho}$$ with $$C=\binom{C_1}{C_2}$$. Then it is easy to see that $$A=C_1D$$ and $$B=C_2D$$. These equations imply $$\rho(A)\leq\rho$$ and $$\rho(B)\leq\rho$$, and together $$\max\{\rho(A),\rho(B)\}\leq\rho=\rho\binom AB$$.
You can prove the second inequality similarly, you have to split $$D$$ into two parts instead of $$C$$.