Linear Algebra Vector Spaces: Prove equivalence

In summary, the three statements, 1. N(A)=0, 2. A is nonsingular, and 3. Ax=b has a unique solution for each b that exists in R^n, are equivalent. This can be proven by showing that 1 implies 2, 2 implies 3, and 3 implies 1. From N(A)=0, it can be implied that A is invertible and the columns of A are linearly independent. This leads to the conclusion that A is invertible. From A being non-singular, it can be shown that A inverse exists and multiplying Ax=b by A inverse leads to a unique solution for x. To prove 3 implies 1, it can be assumed
  • #1
Luxe
10
0

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.
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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?
 
  • #4
What does Ax=0 tell you about the columns of A? Let a1, ..., an be some vector in A, and the components of x are x1,..., xn. Then from Ax=0, you get equations of the form a1x1 + ... + anxn = 0, and from N(A)=0 you get that all the xi's are 0. What can you conclude from this?
 
  • #5
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?
 
  • #6
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?
 
Last edited:
  • #7
And the other directions are probably similar. It's all a matter of shuffling between invertibility, linear dependence, etc.
 
  • #8
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?
 
  • #9
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.
 
  • #10
I still have no idea on the last one. I have looked all through my book...
 
  • #11
Luxe said:
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.
 

1. What is a vector space in linear algebra?

A vector space is a mathematical structure that consists of a set of objects, called vectors, and a set of operations, such as addition and scalar multiplication, that can be performed on these vectors. These operations must follow certain rules, such as closure and associativity, for the structure to be considered a vector space.

2. How do you prove two vector spaces are equivalent?

To prove two vector spaces are equivalent, you must show that they have the same underlying structure and follow the same rules. This can be done by showing that one vector space can be transformed into the other through a series of operations, such as linear combinations and transformations. Alternatively, you can show that the two vector spaces have the same basis, which means they contain the same set of linearly independent vectors.

3. What is the significance of proving equivalence in linear algebra vector spaces?

Proving equivalence in linear algebra vector spaces is important because it allows us to compare different vector spaces and understand their similarities and differences. It also helps us to generalize concepts and properties from one vector space to another, which can be useful in solving problems and making connections between different areas of mathematics.

4. What are some common techniques used to prove equivalence in linear algebra vector spaces?

Some common techniques used to prove equivalence in linear algebra vector spaces include showing that the vector spaces have the same dimension, showing that they have the same basis, or showing that one vector space can be transformed into the other through a series of operations. Other techniques may involve using properties of linear transformations, such as injectivity and surjectivity, or using properties of matrices, such as row and column operations.

5. Can two vector spaces be equivalent but not isomorphic?

Yes, two vector spaces can be equivalent but not isomorphic. This means that they have the same underlying structure and follow the same rules, but they may not have the same basis or the same dimension. In other words, they are equivalent in terms of their structure and properties, but they are not identical.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
340
  • Calculus and Beyond Homework Help
Replies
10
Views
951
  • Calculus and Beyond Homework Help
Replies
8
Views
745
  • Calculus and Beyond Homework Help
Replies
14
Views
533
  • Calculus and Beyond Homework Help
Replies
25
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
422
  • Calculus and Beyond Homework Help
Replies
8
Views
559
  • Calculus and Beyond Homework Help
Replies
2
Views
551
  • Calculus and Beyond Homework Help
Replies
0
Views
419
Back
Top