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In Hoffman and Kunze's Linear Algebra, following proof is given for this corollary: Let A and B be m x n matrices over the field F. Then B is row-equivalent to A if and only if B=PA, where P is a product of m x m elementary matrices.

Proof. (1)Suppose B=PA where P= Es · · · E2E1 and the Ei are m x m elementary matrices. Then E1A is row-equivalent to A, and E2(E1A ) is row-equivalent to E1A . So E2E1A is row-equivalent to A ; and continuing in this way we see that (Es . . . E1)A is row-equivalent to A.

(2)Now suppose that B is row-equivalent to A. Let E1, E2,...,Es the elementary matrices corresponding to some sequence of elementary row operations which carries A into B. Then B = (Es . . . E1)A

Although I get the corollary intuitively, as each elementary matrix is equivalent to elementary row operation, and thus multiplication of matrix with A is equivalent to series of applications of such elementary row operations. But I don't get how it's being proved here.

  1. Particularly, how (1) and (2) are connected? Moreover, in (2) we are supposing what we had to actually prove, so how does it help? Also, it sees quite like re-statement of corollary itself.
  2. In general, what constitutes a mathematical proof? In other words, how do we judge whether it's sufficient and correct? I did go through few questions here, like this one, but how to validate this proof specifically (especially differentiating it with intuition or re-statement)
dheeraj suthar
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    In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$. – angryavian Dec 29 '18 at 20:02
  • @angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer? – dheeraj suthar Dec 29 '18 at 20:24
  • I copied my comment as an answer. – angryavian Dec 29 '18 at 20:40

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In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.

angryavian
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