# Homework Help: Homogenous system

1. Sep 27, 2009

### shiri

If a homogenous system Ax = 0 has infinitely many solutions, then for a non-zero vector b, the associated system Ax = b ____ have _______

In my assignment, the answer I wrote on this question is must have many solutions. However, what I got is wrong.

Can anybody tell me why it is wrong?

Last edited: Sep 27, 2009
2. Sep 27, 2009

### aPhilosopher

if Ay = b, then what can you say about any vector of the form x + y where Ax = 0?

3. Sep 27, 2009

### shiri

Well I am assuming there has to be many solutions since Ax=0 gives infinite solutions.

Plus, no solution or/and one solution sounds less appropriate than many solutions at the moment I answer the question.

4. Sep 27, 2009

### aPhilosopher

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Also, I might be being pedantic but I would say many is a broader term than infinite. I think that I've seen books using 'many' though so if your book or teacher does, just ignore me.

5. Sep 27, 2009

### shiri

Well I don't understand why many solutions is wrong since this tell me how many variables and how many unknowns.

6. Sep 27, 2009

### aPhilosopher

Must is surely wrong as the example shows. The homogeneous equation has infinite solutions. Can you find a solution to the non-homogeneous equation?

Like I said, I was probably just being pedantic with the whole many/infinite thing.

7. Sep 27, 2009

### shiri

Teacher gave me these choices:

A. may have exactly one solution
B. must have many solutions
C. must have either one solution or no solution
D. may have no solution
E. need not satisfy any of the above

8. Sep 27, 2009

### aPhilosopher

Well, you shouldn't have chosen B. Work my example to find out why.

9. Sep 27, 2009

### shiri

I would say many solutions and no solutions. B and D

10. Sep 27, 2009

### aPhilosopher

B is wrong in general. Do you see why? If you want, we can talk about when it's right.

11. Sep 27, 2009

### shiri

So B is wrong in any possible answers?

12. Sep 27, 2009

### aPhilosopher

It's wrong without conditions that weren't assumed in the question. Have you studied dimension yet?

The key thing here is the use of the word "must". I've given you an example already that doesn't so think about it until you realize that it doesn't "must" have infinite solutions.

13. Sep 27, 2009

### shiri

So the correct answer is D only? May have no solutions?

14. Sep 28, 2009

### shiri

the answer is C? is it?

15. Sep 28, 2009

### aPhilosopher

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

16. Sep 28, 2009

### shiri

I see x has a solution and y has no solution.

Am I right?

17. Sep 28, 2009

### aPhilosopher

The second one has no solution, you are correct. How many solutions does the first one have again?

18. Sep 28, 2009

### shiri

the first one has one solution?

19. Sep 28, 2009

### aPhilosopher

No. (1 0) is a solution. (1 1) is a solution. (1 45/77) is also a solution.

How many solutions does $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ have? You can take any one of them, add it to (1 0) and get a solution to

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Right?

20. Sep 28, 2009

### shiri

so are you saying that there must have either one solution or no solution?

21. Sep 28, 2009

### aPhilosopher

I have to go to work now. Maybe somebody else can help you.

22. Sep 28, 2009

### aPhilosopher

Take an homogeneous system, Ax = 0. For example in the following equation,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

The matrix $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ would be called A.

The vector $$\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}$$ would be called x.

We want to find all x1, x2 such that Ax = 0. Towards that end, expand it out into

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.

For any real number x2 that we choose, $$x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

This means that $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ so that the value of x2 doesn't affect the value of Ax at all.

So to solve the above equation, we need to solve 1x1 = 0. But whenever we have two numbers a and b such that ab = 0, we know that either a or b must equal 0. Because we know that 1 is not equal to 0, we can conclude that x1 must be.

This means that $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ when $$\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ x_{2} \end{bmatrix}$$ where x2 is any real number that we choose.

We have shown that any solution of Ax = 0 must be of the form $$\begin{bmatrix} 0 \\ x_{2} \end{bmatrix}$$.

In other words, there is one solution for every real number x2.

Non-Homogeneous System

Let's now consider a non-homogeneous system Ax = r where r is a non-zero vector. Let's take our example as,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ r_{1} \end{bmatrix}$$

Again, we expand this out into,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.

Again, $$x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.

So $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$.

Now, 0x1 = 0. This has the consequence that we can only solve the equation if r2 = 0 because that is all we are ever going to get from A. We must also have 1x1 = r1. So we can only solve the equation if

$$r = \begin{bmatrix} r_{1} \\ 0 \end{bmatrix}$$

Now, as you know, A(v1 + v2) = Av1 + Av2 so if Av1 = r and Av2 = 0 then A(v1 + v2) = Av1 + Av2 = r + 0 = r

This means that the solution set is all vectors of the form

$$\begin{bmatrix} r_{1} \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ x_{2} \end{bmatrix}$$

where r1 is determined by r and x2 is free.

In other words,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ r_{2} \end{bmatrix}$$

has solutions only if r2 = 0. In that case, we have

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} r_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ 0 \end{bmatrix}$$ for all real numbers, x2.

Four more good matrices to solve and really look at would be $$\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} , \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

You probably want to use row reduction or some other method for more complicated systems but you should look at those three like we've looked at the one above.