1
$\begingroup$

Assume that $F$ is a field. Consider a matrix like $A \in M_{m,n}(F)$.

If we look at each row of $A$, as a member of $F^n$, the subspace produced by rows of $A$ is called the row-space of $A$.

a) Prove that elementary row operations on $A$ don't change the row-space of $A$. Then, Conclude that foreach matrix like $B$, if $A$,$B$ are row equivalent, then they have the same row-space.

b) Prove that $A \in M_{n}(F)$ is invertible iff the row-space of it is $F^n.$

Note : I'm confused with the definitions ... I have no idea how to solve this problem.

$\endgroup$
  • $\begingroup$ Are you comfortable with expressing row operations by matrix multiplication? $\endgroup$ – Ben Grossmann Nov 3 '16 at 20:47
  • $\begingroup$ @omnomnomnom yes why not :) $\endgroup$ – Arman Malekzadeh Nov 3 '16 at 21:03
  • $\begingroup$ Please pick a title that makes sense! $\endgroup$ – Mariano Suárez-Álvarez Nov 4 '16 at 2:41
  • $\begingroup$ @MarianoSuarez-Alvarez you propose a title :) i'll change mine :) $\endgroup$ – Arman Malekzadeh Nov 4 '16 at 6:37
  • $\begingroup$ @IStillHaveHope, no, it does not work like that. You have to come up with a reasonable title: it is part of the (rather little) effort that you are expected to put in the site in order to get others to help you. $\endgroup$ – Mariano Suárez-Álvarez Nov 4 '16 at 6:43
1
$\begingroup$

Let $v_1,\ldots,v_n$ denote the rows of $A$, and let $V$ be the row-space of $A$. In other words, $$ V=\text{span}\,\{v_1,\ldots,v_n\}. $$ Elementary operations could consist of a permutation of rows, which amount to permuting some $v_j$ and $v_k$ above; such operation will not change the span of $v_1,\ldots,v_n$. The same happens with multiplying by a nonzero number: $v_1,\ldots,v_n$, and $v_1,\ldots,\lambda v_k,\ldots, v_n$ span the same subspace.

The last kind of elementary operation consists of replacing $v_k$ with $v_k+\lambda v_j$. In this case, we can write $$ \alpha_1v_1+\cdots+\alpha_nv_n=\alpha_1v_1+\cdots+\alpha_{k-1}v_{k-1}+\alpha_k(v_k+\lambda v_j)+\alpha_{k+1}v_{k+1}+\cdots (\alpha_j-\alpha_k\lambda)v_j+\cdots+\alpha_nv_n. $$ So every linear combination of $v_1,\ldots,v_n$ is also a linear combination of $v_1,\ldots,v_n$ with $v_k$ replaced by $v_k+\lambda v_j$.

In summary, after doing any elementary operation to $v_1,\ldots,v_n$, the span doesn't change. It follows directly that if $A$ and $B$ are row equivalent, since the rows of $B$ can be obtained by elementary operations from the rows of $A$, the spans of their rows are equal.

If $A$ is invertible, then it is row equivalent to $I$, and so its row space is $F^n$. Conversely, if the row space of $A$ is $F^n$, then by writing each of $v_1,\ldots,v_n$ in terms of the canonical basis, we get a recipe to go from $I$ to $A$ by elementary operations, and so $A$ is invertible.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ i'm not familiar with "canonical basis". would you please explain it? or maybe use a more simple word? $\endgroup$ – Arman Malekzadeh Nov 4 '16 at 7:09

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.