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I have shown that two of the three elementary operations will not change the image of the row space of the matrix: given a row vector $\vec{v}$, $k\vec{v}$ will span the same (scalar multiplication), and switching the order of the rows won't change anything because the span of the row vectors will be the same (vectors will be in a different order). However, how would one show that adding/subtracting vectors does not change the row space?

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up vote 4 down vote accepted

HINT $\ \ $ If $\rm\ v\in V\ $ then $\rm\ w\in V\ \iff\ w+c\:v\in V\:.\ $ Thus $\rm\ span(v,w)\ =\ span(v,w+c\:v)\:.$

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Suppose the rows of the matrix are $A=(v_1 ,v_2 ,...,v_n)$, then switching two rows, is just switching the order of the vectors, and as you said, it doesn't change the span of this set. The same is true for multiplying by a scalar ($\neq 0$) one of the rows.

Suppose you want to change $v_2$ to the vector $v_2 + \alpha v_1$ which is the last elementary operation. The span of $B=(v_1 ,v_2 + \alpha v_1 ,...,v_n)$ will contain $(v_2 + \alpha v_1) - \alpha v_1 = v_2$ so $A \subseteq span(B) \Rightarrow span(A) \subseteq span(B)$. In the same way $v_2 + \alpha v_1 \in span(A)$ so $B\subseteq span(A) \Rightarrow span(B) \subseteq span (A)$.

now you have that $span(B)=span(A)$ so adding one row (times a scalar) to another doesn't change the row space.

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Yet another attempt to convey the answer succinctly:

By definition, the rowspan is the collection of vectors you can get when you take (the sums of) all possible combinations of the row-vectors (together with scalar multiples).

When you add one vector to another, you may think of this as "doing some of the combining ahead of time". You're not going to change the collection of "all possible combinations of the rows" by doing some of the combining ahead of time.

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Consider two vectors, X and Y, in the plane, beginning at the origin and enclosing a nonzero angle. How many times you add Y to X, geometrically continue the vector X into the direction of Y , meaning: Z = X + k*Y - the angle between Y and Z diminuishes, but never vanishes.

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The main point is that elementary operations are invertible, so you can go back to the original set of vectors.

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You could go back to the original vectors of column-space through invertible elementary matrix operations, but column-space of the original matrix and of the rref matrix are not the same, while row-space of the original matrix and of the rref are (if that's the argument that you are making). – InterestedGuest Feb 3 '11 at 20:32
@InterestedQuest: I assume the poster meant row operations. – lhf Feb 4 '11 at 2:04
indeed, I am trying to say that invertibility of the row operations (and being able to go back to the original set of vectors) is not necessarily a guarantee of the same Image (it is in case of row-space, but not in case of column-space). – InterestedGuest Feb 4 '11 at 2:38

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