Linear Algebra Vector Spaces: Prove equivalence

[Luxe]

Homework Statement

Prove that the following are equivalent:
1. N(A)=0
2. A is nonsingular
3. Ax=b has a unique solution for each b that exists in R^n.

Homework Equations

The Attempt at a Solution

I think you prove this by showing that 1 implies 2, 2 implies 3, & 3 implies 1.
But after that I don't know how to prove that.

 

[VeeEight]

Start by trying one of the three implications you must show.
For example, for 1 implies 2, what does N(A) = 0 tell you about the equation Ax=0? Can you imply this to determine that A is invertible?

 

[Luxe]

if N(A)=0 then in the equation Ax=0, x equals 0. But then how do you show that A is invertible from that?

 

[VeeEight]

What does Ax=0 tell you about the columns of A? Let a₁, ..., a_n be some vector in A, and the components of x are x₁,..., x_n. Then from Ax=0, you get equations of the form a₁x₁ + ... + a_nx_n = 0, and from N(A)=0 you get that all the x_i's are 0. What can you conclude from this?

 

[Luxe]

Ok I think I figured the first part out:
The columns of A are linearly independent.
So, A is row equivalent to the identity matrix, and Ax=0 and Ix=0 have only the solution, x=0. So, the A=E1E2...Ek, which says that the product of invertiable matrices is invertiable and E is invertiable, so A is invertable.

So, then How do you imply that 2 equals 3, and 3 equals 1?

 

[VeeEight]

Give them a try, pick one of the two. What have you done so far?

For example, 2 to 3:
A is non singular, So A^-1 exists. Then multiply Ax=b by A^-1 . What does this imply? Is this unique?

 

[VeeEight]

And the other directions are probably similar. It's all a matter of shuffling between invertibility, linear dependence, etc.

 

[Luxe]

ok, i think that for 2 implies 3. A being nonsingular says that A inverse exists. So, then you can show that x= (inverseA)*B which proves #3. Is this right?

Now, the one I am stuck on is 3 implies 1. Any hints to get me started?

 

[VeeEight]

Yes, you are correct and it is easy to understand that this is in fact a unique solution (since for all other solutions you can simply left multiply by A^-1 to get the same conclusion.

I have not yet thought about 3 to 1. Think about it. Maybe there is some trick you can use from what you already proved in this problem.

You are assuming that Ax=b has a unique solution. Then b=0 also has a unique solution, and you know what it is.

 

[Luxe]

I still have no idea on the last one. I have looked all through my book...

 

[vela]

ok, i think that for 2 implies 3. A being nonsingular says that A inverse exists. So, then you can show that x= (inverseA)*B which proves #3. Is this right?

That shows that Ax=b has a solution. You should prove that this solution is unique.