Why is the rank of a matrix invariant under row operations?

Question

Prove that the rank of a matrix ($m\times n$) doesn't change if we apply row operations. For example if we multiply a row with a nonzero number $k$.

Answer 1

Hint: Let $A$ be a matrix whose columns are $v_1,\dots,v_n$. Applying a row operation to $A$ gives us the matrix $RA$ for some invertible matrix $R$. Note that the columns of $RA$ are $Rv_1,\dots,Rv_n$.

Show that a set of vectors $\{v_{k_1},\dots,v_{k_r}\}$ is linearly independent if and only if $\{Rv_{k_1},\dots,Rv_{k_r}\}$ is as well.

Answer 2

Considering a matrix $X$ as a linear map $A:\mathbb{R}^n \to \mathbb{R}^m$, the rank of $X$ is just the dimension of its image. A row operation takes $X$ to $AX$ for some invertible matrix $A$. (For example, multiplying a row of $A$ by a scalar $k\not =0$ corresponds to $A = \operatorname{diag}(1, \dots, 1, k, 1, \dots, 1)$.) Since $A$ is invertible, $\operatorname{rank} AX = \dim \operatorname{im}(AX) = \dim \operatorname{im}(X) = \operatorname{rank} X$.