Recent content by tamintl

Consider the 4x4 matrices A = (1 2 3 4) (5 6 7 8) (9 10 11 12) (13 14 15 16)B= (1 2 3 4) (8 5 6 7) (11 12 9 10) (14 15 16 13) The question I was asked was the following: Show that there does not exist an endomorphism f: ℝ4 -> ℝ4 and basis 'a' and 'b' of R^4, such that A = a[f]a and B=b[f]b...

ok - taking a=b=c=d=1. our basis would be:{1,x,x3+x2} I am not sure where this is going? the question asked for a basis consisting of 4 vectors. Would an alternative method to show that we have a basis be the following: Take {1,x,x3+x2,x3} which is linear independent. Now we have...

If ad=bc, then we have linear independence, thus a basis.

Nice. Can't believe I didn't see that. It is clearly linearly independent. Is it enough to say that since \{1,x,x^3+x^2,x^3\} is a linear combination of pi that it will span the vector space? Thanks

Homework Statement Can you find a basis {p1, p2, p3, p4} for the vector space ℝ[x]<4 s.t. there does NOT exist any polynomials pi of degree 2? Justify fully.Homework Equations The Attempt at a Solution We know a basis must be linearly independent and must span ℝ[x]<4. So intuitively if there...

Thanks for your reply. If I were to take the transpose so it becomes: ({1,2,3,4},{2,3,4,5},{3,4,5,1},{4,5,1,2}, {5,1,2,3}) Would you see any quick way to determine linear independence? Many thanks

Hi, I was wondering what the quickest set of vectors in higher dimensions, say ℂ5? I have a set of vectors { (1,2,3,4,5), (2,3,4,5,1), (3,4,5,1,2), (4,5,1,2,3) } Clearly this is linearly dependent but how could I justify this quickly? What are the quickest ways to determine if this is...

[/itex]Homework Statement Is the following statement true or false: 'if (cn) and (dn) are bounded sequences of positive real numbers then: lim sup (cndn) = (lim sup cn)(lim sup dn)Homework Equations The Attempt at a Solution for all n in the positive reals. cn and dn are bounded. Since cn...

okay thanks.. I got it

Homework Statement Let F = ( -y/(x2+y2) , x/(x2+y2) ) Show that this vector field is irrotational on ℝ2 - {0}, the real plane less the origin. Then calculate directly the line integral of F around a circle of radius 1.Homework Equations The Attempt at a Solution To show F is irrotational we...

Yeah forget about the 'f's.. yeah that makes sense. Deveno, if I sent you the question sheet it may be easier for both you and I to understand. Of course, only if you are happy to help. Would that be okay? The reason I ask is that it is hard for me to get my points across since I don't know...

Yes, I think... Edit: taking λ = 1/2 f(x+y) = f(1/2(2x)) + f(1/2(2y)) = 1/2 [ f(2x) + f(2y) ] taking 2 out gives: = f(x) + f(y) Hence closed under addition Is that sufficient? Thanks

bump?

Take r=x and s=0 so since we know (lambda)x is in M, x+0 is in M. Or could we use the M+a proof?

Oh okay. If we take y=0 (using the condition 0€M) Then we get (lambda)x + (1-lambda)(0) which is just (lambda)x So we know for any x and lambda that it will be in M. So that is iii done. What about ii Ps: I'm on my phone so sorry for weak notation. Thanks micro