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so I understand row space of a matrix X to mean the subspace of $R^n$ spanned by the row vectors of A

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Can the rows of B be written as a linear combination of A just because they are row-reduced from A?

Say we have matrix:

$$\begin{bmatrix} 1 & 2 \\ 3 & 8 \end{bmatrix}$$ $$ -> \begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}$$

How is the row-reduced matrix a linear combination of A?

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closed as unclear what you're asking by GNUSupporter 8964民主女神 地下教會, Yanior Weg, max_zorn, Shailesh, Shogun Jun 15 at 1:39

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In your example, $(1,2)=(1,2)$ and $(0,2)=(3,8)-3(1,2)$. (You can see why this is always true from the definition of elementary row operations.)

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Can the rows of B be written as a linear combination of A just because they are row-reduced from A? Yes

Here is a proof($RS(A)$ I meant row space of $A$, and $M_{m\times m}(\mathbb F)$ is a vector space over $\mathbb F$ consists of $m$-by-$m$ matrices with entries in $\mathbb F$, and you can think of it as a set of all $m$-by-$m$ matrices.):

Let $A\in M_{m\times n}(F),$ and $U=PA,$ where $U$ is row equivalent to $A$ and $P$ is invertible(i.e. $P$ is your row operations). Now $\forall y\in RS(A), \exists x\in M_{1\times m}(F),$

$$y=xA=x(P^{-1}P)A=(xP^{-1})PA=(xP^{-1})U\implies y\in RS(U),$$

The other way $RS(U)\subseteq RS(A)$ should be easy you should try yourself and get the conclusion $RS(A)=RS(U),$ which may solve your confusion.

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  • $\begingroup$ What is F? What is $M_{mxn}(F)$ $\endgroup$ – Jwan622 Sep 11 '18 at 13:55
  • $\begingroup$ @Jwan622: Apologies, $F$ means field, in your case it is $\mathbb R$. And $M_{m\times m}(\mathbb R)$ means a vector space of $m$-by-$m$ matrices over $\mathbb R$. You can think $F$ means the scalar type when you do scalar-multiplication to a matrix. $\endgroup$ – Ning Wang Sep 11 '18 at 14:30

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