# Span of a matrix

Let $A$ be an $m \times n$ matrix, and let $A'$ be the result of a sequence of elementary row operations on $A$. Prove that the rows of $A$ span the same space as the rows of $A'$.

I know how to prove that A and A' have both the same rank (rank (A)= dim(Vect(A)) ), but I am no sure how to prove that they both span the same space (ie they have the same Vect)

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It is enough to show this for a single elementary row operation, so let $r_1,\cdots,r_m$ be the rows of A.

1) If $A^{\prime}$ is formed by interchanging two rows of A, then A and $A^{\prime}$ have the same rows and therefore the same row space.

2) If $A^{\prime}$ is formed by changing $r_i$ to $cr_i$ for some constant $c$, then all the rows of $A^{\prime}$ are in the row space of A; so the row space of $A^{\prime}$ is contained in the row space of A.

3) If $A^{\prime}$ is formed by replacing $r_j$ by $cr_i+r_j$ where $i\not=j$, then all the rows of $A^{\prime}$ are in the row space of A; so the row space of $A^{\prime}$ is contained in the row space of A.

Since A is also formed by performing an elementary row operation on $A^{\prime}$, the row space of A is contained in the row space of $A^{\prime}$; so the two row spaces must be the same.

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Suppose that $v_1,\ldots, v_m$ are the rows of $A$, in $\Bbb R^n$. You have to show that $$\langle v_1,\ldots,v_i,\ldots,v_i+v_j,\ldots,v_m\rangle=\langle v_1,\ldots,v_i,\ldots,v_j ,\ldots,v_m\rangle$$

$$\langle v_1,\ldots,\lambda v_i, \ldots,v_m\rangle=\langle v_1,\ldots,v_i,\ldots,v_m\rangle \;\;,\;\lambda \neq 0$$

$$\langle v_1,\ldots,v_i, \ldots,v_j,\ldots,v_m\rangle=\langle v_1,\ldots,v_j,\ldots,v_i\ldots,v_m\rangle$$

Think why this is true by looking at linear combinations of the $v_i$. For example, if $v$ can be written as $v= \mu_1 v_1+\mu_2v_2+\cdots+\mu_iv_i+\cdots+\mu_jv_j+\cdots+\mu_mv_m$ then it can be written as $$v = {\mu _1}{v_1} + {\mu _2}{v_2} + \cdots + \left( {{\mu _i} - {\mu _j}} \right){v_i} + \cdots + {\mu _j}\left( {{v_j} + {v_i}} \right) + \cdots + {\mu _m}{v_m}$$ and conversely.

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And vice-versa for all! – Ted Shifrin Sep 16 '13 at 22:13
@TedShifrin Yes. – Pedro Tamaroff Sep 16 '13 at 22:13
Feedback on the downvote? – Pedro Tamaroff Sep 16 '13 at 22:29