Finding rank of a matrix using elementary column operations

Question

I am learning about finding the rank of a matrix and in my textbook, I have only seen elementary row operations being used. For example, given any matrix, to find its rank, we need to simply just use elementary row operations to reduce the matrix into row echelon form and the rank is then just the number of non-zero rows. But, can we also use elementary column operations to find the rank? For example, say I have a $3 \times 4$ matrix:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$$ and after using elementary column operations, say I reduce it into:

$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Can I conclude that the rank is $3$ because there are three non-zero rows?

Answer 1

Every elementary row operation corresponds to left multiplication by an invertible matrix while every elementary column operation corresponds to right multiplication by an invertible matrix. Thus, if you started with a matrix $A$ and after $k$ elementary row operations and $\ell$ elementary column operations, you obtain a matrix $B$, then

$$B = E_1 \cdots E_k A F_1 \cdots F_\ell$$

where each $E_j$ is an elementary row operation and each $F_j$ is an elemtnatary column operation. Multiplying by an invertible matrix doesn't change the rank, so $B$ has the same rank as $A$.

In your particular example, this does indeed imply that the rank is $3$.

See this article for more detail.

Answer 2

If you've done the column reduction correctly, yes, the rank is 3. A matrix has the same rank as its transpose, so using column operations or row operations (or mixing the two) works either way, since you may view column operations as row operations on the transpose.