Alternating row and column operations for block Gaussian elimination to determine rank.

Question

I am trying to determine the rank of a 6x6 symbolic matrix. The matrix can be represented as follows:

$$ M = \begin{bmatrix} A_{ 3 \times 3} & R_{ 3 \times 3}A_{ 3 \times 3}\\ B_{ 3 \times 3} & E_{ 3 \times 3}\\ \end{bmatrix} $$

Each element of this matrix is a 3x3 block. It is relatively easy to determine the rank of individual blocks (analytically). I am using block Gaussian Elimination. I proceed as follows:

1. Column 2 = Column 2 - $R_{3 \times 3}$ Column 1

$$ \begin{bmatrix} A_{ 3 \times 3} & 0_{ 3 \times 3}\\ B_{ 3 \times 3} & E_{ 3 \times 3} - R_{ 3 \times 3} B_{ 3 \times 3}\\ \end{bmatrix} $$

2. I prove (analytically) that $A_{3 \times 3}$ is always full rank and then proceed to eliminate block $ B_{3 \times 3}$.

$$ \begin{bmatrix} A_{ 3 \times 3} & 0_{ 3 \times 3}\\ 0_{ 3 \times 3} & E_{3 \times 3} - R_{ 3 \times 3}B_{ 3 \times 3}\\ \end{bmatrix} $$

3. Next I prove that $E_{3 \times 3} - R_{ 3 \times 3}B_{3 \times 3}$ is full rank (analytically).

From above steps I conclude that matrix $M$ is full rank.

My questions/conjectures are:

1. Is it valid to alternate between row and column operations since I am only interested in the rank of the matrix ?
2. In step 1, where I per-multiply by $R_{3 \times 3}$, it is necessary that the matrix $R_{3 \times 3}$ be full-rank. If not, then I am altering the rank of the column 2.

Answer 1

As row rank and column rank are equal you may use steps which leave either row rank unchanged or column rank unchanged: they are the same thing.

In step (1) you are postmultiplying the given matrix by the non-singular $\begin{bmatrix} I & -R\\O & I\end{bmatrix}$ and that does not change the rank. (Note that this depends in no way on the rank of $R$, we could even have $R=O$. It may change the rank of the top right block, but it does not change the rank of $M$ and that's all that matters.)

In short, what you are doing is correct.