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: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.
Probably not, but it's not too far off.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.
JPanthon said:Any help would be so appreciated, I'm completely stuck here!
Thank you
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.
The notation in the problem means that both j and k are between 1 and n.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.
JPanthon said:Should I mention this at all? What is the significance?
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.
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.
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.
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.
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.