# Vector Space

1. Oct 28, 2014

### Jaglowsd

• OP warned about not using the homework template
Let B be a non-zero mx1 matrix, and let A be an mxn matrix. Show that the set of solutions to the system AX=B is not a vector space.

I am thinking that I need to show that the solution is not consistent. In order to do so would I need to show that B is not in the column space of A?

2. Oct 28, 2014

### Dick

It's a lot simpler than that. A vector space is supposed to contain a zero vector.

3. Oct 28, 2014

### Jaglowsd

So unless B is a zero vector than in any case AX=B can not be a vector space? It must be AX=0, or the null space?

4. Oct 28, 2014

### Dick

That's kind of confusing, but if X=0 doesn't solve your system then the set of solutions isn't a vector space.

5. Oct 28, 2014

### Jaglowsd

Sorry, I am still trying to wrap my head around vector spaces. Could you elaborate why X=0 must solve the system to be a vector space.

6. Oct 28, 2014

### Dick

When does a set of vectors constitute a vector space? You'll need to look up the definition if you can't recall. The answer is there.

7. Oct 28, 2014

### HallsofIvy

Staff Emeritus
If x and y both satisfy Ax= B, Ay= B, then A(x+ y)= Ax+ Ay= B+ B= 2B. Unless B= 0, x+ y does NOT satisfy the equation A(x+ y)= B so is NOT in this set. The set is not closed under addition, so is not a vector space.

8. Oct 28, 2014

### Jaglowsd

1) Vector addition of vectors u,v
2) Scalar multiplication of a real number a, and u
3) A vector space has to have a zero vector

9. Oct 28, 2014

### Jaglowsd

Since AX=B does not satisfy 3 because if X=0 then B must be a zero vector then it is not a vector space. Am I correct?

10. Oct 28, 2014

### Dick

Yes, you are correct. Some other properties don't work either as Halls pointed out. But that's the easiest one to check.

11. Oct 28, 2014

### Jaglowsd

Much appreciated for the help from the both of you.