Recent content by blashmet

Can you have a tensor that isn't multi-linear? Thanks!

Are all tensors multi-linear functionals? Do the arguments of all tensors have to be vectors? That is, are these necessary conditions of being a tensor? Thanks for the help!

Is it correct to view tensors as multi-variable functions? For example, it seems the permutation tensor is a function of three variables and the metric tensor is a function of two variables. Of course, these "functions" turn into constants when i,j, and k (the indices) are known, but it seems...

Those are exactly my thoughts. It makes sense intuitively, I just don't know how to put it down in "math speak" haha. If anyone can help that'd be sweet...

(x_1 + x_2 +... + x_k) is greater than (x_1 + x_2 +...+ x_k)/k ? thanks for the help btw :)

Homework Statement http://img440.imageshack.us/img440/9242/captureyeq.jpg Uploaded with ImageShack.us Homework Equations Given in problem statement. The Attempt at a Solution Basically I've come up with the contrapositive, but I'm unsure of how to proceed with the proof. Can...

Ok great! Thanks Serena!

Does the following suffice as a complete, formal proof? "d1...dn are the eigenvalues and q1...qn are the eigenvectors, since they satisfy Av=λv. So each column of Q is an eigenvector of A." Thanks Serena! :)

It looks like the eigenvalues are dn (the diagonal components of the matrix D). So I suppose the eigenvectors would be qn (column vectors). I guess my problem is this: I need to write out a formal proof from beginning to end, and I'm not sure what to add to what I've already written (or...

Ahh Deveno! lol... I didn't know you were on this forum too. I was going to have another look at our convo about this problem on MHF, but it's down apparently. Anyways, why do you think the matrix will be ugly? The solution is posted as...

Homework Statement http://img847.imageshack.us/img847/2783/77597781.jpg Homework Equations See below. The Attempt at a SolutionI get all the way to the last step, but I'm not sure how to perform it.I get to the point where I have: [(T^-1)(ax^2+bx+c)]\alpha= [T^-1]\alpha\beta*[v]\beta= [ a/4...

Hi deluks917, i'm not sure how you'd write it, but i know that AQ and QD are nxn square, so it's probably just a shorthand notation for the general element form of the multiplication of 2 nxn square matrices. can you answer my other questions? (it's ok if you don't know, i just wasn't...

Hi Serena! In a formal proof of this, do I need to have all of this first?: Q(Q^-1)AQ=QD iff AQ=QD=diag[d1,...,dn] AQ=A[q1,...,qn]=[Aq1,...,Aqn] QD=[d1q1,d2q2,...,dnqn] so AQ=QD iff [Aq1,...,Aqn]=[d1q1,...,dnqn] Now, is there where I show q1 is an eigenvector for A? If so...

Thanks for the help guys! This is what I have so far: Q(Q^-1)AQ=QD iff AQ=QD=diag[d1,...,dn] AQ=A[q1,...,qn]=[Aq1,...,Aqn] QD=[d1q1,d2q2,...,dnqn] so AQ=QD iff [Aq1,...,Aqn]=[d1q1,...,dnqn] I'm stuck at this point. How do I proceed from here? Thanks! :)

Homework Statement Prove: if (Q^-1)AQ=D, then each column of Q is an eigenvector of A. Homework Equations A vector v is an eigenvector of A iff there exists a scalar λ such that: Av=λv The Attempt at a Solution Suppose (Q^-1)AQ=D. We need to show each column of Q is an...