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I have seen a proof that B is row equivalent to A iff exist invertible matrix C such that : B = CA , because C is elementary matrix , but i cant find the next step . can B "used" as the elementary matrix that lead from B to BA without changing the row space of A , so A and BA have same rowspace?

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  • $\begingroup$ This might help. $\endgroup$ – Itay4 Jul 18 '17 at 7:12
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if $B$ is an invertible matrix, its reduced row echelon form is the identity matrix.

That is I can write I can elementary matrices $E_i$ such that

$$E_m \ldots E_1 B = I$$

$$B=E_1^{-1}\ldots E_m^{-1}$$

Hence I can write $B$ as a product of elemntary matrices.

$$BA=E_1^{-1}\ldots E_m^{-1}A$$

Note that elementary matrices doesn't change the row space.

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Well, since $B=CA$, you see that rows of $B$ can be written as linear combinations of rows of $A$. (To be more explicit, write $$B=\begin{bmatrix}\mathbf{b_1\\b_2\\ \vdots \\ b_m}\end{bmatrix},\quad A=\begin{bmatrix}\mathbf{a_1\\a_2\\ \vdots, \\ a_m}\end{bmatrix},\quad C=\begin{bmatrix}c_{11} &\cdots & c_{1m}\\ \vdots & \ddots&\vdots\\ c_{m1} & \cdots & c_{mm} \end{bmatrix}$$ and perform partitioned matrix multiplication. Here letters in boldface denote row vectors comprising matrices.)

Likewise we note $A=C^{-1}B$.

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