1

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 $k$.

Asaf Karagila
  • 405,794
jstack
  • 11
  • Are you defining rank has the dimension of the column space? – Git Gud Feb 09 '15 at 16:58
  • 1
    What is your definition of rank? – Git Gud Feb 09 '15 at 17:04
  • @jstack: What is your definition of "the rank of vectors"? – hmakholm left over Monica Feb 09 '15 at 17:09
  • And I presume you mean "with a nonzero number $k$" in your example of a row operation; multiplying the first row of the 2x2 identity matrix by 0, for instance, changes the rank from 2 to 1. – John Hughes Feb 09 '15 at 17:10
  • @JohnHughes k is a nonzero number – jstack Feb 09 '15 at 17:14
  • @HenningMakholm rank is a number r where r is the number of linearly independent vectors and r+1 vectors are linearly dependent – jstack Feb 09 '15 at 17:15
  • 1
    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. – John Hughes Feb 09 '15 at 17:49
  • 1
    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. – Bart Feb 19 '15 at 14:50
  • Follow the advice here: http://math.stackexchange.com/help/deleting-account However, you can't take your content with you. That is here to stay. – Bart Feb 19 '15 at 14:52
  • Related: http://math.stackexchange.com/questions/20204/why-do-elementary-matrix-operations-not-affect-the-row-space-of-a-given-matrix – Martin Sleziak Feb 19 '15 at 15:02

2 Answers2

3

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.

Ben Grossmann
  • 234,171
  • 12
  • 184
  • 355
  • 1
    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 $R = \begin{bmatrix} 0 & 1 \ 1 & 0\end{bmatrix}$. – John Hughes Feb 09 '15 at 17:47
  • @jstack More than being represented by multiplication, it's exactly what they are, that's how you define elementary operation. – Git Gud Feb 09 '15 at 17:50
  • @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. – hmakholm left over Monica Feb 09 '15 at 18:26
0

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, $\operatorname{rank} AX = \dim \operatorname{im}(AX) = \dim \operatorname{im}(X) = \operatorname{rank} X$.

anomaly
  • 26,475