Recent content by Daron

Suppose an nxn matrix has n distinct eigenvectors vi when treated as a linear operator over ℝn. What is the relationship between these and the eigenvectors of the matrix when treated as a linear operator over ℝnxn, the space of nxn matrices? Since a matrix L acting on one with columns a1, a2...

I'm not sure exactly what you mean here by "smooth it out". But all of the functions of this type that I've seen go to infinity at the origin, which is prohibited by this one being C∞. This function is bounded from above by 1/2, unless you are proposing to make it 1 on [1/4, 1/2]; 2 on...

Suppose I have a C∞ function, which I wish to prove attains its maximum/minimum. First I must prove that the function is bounded at all. If I determine R, the region (of the plane in this case) where the function is strictly positive, and integrate over R to find a finite answer, can I say the...

I know the Cartesian product for an algebraic structure: A x B = {(a,b): a ∈ A, b ∈ B} Which naturally gives An = {(a1, a2, ... , an): ai ∈ A ∀ i} Some of the time, at least we can also have a non integer n. For example [A x A x A]2/3 = A x A. Is there any way of continuing the...

Homework Statement Verify that the following mappings are isometries on R^2 Reflection Through the Origin Translation RotationHomework Equations Qualities of a metric: d(x,y) = d(y,x) d(x,x) = 0 d(x,y) = 0 <=> x = y d(x,y) =< d(x,z) +d(z,y) The Attempt at a Solution As a metric hasn't...

Does this make sense? Every element of {x ∈ G: order of x ≥ 3} is not equal to its inverse. So for every element of the set, we can find a distinct element of the set which is its inverse. As inverses are unique, these two elements will be inverses ONLY to each other. In this way the set can be...

Homework Statement Let G be a group of finite cardinality. By considering the size of the set {x ∈ G: x is of order ≥ 3} show that |G| is even iff there is an element of G with order 2. Homework Equations Perhaps Lagrange's Theorem The Attempt at a Solution It's obvious that if g...

They're equal up to a scalar multiple. So for every eigenvalue λ with multiplicity m, we will get a system of m linear equations of the form Bxi = aixaj which define an eigenspace that is invariant under B. And because B has an orthonormal basis of eigenvectors, we may consider B acting...

Is the idea that every eigenvector of B with eigenvalue λ can be formed from a linear combination of the eigenvectors of A with eigenvalue λ, and that these combinations are still eigenvectors of A due to linearity?

If the multiplicity is 1, then A(Bx) = λBx, so Bx is an eigenvector of A with eigenvalue λ, but since there is only one eigenvector with eigenvalue 1. then Bx = x. I've moved on to trying to prove how commuting matrices share a basis of eigenvectors, which implies that both are diagonal in...

Homework Statement A and B are commuting diagonalizable linear operators. prove that they are simultaneously diagonalizable. Homework Equations AB = BA The Attempt at a Solution We deal with the problem in the Jordan basis of A, where A is diagonal, as Jordan forms are unique. Then by...

I figured it out.

Homework Statement A and B are commuting diagonalizable matrices. Prove that they are simultaneously diagonalizable. Homework Equations AB = BA The Attempt at a Solution I have what looks like a proof, but I'm not very happy with it. Is there anything wrong here? AB = BA B = ABA-1 Every...