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Let $\text{Row}(A)$ denote the row space of a matrix $A$, and $\text{Col}(A)$ the column space.

Let $\text{rref}(A)$ denote the "reduced-row echelon form" of $A$.

Why is it the case that

$$\text{Row}(A)=\text{Row}(\text{rref}(A))$$

but

$$\text{Col}(A)\neq\text{Col}(\text{rref}(A))\,?$$

symplectomorphic
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Yilun Liu
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  • I'm not sure what "rref" means (row-reduced echelon form, perhaps?) but the corresponding thing on the right hand side should probably by "cref" (column-reduced echelon form), which would then give you a true statement. – John Hughes Oct 10 '18 at 18:01
  • It is quite unclear what you are asking, but here is my guess. Do you want to know why the row rank of $A$ equals the row rank of the row-reduced echelon form of $A$, but the column rank of $A$ does not equal the column rank of the row-reduced echelon form of $A$? –  Oct 10 '18 at 18:02
  • rref is a reasonably standard abbreviation for reduced row-echelon form. Although I think it should be rcef on the right – HackerBoss Oct 10 '18 at 18:02

1 Answers1

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Row operations preserve the row space.

Therefore, if two matrices $A$ and $B$ are row-equivalent, their row spaces are equal. Since $\text{rref}(A)$ is row-equivalent to $A$, we have $\text{Row}(A)=\text{Row}(\text{rref}(A))$.

But row operations do not preserve the column space.

For a counterexample, take $[1, 2]^T$. This column is row equivalent to $[1,0]^T$. But the column spaces of these matrices are not equal.

Thus, just because $A$ and $B$ are row-equivalent, it does not follow that $\text{Col}(A)=\text{Col}(B)$.

symplectomorphic
  • 19,077
  • To expand on the first part, the point is that the row operations (exchange, multiply by non-zero scalar, and add multiple of one row to another) produce a new row which is easily seen to be in the span of the original rows. Thus the new row space is a subspace of the original. Since each of the row operations is reversible as a row operation, the same reasoning implies the old row space is a subspace of the new one. So the rows of both original and modified matrix span the same subspace. – Ned Oct 10 '18 at 18:14
  • @Ned: My answer doesn't need expansion. You're just literally repeating what the answer in the first link says. – symplectomorphic Oct 10 '18 at 18:15
  • Sorry I didn't click the link, I'm a little lazy, I guess. – Ned Oct 10 '18 at 18:18