We begin by considering that A and B are two row equivalent matrices, which means that there exists an invertible matrix P such that: B=PA .

A1= all the rows in A.

B1 = all the rows in B.

Let's also assume we know that every row in B is a linear combination of the rows of A if the two matrices are row equivalent (This part can be proven with some algebra).

Therefore every vector in B1 is a member of Span(A1). Span(B1) is the smallest vector space that contains B1. Therefore Span(B1) is contained or equal to Span(A1).

We can also say that A=(P^-1)B, so we can do the same process and say that Span(A1) is contained or equal to Span(B1). So Span(A1)=Span(B1). which means that the row space of A is equal to the row space of B.

Is this correct? Any suggestions to improve? Thanks!

(Sorry for not using MathJax, its the first proof I wrote in this site)


Looks ok to me, although I believe that in your proof you are using "B1" for the set you previously named "A2".

  • $\begingroup$ Thanks, I will change it to B1 $\endgroup$ – ExtraFlash Nov 12 '18 at 7:15

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