Proof that the nullspace is closed in addition/multiplication

[JPanthon]

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

 

[JPanthon]

Anyone? Please?

 

[Mark44]

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.

a_kj 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.

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, ..., b_n> and c = <c₁, c₂, ..., c_n> be vectors in N(A).

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

And similar for scalar multiplication.

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

Thank you

 

[JPanthon]

a_kj 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, ..., b_n> and c = <c₁, c₂, ..., c_n> 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

 

[Mark44]

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.

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).