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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 $k$.

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  • $\begingroup$ Are you defining rank has the dimension of the column space? $\endgroup$
    – Git Gud
    Feb 9 '15 at 16:58
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    $\begingroup$ What is your definition of rank? $\endgroup$
    – Git Gud
    Feb 9 '15 at 17:04
  • $\begingroup$ @jstack: What is your definition of "the rank of vectors"? $\endgroup$ Feb 9 '15 at 17:09
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    $\begingroup$ 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. $\endgroup$ Feb 9 '15 at 17:49
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    $\begingroup$ Please don't vandalize your questions. That is not appropriate and will be reversed. If there is any relevant reason to not want your content associated to you, you can ask to be disassociated from it by flagging the post for moderator attention. $\endgroup$
    – Bart
    Feb 19 '15 at 14:50
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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.

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    $\begingroup$ 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 $R = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$. $\endgroup$ Feb 9 '15 at 17:47
  • $\begingroup$ @jstack More than being represented by multiplication, it's exactly what they are, that's how you define elementary operation. $\endgroup$
    – Git Gud
    Feb 9 '15 at 17:50
  • $\begingroup$ @GitGud: Some texts define it that way. Other texts describe elementary operations as something concrete you do to a matrix, and merely represent these operations with elementary matrices. $\endgroup$ Feb 9 '15 at 18:26
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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, $\operatorname{rank} AX = \dim \operatorname{im}(AX) = \dim \operatorname{im}(X) = \operatorname{rank} X$.

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