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

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 $k$.

• Are you defining rank has the dimension of the column space? Feb 9 '15 at 16:58
• What is your definition of rank? Feb 9 '15 at 17:04
• @jstack: What is your definition of "the rank of vectors"? Feb 9 '15 at 17:09
• I've edited your question to include "nonzero". You, too, could have made this edit, and when people ask clarifying questions, it's a good idea to improve the question this way, saving later readers the trouble of trying to figure out what the real question is. Feb 9 '15 at 17:49
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– Bart
Feb 19 '15 at 14:50

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.
• Yes, and as @Om said, EVERY such row operation can be represented by multiplication by an invertible matrix. Exchanging the first and second row in a $2 \times 2$ matrix, for instance, is effected by $R = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$. Feb 9 '15 at 17:47
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, $\operatorname{rank} AX = \dim \operatorname{im}(AX) = \dim \operatorname{im}(X) = \operatorname{rank} X$.