I have two proofs I do not know how to start:

Q1: Prove that if A is row-equivalent to B and B is row-equivalent to C, then A is row-equivalent to C.

Q2: Let A be a nonsingular matrix. Prove that if B is row-equivalent to A, then B is also nonsingular.

Q1 comes with some advice:

"Getting started: To prove that A is row-equivalent to C, you have to find elementary matrices $$E_1E_2… E_k$$ such that A = $$E_k… E_2E_1C$$.

I. Begin by observing that A is row-equivalent to B and B is row equivalent to C.

II. This means that there exist elementary matrices $$F_1F_2… F_n$$ and $$G_1 G_2… G_m$$ such that A = $$F_nF_{n-1}… F_1B$$ and B = $$G_mG_{m-1}… G_1C$$.

III. Combine the matrix equations from step II."

I am unsure what row equivalence means and have no idea how A having an inverse makes B have an inverse if it is row equivalent; I've always froze up at proof questions for some reason so please do not be afraid to be verbose; I'd like to be able to explain and understand this stuff someday such that I will be able to write proofs properly.

  • $\begingroup$ for the first one, Do I literally just make $$A = F_n...F_2F_1B$$ into $$A = F_n...F_2F_1 G_m...G_2G_1C$$? Wouldn't the elementary matrices being multiplied to themselves once more do something different from what we might predict to happen? Also row-equivalence just means that the matrices have the same numbers in the same places instead of meaning just one row matches right? $\endgroup$
    – Charlatan
    Sep 18 '18 at 6:15
  • $\begingroup$ Ok I figured out row equivalence means more that 2 matrices share the same values and does not necessary look the same but give the same information that could be shown in different ways. Still unsure on how step 3 wants me to figure out A = C if A = B and B = C by elementary operations. $\endgroup$
    – Charlatan
    Sep 18 '18 at 6:38
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    $\begingroup$ As you suggested in your first comment, you simply have $A=FGC$ which implies that $A$ is row equivalent to $C$ $\endgroup$
    – user418131
    Sep 18 '18 at 8:49
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    $\begingroup$ It simply means that by performing elementary row operations, we can convert one to another $\endgroup$
    – user418131
    Sep 18 '18 at 9:12
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    $\begingroup$ That makes things much easier to understand; thank you @AnotherJohnDoe $\endgroup$
    – Charlatan
    Sep 19 '18 at 23:58




This shows that matrix A is equal to matrix C as B is equal to C through elementary row operations.

As for proving that if A is nonsingular, B is also nonsingular if row-equivalent to A...

B ~ A

$$T_n...T_1B = A$$ $$T^{-1}(T_n...T_1B) = T^{-1}(A)$$ $$B = T_n^{-1}...T_1^{-1}A=E_1...E_nA$$

Should show that B is also row-equivalent.


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