Recent content by unfunf22

Well, if I didn't then I wouldn't be learning anything about them. I mean, most of these proofs I don't see how to use similar matricies to figure them out... There's a huge number: a) If A can be diagonalized, then its eigenvalues are distinct. I said this is false. I just used a counter...

Homework Statement If A is nonsingular, prove that the eigenvalues of A-1 are the reciprocals of the eigenvalues of A. *Use the idea of similar matrices to prove this. Homework Equations det(I\lambda - A) = 0 B = C-1AC (B and A are similar, and thus have the same determinants) The...

Oh, alright, well if I have DT([1,1,1,1]) = [1,4,9,0] then that would be the matrix would be [1,0,0,0] [0,4,0,0] [0,0,9,0] [0,0,0,0] As [1,0,0,0]*[1,1,1,1] = [1] (corresponding to 0 degree poly) [0,4,0,0]*[1,1,1,1] = [4] (corresponding to x) [0,0,9,0]*[1,1,1,1] = [9] (corresponding to x^2)...

So then I should be showing that D(T([1,1,1,1])) -> [0,1,4,9] ? Wouldn't this still have the same problem as before? Can I not consider these as column vectors? Maybe that's the issue.

I meant that [1,x,x^2,x^3] is a column-vector, so it would look like |1| |x| |x^2| |x^3| M multiplying that, so it would be 4x4 matrix times a 4x1 amatrix, to give out a 4x1 matrix.

Homework Statement Find m(DT), that is, find the matrix for the transformation DT where D is the derivative operator and T: V -> V , T(p(x)) = xp'(x). The polynomial is of degree <= 3, and the basis for it is (1,x,x^2, x^3).Homework Equations Basic matrix multiplication needs to be...

I was able to prove it thanks to the general formula for the product rule of n functions. Thanks!

Alright. Well I am using Leibniz to informally argue it at the moment. I'll see if that pans out.

So you are saying use leibniz and match its behavior to that of the product rule? Is there not a way to do this inductively?

Homework Statement Given n2 functions fij, each differentiable on an interval (a,b), define F(x) = det[fij] for each x in (a,b). Prove that the derivative F'(x) is the sum of the n determinants, F'(x) = \sum_{i=0}^n det(Ai(x))$. where Ai(x) is the matrix obtained by differentiating the...

I'll have to think about that more. I can't see the connection right now. It's 5 am where I am so I need to sleep soon. I figured out that (-A + I) is an inverse of A + I using lane's hint. Thanks lane.

Ah, that is an interesting insight. So A + I is close to 1 as I is the identity, 1 (as I^n = I). But what does being close to the identity have to do with being invertible? I can't make the connection, is there some property revealed by A+I being "close" to 1? What does that even mean, exactly...

Oh yea, I tried that earlier. You mean A*A + A*I = A*C -> 0 + A*I = A*C -> A*I = A*C The only way to get the identity from that is if I could invert A, and show that I = C. But nothing says that A is nonsingular. So I can't just invert it. In other words, I could have the inverse (A^-1)A, which...

I can't use determinants yet. That is the next chapter in the book and the professor does not want us using material not yet gone over in our proofs. :( Is there a way to think about it without resorting to determinants (I already figured out how to trivially prove most of these questions using...

1. Matrix A^2 = 0, then A + I is nonsingular. Prove this or show a counter-example Homework Equations I suppose you must know Gauss-Jordan and matrix multiplication. The Attempt at a Solution I first thought this was a false statement, so I tried to provide a counter-example. I...