Recent content by sam0617

Well, thanks anyway. I got help from the wonderful people at yahoo answers. Thank you for trying to work with me.

Sorry, I think you're losing patience with me but I did what you asked and here it is. Please correct me if I'm incorrect. <u,v>2≤∥u∥2∥v∥2 = (u1v1+u2 v2)(u1v1+u2 v2) ≤ √(u12+ u22) √(v12 + v22)

I see that a2 + b2 would be a12a12+b12b12

Oh okay, that makes sense. So just to clarify, because if <v, v > = 0 if and only if v = the zero vector but we don't know v2 due to how <u, v > is defined so that means v2 could be a non-zero number. Correct me if my logic is wrong. Thank you again, HallsofIvy.

<u,v> in R2 would be u1v1 + u2v2 I believe.. and ||u|| in R2 is √u2 I'm terribly sorry but I'm not sure what you're trying to get at. I know and I also don't want to be spoon fed the answer.

Homework Statement Let u = (u1, u2, u3) and v = (v1, v2, v3) Determine if it's an inner product on R3. If it's not, list the axiom that do not hold. Homework Equations the 4 axioms to determine if it's an inner product are (all letters representing vectors) 1. <u,v> = <v,u> 2...

Homework Statement Use the Cauchy-Schwarz inequality to prove that for all real values of a, b, and theta (which ill denote as θ), (a cosθ + b sinθ)2 ≤ a2 + b2 Homework Equations so the Cauchy-Schwarz inequality is | < u,v>| ≤ ||u|| ||v|| The Attempt at a Solution I'm having...

Which of the transformations are onto? 1) T:R2 -> R2, where T(x,y) = (5x-y, 0) I don't know if I'm understanding this correctly but this transformation is NOT onto because if I let 5x-y = a 0 = b this means that b doesn't cover all the range of T? Could someone explain it better if...

Deveno, thank you so much for your help. It's much clearer now. Really, thank you again and Halls for really helping me out.

Yes, sorry. I wrote the question wrong. Apologies. I guess I'm having a very difficult time understanding why the book used RREF (reduced row echelon form) to get the basis. To take a guess, would it be because a basis are vectors that are linearly independent, and so the author puts it in...

Hello. First, I'd like to apologize because I don't know where to go ask for homework on linear algebra on the forums so if anyone could please let me know, that would be appreciated. Here's the question: Find a basis for the subspace of R^4 spanned by the given vectors Here's the answer...

Thank you. Sorry for the confusion everyone.

T: Mnn => R, where T(A) = tr(A) Attempt: 1) T(kA) = tr(kA) = k tr(A) = k T(A) 2) T(A+B) = tr (A + B) = tr(A) + tr(B) = T(A) + T(B) so it's linear transformation. Am I correct?

I tried googling this first and of course looked at my notes but the questions asked pertaining to dimensions were confusing me more so I thought I'd ask directly. Sorry for asking the same question I'm sure someone else has asked. Okay. So, to find the dimension, do I first make sure it's a...

I'm sorry, I don't understand what the x=z=a and y=x then it wouldn't satisfy above. Could you explain more? EDIT: Then to make question 1 transitive, all I would have to add is (b,b) ?