Showing the linear independence between two row equivalent matrices

Question

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.

score 2 · Accepted Answer · answered Nov 22 '13 at 14:55

2

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.

answered Nov 22 '13 at 14:55

egreg

238,574

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

score 0 · Answer 2 · answered Mar 20 '16 at 11:48

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.

Showing the linear independence between two row equivalent matrices

2 Answers2