3
$\begingroup$

I've just begun to re-learn linear algebra because is so important, the book that I chose is naturally the Hoffman's for a lot of reason.

Well, In the first chapter I'm stuck with the following, because is at the very beginning of the book so, concept as nullity and rank are not define yet, neither linear independence; using this it was my first idea, analyze by cases when the solution is trivial and when is not, but for that I need the concept of linearly independent. I'd to avoid a posterior knowledge.

Definition: Two systems of linear equation are said to be equivalent: if each equation in every system is a linear combination of the equations in the other system.

Prove that If two homogeneous systems of linear equation in two unknowns have the same solution, then they are equivalent.

Proof: Let us consider (1) $A_{i1}x_1+A_{i2}x_2=0$ ( $\,1\le i \le n$ ) and (2) $\,B_{k1}x_1+B_{k2}x_2=0\,$ ( $\,1\le k \le m$ ), and suppose that both homogeneous systems have the same solution. We have to show that (1) is equivalent to (2), i.e., each equation in (1) can be written as a linear combination of the equations in (2).

Suppose we select $m$-scalars $\,c_1,...,c_m \in \mathbb{F}$, multiply the $j^{th}$ equation in (2) by $c_j$ and then add. So, we have $\sum _{k=1}^m c_k B_{k1} x_1+ c_k B_{k2} x_2=0$.

And here is where I'm stuck, I know I have nothing. I think, maybe it's possible to use induction... Any suggestion? whatever would help

PS: I've already searched here and of course there are similar questions, but in every one uses a posterior knowledge and at this stage the book does not define rank, nullity, linear independence, etc. (I know is almost trivial with that, but I really like to know how the author thought in the answer).

$\endgroup$
  • $\begingroup$ How can you conclude that $A_{i1} = \sum_{k=1}^mc_kB_{k1}$? With this you conclude that $(1)$ consists of a single equation since $A_{11}=A_{21}=A_{31} =\cdots = \sum_{k=1}^mc_kB_{k1}$. In any case, it's not very clear what you've done. $\endgroup$ – EuYu Oct 24 '13 at 2:20
  • $\begingroup$ @ EuYu: OK, I understand the problem. But I'm not sure of what to do, because I have nothing more than linear combination and the meaning of equivalence (even in this, I'm not completely comfortable since there is not proof yet that is indeed an equivalence relations, this will be the next thing that I want to do). $\endgroup$ – Jose Antonio Oct 24 '13 at 2:30
  • $\begingroup$ I'm not sure why you're going out of your way to avoid concepts such as linear independence if you already know them. The way I see it, any way you solve the problem will involve using those concepts, the only difference being that you haven't formally defined and recognized what they are. $\endgroup$ – EuYu Oct 24 '13 at 2:36
  • $\begingroup$ I'd like to do it with what already the book shows at this stage; I think is possible but I cannot figure out how :( $\endgroup$ – Jose Antonio Oct 24 '13 at 2:42
1
$\begingroup$

Let's work out an easier case to get started. Suppose $ax+by=0$ and $cx+dy=0$. Think of these calculus III style. These equations say $\langle a,b \rangle \cdot \langle x,y \rangle =0$ and $\langle c,d \rangle \cdot \langle x,y \rangle =0$. This means $\langle a,b \rangle$, $\langle a,b \rangle$ are both perpendicular to $\langle x,y \rangle$. Therefore, it is geometrically clear that there exists $k \in \mathbb{R}$ such that $\langle a,b \rangle =k\langle c,d \rangle$ thus $ax_1+bx_2 = k(cx_1+dx_2)$ (which is what you want to show for two equations).

Generally, I do agree that time is better spent on structure here. Much more can be gained by using LI and matrix structure. That is after all the difference between now and 200 years ago. We have the benefit of not just thinking in equations.

$\endgroup$
  • $\begingroup$ Thanks is very intuitively. I think I can use induction on $n$. $\endgroup$ – Jose Antonio Oct 24 '13 at 2:47
  • $\begingroup$ Right, the point is the coefficients in the space appear as two dimensional vectors. That said, I doubt this is what the authors had in mind... $\endgroup$ – James S. Cook Oct 24 '13 at 3:10
  • $\begingroup$ Maybe sounds too idiot but when $A_{11} = \sum_{k=1}^mc_kB_{k1}$ and $A_{12} = \sum_{k=1}^mc_kB_{k2}$ is when we can use induction on $m$. If we let the base case when $m=1$ so the result is clearly trivial. Now we assume for some arbitrary $m\ge 1$ and to show that holds for $m+1$, just we get $c_{m+1}=0$ and the result follows by the inductive hypothesis. And we can use the same argument for every equation in (1) and interchanging the roles of (1) and (2) we conclude the proof. The only thing is assume that each $A_{i,j}, B_{i,j} \not=0$. But I'm not sure if is really sound. $\endgroup$ – Jose Antonio Oct 24 '13 at 3:30
  • $\begingroup$ @JoseAntonio I'm not clear on how you get $c_{m+1}=0$. That said, induction does make sense as a route to make this rigorous to arbitrary order. $\endgroup$ – James S. Cook Oct 24 '13 at 4:03
  • $\begingroup$ Sorry It have to said: we set $c_{m+1}=0$. Sorry to be too annoying but how do you argue rigorously that the case when $m=1$ holds? intuitively using the argument that you it's kinda obvious $\endgroup$ – Jose Antonio Oct 24 '13 at 4:40

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.