4
$\begingroup$

I can prove that if A and B are row equivalent matrices, then the column vectors of A are linearly independent iff the column vectors of B are linearly independent.

However, does this result also hold for row vectors? That is, is it true that if A and B are row equivalent matrices, then the row vectors of A are linearly independent iff the row vectors of B are linearly independent? How exactly do you prove this?


I know how to prove that elementary row operations do not change the row space of a matrix, but I'm not sure if that's any use here.

Solution: Let $A$ be an $n$ by $m$ matrix. If we assume that the $n$ row vectors of $A$ are linearly independent, then they form a basis for the row space of $A$ since they span the row space by definition. So we know that the dimension of the rowspace of $A$ is $n$. Now $B$ also has $n$ row vectors, since elementary row operations do not change the rowspace of a matrix, then the row vectors of $B$ also span the same rowspace of $A$. Thus, the row vectors of $B$ also forms a basis for the common rowspace. Hence, the row vectors of $B$ are linearly independent.

To prove the converse, note that we can go from matrix $B$ to $A$ by using inverse elementary row operations, hence the same argument can be used.

$\endgroup$
1
$\begingroup$

Hint: If the rows of $A$ are linearly independent, they form a basis of the row space; the rows of $B$ span the same subspace.

$\endgroup$
  • $\begingroup$ Thanks for that! I think I got it, I've edited my initial post with my solution. $\endgroup$ – Trts Nov 23 '13 at 14:35
0
$\begingroup$

Another way to prove this can be- Let R1, R2, R3, .... Rn be the rows vectors of matrix A which are linearly independent. => x1R1+x2R2....+xnRn=0 => x1=x2=....=xn=0 Now suppose matrix B is obtained by performing row operation on A( Ri=aRj+Ri). Again considering the linear combination of row vectors of B y1R1+y2R2...+yiRi...+ynRn=0 y1R1+y2R2...+yi(aRj+Ri)...+ynRn=0 y1R1+y2R2+..(yj+a)Rj...+yiRi...+ynRn=0 Since R1, R2...Rn are linearly independent. =>y1=y2=....=yj+a=...yi=..yn=0

Hence row vectors of B are linearly independent. The reverse implication is true because A can be obtained by finite number of row operation of B.

$\endgroup$
  • $\begingroup$ Please use MathJax formatting $\endgroup$ – R_D Mar 20 '16 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.