Question

Prove that a system of linear equations cannot have exactly two solutions.

Answer 1

Can you tell whether a line may intersect at more than one point?

Answer 2

Hmm, can we prove it in a more formal way? Intuitively, I know we can only have exactly one solution, no solutions, or infinitely many solution. Though, we need a more formal proof.

Answer 3

I will try first and recite if I understand this problem correctly, the way I read it they say that the system of equation given as \[\large Ax=b \] has one and only one solution lets call it \(x_0\) right?

Answer 4

Because the condition that a system has only one solution will give you that the Kern is zero, otherwise the Kernel/Nullspace will have a dimension greater or equal than one and you can have as many solutions as you want.

Answer 5

Spacelimbus, we should prove it more generally.

Answer 6

RolyPoly, I have got a more general solution to your problem.

Answer 7

@silversurfer , this would be general. You can show uniqueness of a solution space given that:
\[\large Ax_0=b \] let \(Ay_0=b \) be another solution, show that \(y_0=x_0\) will be as general as it can get. \[A(x_0-y_0)=b-b=0 \implies x_0-y_0 \in \ker A = \lbrace 0 \rbrace \implies x_0 = y_0 \]

Answer 8

How do you assume that it passes through the origin@Spacelimbus?

Answer 9

General solution for a system of linear equations:
\[\large x= x_p + \ker A \]
where \(x_p\) is a particular solution to the system and the Kernel is the vector space (subspace) that has the property that \(Ax=0\). If the Kernel is zero you're left with the statement that \(x=x_p\), which is a single vector.

Answer 10

Isn't that a too high level to solve this? I mean,we could prove easily that the solutions are equal to each other and hence unique using the standard ax+by+c=0 form.

Answer 11

@RolyPoly works in \(n\) dimensional Euclidean Space. The Standardspace \(\mathbb{R}^n\). You would be right for the \(ax+by+c=0\) plane case, but in \(\mathbb{R}^n\) you're dealing with hyperplanes rather than with two dimensional planes in 3 dimensional space.

However, please feel free to share your solution here, I hope I did not leave the impression that I meant to silence you, I am always interested in elementary proofs :-)

Maybe @RolyPoly will tell us a bit more about the problem and his perspective on it, it could be possible that I misunderstand it from the beginning, because of the "exactly two", exactly two can usually be disproved by showing uniqueness of solution.

Answer 12

Hmm, yes, I am working in n dimensional Euclidean Space. Showing the uniqueness of solution is, I think, a way to disprove that a system of linear equations cannot have exactly two solutions.

Answer 13

This is a bit off-topic, but if we can prove the statement in 2-D space, can we generalize the result in n dimensional Euclidean Space? I remember in some other lesson, I heard my professor saying the phrase "without the loss of generality, ...". Can this "principle" be applied here? If not, why?

Answer 14

Without the loss of generality is a bit of a weird concept.

Basically it means the following: You use your assumptions and narrow them to some special case, a case which might be easier to prove, it then has to imply that the proof for all other cases can be easily adapted and applied to all other cases, in an equivalent manner.

Sometimes without loss of generality can also mean that you pick a special case, and then all the remaining cases are "trivial" to show. However, as you can maybe see already, this is a bit of a philosophical topic. Because what might seem trivial to some, seems complicated to others.

The way I think about it for this problem. Showing that a 2x2 well defined system of equations has exactly one and one solution only (for that case), would not imply to me that this is general for the n x n or n x m case, because it is hard to visualize hyperplanes. Their geometry is very weird.

So, here is how I would have done it:

- Show that the general solution to \(Ax=b\) is of the form \(x=x_p+ \ker A\) where \(x_p\) is a particular solution and \(\ker A\) is the Nullspace.

As soon as your Nullspace has a dimension greater or equal than 1, you're granted to have infinite many solutions to your equation \(Ax=b\), therefore not exactly two.

Next set your Kernel to have dimension zero, now your solution becomes to \(x=x_p\), which is "a particular solution". Show that there is not another particular solution, meaning show that this \(x_p\) is unique.

Answer 15

Hmm, you are forcing me to learn ahead so as to understand your idea. I haven't learnt null space, and that ker yet. It will take some time to understand it.
Just another question, you have shown that in the form of the general solution, we can get infinitely many solutions to the system when ker A ≥ 1 or a unique solution to the system when ker A = 0, but in what case will we get no solutions?

Answer 16

No solutions means \( \not \exists x_p \text{ such that } Ax_p=b \)
the null space can still hav ea dimension because \(Ax=0\) always has the trivial solution solution \(x=0\), but \(Ax=b\) requires at least one particular solution to be solvable, there could of course be more, dictated by the Kernel.

I am sorry, it's often difficult to know which level a class is at at linear algebra. The way I have shown above of course requires the introduction of the Kernel (Synonym is Nullspace) and also dimensions (therefore Basis and discussions of subspaces)

In a more elementary way I would have to think myself a bit more too. Maybe I can come up with another idea. One that is floating around in my head is using the idea that \(Ax=b\) is a monomorphism, which is basically the same idea as above, but yet, not elementary.

Answer 17

It's alright. I'll try to do the proof and type it here by tomorrow. Thanks for your help so far!

Answer 18

use matrix

Answer 19

^ ya allah :p try elaborating you :D

Answer 20

ur solving it nw ?

Answer 21

@ikram002p May I know how you use matrix to prove it?

Answer 22

ill show it ltr , g2g nw
but u can try its not hard

Answer 23

Suppose \(x_1\) is any particular solution to the system \(Ax=b\) of linear equations, \(x_0\) be some solution to the homogeneous equation \(Ax=0\) such that for every solution \(x_2\) to \(Ax=b\), it has a form of \(x_2=x_0+x_1\).

Suppose \(x_2\) is also a solution to \(Ax=b\). Then, \(Ax_2=b\).

Write \(x_0= x_2-x_1\), then we have
\(Ax_0 = A(x_2-x_1) = A(x_2) - A(x_1) = b-b=0\).

So, \(x_0\) is a solution to the homogeneous system \(Ax=0\), and we have \(x_2=x_0+x_1\)

It's a bit odd. :\