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.


1 Answer 1


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 $C$.

  • $\begingroup$ Thanks, it was easier than I expected. $\endgroup$
    – Marcos
    Feb 13, 2022 at 13:43

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