1
$\begingroup$

1) Suppose the homogeneous system $Ax=0$ has non-trivial solution. Show that the linear system $Ax=b$ has either no solution or infinitely many solutions.

Let $A$ and $B$ be $m×n$ and $n×p$ matrices respectively.

2) Suppose the homogeneous linear system $Bx=0$ has infinitely many solutions. How many solutions does the system $ABx=0$ have?

3) Suppose $Bx=0$ has only the trivial solution. Can we tell how many solutions are there for $ABx=0$

My Attempts:

1) If $Ax=0$ has the non-trivial solution, we expect the Reduced Row Echelon Form of $A$ to contain at least 1 non-pivot column, this implies that $Ax=b$ has infinitely many solutions as the variable in the non-pivot column is set as the arbitrary parameter.

Assume that $b\in\mathbb R^n -{0}$, essentially telling us that no entry in $b$ is equal to 0. If $Ax=0$ has the non-trivial solution, we can also expect that the Reduced Row Echelon Form of $A$ to contain at least 1 zero row as well. However, this will lead to the linear system being inconsistent, meaning that $Ax=b$ will have no solution.

I am pretty much stuck at questions 2 and 3.... Is my proof for question 1 complete?

Any help is much appreciated, thanks!

$\endgroup$
1
$\begingroup$

Your proof of $1$ isn't particularly good, actually. You are jumping over a lot of points. For example, you start with

If $Ax=0$ has the non-trivial solution, we expect the Reduced Row Echelon Form of $A$ to contain at least 1 non-pivot column, this implies that $Ax=b$ has infinitely many solutions as the variable in the non-pivot column is set as the arbitrary parameter.

Which seems like a proof that the system $Ax=b$ will always have infinitely many solutions, and this is simply not true.

You then assume that no entry in $b$ is equal to $0$, which is a strong assumption which is also not needed. And your second paragraph directly contradicts the first paragraph, since the first "proves" that there (always) exist infinitely many solutions, and the second "proves" that there is no solution.


My advice is to ditch the row echelon form altogether and think about the linear operators alone.

You have a matrix $A$, and a vector $x_0\neq 0$ for which $Ax_0=0$.

Now, you want to prove the statement

The system $Ax=b$ either has no solution or it has infinitely many.

First of all, look at this from a logical standpoint. A system can have (1) no solution, (2) finitely many solutions or (3) infintelly many solutions.

Basically, all you have to prove is that (2) is impossible, and you don't have to show that (1) is possible. All you really have to do is to show that if (1) is not true, then (3) must be true. In other words, you have to prove the (equivalent) statement

If the system $Ax=b$ has some solution, then it has infintielly many solutions.

To do that, here's a hint:

  • Can you find infintelly many solutions to the equation $Ax=0$? Remembere, $A\lambda x = \lambda Ax$!
  • Once you are done there, if you know that $Ax=b$, what is $A(x+x_0)$?

For the second question, all you need to look at is this:

if $x$ is a solution to $Bx=0$, then what is $ABx$ equal to?


For the third question, after answering the second question, much more should be clear. If not, comment and I'll elaborate.

$\endgroup$
  • $\begingroup$ If $Ax=0$ has a non-trivial solution, then there exists $\lambda\in \mathbb R$ such that $A\lambda x=0$. Since, $A\lambda x=\lambda Ax$, $\lambda Ax=0$ is true for $\lambda\in \mathbb R$. Therefore $Ax=0$ has infinitely many solutions. Let $Ax_0=0$ and $Ax=b$, this implies that $A(x+x_0)=Ax+Ax_0=b.$ Thanks for your help, but I am not sure how am I supposed to continue on to show that $Ax=b$ has no finitely many solutions $\endgroup$ – Derp Apr 3 '17 at 14:11
  • $\begingroup$ As for Question 2: If $Bx=0$ has infinitely many solutions. Then there exists $\lambda\in \mathbb R$ such that $B\lambda x=0$. $B\lambda x=\lambda Bx=0.$ Therefore $ABx=\lambda ABx=0.$ which implies that $ABx=0$ has infinitely many solutions. As for Question 3: Is it true that if $x$ is the solution such that $Bx=0$, then $x$ has to be a zero matrix, which means that this system only has a trivial solution. This then implies that $ABx=0$ also has only the trivial solution, which is 0. Am I doing okay for Question 2 and 3 so far? Am I lacking anything important? $\endgroup$ – Derp Apr 3 '17 at 14:22
  • $\begingroup$ Any other hints that you can provide me in order to finish this please? $\endgroup$ – Derp Apr 4 '17 at 14:00

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.