$A,B \in M_{3}(\mathbb{R})$ if a series of ``Elementary Row Operations" are performed on them how do we show they have the same rank? Where I believe rank is the dimension of the span of the rows?


By construction/definition, ERO's preserve the solution set of a system Ax=b (trivial to see for exchanges and for scaling, a little harder for shears ), so that, in particular, ERO's would preserve the set of solutions to Ax=0, i.e., the kernel, so, by rank-nullity, they would preserve the column space, and therefore its dimension, i.e., the rank of A.


Another way to think about the problem: elementary row operations are linear combinations of row vectors that do not change the span of the rows.

For instance, let

$$ A = \begin{pmatrix} a^1_1 & \cdots & a^1_n \\ \vdots & \ddots & \vdots \\ a^m_1 & \cdots & a^m_n \end{pmatrix} = \begin{pmatrix} a^1 \\ \vdots \\ a^m \end{pmatrix} $$

Then, if you interchange rows $i$ and $j$, obviously the span of the row vectors $[a^1 , \dots , a^m]$ doesn't change:

$$ [\dots , a^i , \dots , a^j, \dots] = [\dots , a^j , \dots , a^i, \dots] \ . $$

The same happens if you multiply one row by some $\lambda \neq 0$:

$$ [\dots , \lambda a^i , \dots ] = [\dots , a^i , \dots] $$

because every linear combination you can do with the vectors on the right is also a linear combination with the vectors on the left and vice versa:

$$ \dots + \alpha_i (\lambda a^i ) + \dots = \dots + (\alpha_i \lambda ) a^i + \dots $$


$$ \dots + \alpha_i a^i + \dots = \dots + \frac{\alpha_i}{\lambda} (\lambda a^i ) + \dots $$

Finally, if you add two rows, you also have

$$ [\dots , a^i , \dots , a^j, \dots ] = [\dots , a^i , \dots , a^i + a^j, \dots ] \ . $$

Exercise: why?

  • $\begingroup$ Just wondering, are the square braces standard notation for denoting the span of a set of vectors? I often see angle brackets instead, or even just $span\{\mathbf{a}_1,\cdots\}$ $\endgroup$
    – Josh Chen
    Jun 13 '11 at 9:21
  • $\begingroup$ Yes. I mean: it's standard for me, maybe. :-) Perhaps this is not standard in the USA or Anglo-Saxon countries? $\endgroup$ Jun 13 '11 at 9:28

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