Proof that the nullspace is closed in addition/multiplication

In summary, The problem asks to show that for a matrix A, if b and c are two vectors in the null space of A, then their sum and scalar multiple are also in the null space of A. This can be proven by using the properties of matrix multiplication and the fact that the null space of A is defined as the set of all vectors that when multiplied by A, result in the zero vector.
  • #1
JPanthon
20
0

Homework Statement



It is number three on the following page.
http://people.math.carleton.ca/~mezo/A3math1102-11.pdf

Homework Equations



No idea.

The Attempt at a Solution



I have no idea how to incorporate the kj.

Best I could reason through this is supposing: b1 ∈ N(A) , c1 ∈ N(A)

Ab1 + Ac1 = 0
A(b1 + c1) = 0

=> (b1 + c1) ∈ N(A)

This clearly isn't as rigorous as is desired though.

Any help would be so appreciated, I'm completely stuck here!

Thank you
 
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  • #2
Anyone? Please?
 
  • #3
JPanthon said:

Homework Statement



It is number three on the following page.
http://people.math.carleton.ca/~mezo/A3math1102-11.pdf


Homework Equations



No idea.

The Attempt at a Solution



I have no idea how to incorporate the kj.
akj is nothing more than the entry in row k, column j of the matrix. Each number in the matrix is an element of the field F.
JPanthon said:
Best I could reason through this is supposing: b1 ∈ N(A) , c1 ∈ N(A)

Ab1 + Ac1 = 0
A(b1 + c1) = 0

=> (b1 + c1) ∈ N(A)

This clearly isn't as rigorous as is desired though.
Probably not, but it's not too far off.
Let b = <b1, b2, ..., bn> and c = <c1, c2, ..., cn> be vectors in N(A).

Show that b + c is in N(A).

And similar for scalar multiplication.
JPanthon said:
Any help would be so appreciated, I'm completely stuck here!

Thank you
 
  • #4
Mark44 said:
akj is nothing more than the entry in row k, column j of the matrix. Each number in the matrix is an element of the field F.
Probably not, but it's not too far off.
Let b = <b1, b2, ..., bn> and c = <c1, c2, ..., cn> be vectors in N(A).

Show that b + c is in N(A).

And similar for scalar multiplication.



Thank you so much for your reply!

b = <b1, b2, ..., bn>
c = <c1, c2, ..., cn>

b + c = < (b1 + c1), (b2 + c2), ..., (bn + cn)>

We know:

Ab = 0
s.t. 0 = Ab1, Ab2, ..., Abn

Ac = 0
s.t. 0 = Ac1, Ac2, ..., Acn



=> Ab + Ac = <(Ab1 + Ac1, Ab2 + Ac2, Abn + Acn)>
=> A(b+c) = <(A(b1 + c1), A(b2 + c1), A(bn + cn)>

So, addition is closed in the null space of A


Does this prove it?
Other info is given to the nature of the matrix i.e. k ≥ 1, j ≤ n, etc.
Should I mention this at all? What is the significance?

Thanks again
 
  • #5
JPanthon said:
Thank you so much for your reply!

b = <b1, b2, ..., bn>
c = <c1, c2, ..., cn>

b + c = < (b1 + c1), (b2 + c2), ..., (bn + cn)>

We know:

Ab = 0
s.t. 0 = Ab1, Ab2, ..., Abn

Ac = 0
s.t. 0 = Ac1, Ac2, ..., Acn



=> Ab + Ac = <(Ab1 + Ac1, Ab2 + Ac2, Abn + Acn)>
=> A(b+c) = <(A(b1 + c1), A(b2 + c1), A(bn + cn)>

So, addition is closed in the null space of A


Does this prove it?
Other info is given to the nature of the matrix i.e. k ≥ 1, j ≤ n, etc.
The notation in the problem means that both j and k are between 1 and n.
JPanthon said:
Should I mention this at all? What is the significance?

I don't think what you have above is what they're looking for. By telling you the entries of the matrix, I believe that they are expecting you to use them.

For example, with b = <b1, b2, ..., bn>, you are given that b is in N(A). This means that Ab = <0, 0, ..., 0>.

I believe that you're supposed to write something that represents the multiplication of A and b. In other words, you should get an n x 1 vector whose i-th entry is the dot product of the i-th row of the matrix and the vector b. Since Ab = 0, each of those dot products is equal to 0.

Same thing for c. Then use this information to show that b + c is in N(A).
 

FAQ: Proof that the nullspace is closed in addition/multiplication

1. What is the nullspace in linear algebra?

The nullspace, also known as the kernel, of a matrix is the set of all vectors that when multiplied by the matrix result in a zero vector. In other words, it is the set of all solutions to the homogeneous equation Ax = 0, where A is the matrix and x is a vector of unknowns.

2. Why is it important to prove that the nullspace is closed under addition and multiplication?

Proving that the nullspace is closed under addition and multiplication is important because it allows us to ensure that the solutions to linear equations are consistent and can be combined to create new solutions. This property is necessary for many applications in science and engineering.

3. How is the closure of the nullspace proven in addition?

To prove that the nullspace is closed under addition, we must show that if x and y are vectors in the nullspace, then their sum x + y is also in the nullspace. This can be done by substituting x and y into the homogeneous equation Ax = 0 and showing that the resulting equation is also equal to 0.

4. Is the closure of the nullspace also proven in multiplication?

Yes, the closure of the nullspace is also proven in multiplication. This means that if x is a vector in the nullspace and c is a scalar, then the product cx is also in the nullspace. This can be shown by substituting cx into the homogeneous equation Ax = 0 and showing that the resulting equation is also equal to 0.

5. What are the implications of the nullspace being closed under addition and multiplication?

The closure of the nullspace under addition and multiplication allows us to perform operations on linear equations and their solutions with confidence that the solutions will remain consistent. This property is essential in many areas of mathematics, science, and engineering, including solving systems of linear equations, finding eigenvalues and eigenvectors, and understanding the behavior of linear transformations.

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