Recent content by Gotejjeken

It's been a while, but looking at my old notes I had (these are free body diagram equations): Mass1: (Fnet)y: T1 - m1*g = m1*a(m1)y Mass2: (Fnet)y: T2- m2*g = m2*a(m2)y Then just solved those for T1 and T2 respectively. After that it looks like an application of the torque equations and...

I'm not sure how to think of this then. I tried thinking of the kernel as a nullspace: t^2(3a,b,c) + t(0,b,0) = 0 If I do this though, I obviously get that t^2 and t must both be zero. I thought it would suffice to say that any scalar multiplied by the vector (-1/3,0,-3) would yield a...

Using this, any vector of the form y * (-c/3, 0, -3a) (y is a scalar) will fall in the kernel of L. I cannot seem to find other polynomials that are not multiples of the vector above...so is that a basis for the kernel? I feel I may be getting close, however I am having a few syntax issues.

If a = 1, b = 0, and c = -3, then a*t^2 + b*t + c => t^2 -3 and: L(t^2-3) = t*(2*t) + t^2*(1-3) = 2*t^2 - 2*t^2 = 0. So t^2 - 3 is one non-zero polynomial in the kernel.

After looking it over again (and writing my basis more carefully), I come up with a basis of: ker(L) = {(-c/3)*t^2, 0, -3*a} I tested a few values and this seems to check out fine...although knowing me I'm probably overlooking something again. In the same vein, is a basis for the range...

Substituting p(t) in, I get: L(p(t)) = t*(2*a*t + b) + t^2*(a+b+c) = 3*a*t^2+b*t^2+c*t^2 + b*t = t^2*(3*a + b + c) + t*(b) so: t^2*(3*a + b + c) + t*(b) = 0. It appears as if b must be 0 at the least. If 3*a + b + c = 0 => a = -c/3 and c = -3*a will make this term zero, so would a...

Homework Statement L(p(t)) = t*dp/dt + t^2*p(1) If p(t) = a*t^2 + b*t + c, find a basis for the kernel of L. Homework Equations None. The Attempt at a Solution I know that L(a*t^2 + b*t + c) = 0, so that would mean that the derivative needs to be zero and p(1) needs to be zero. This...

Oh, wow. Thank you, it seems I was over-complicating things...the wording of that problem had my brain in knots.

Homework Statement The following vectors form an ordered basis E = [v1, v2] of the subspace V = span(v1,v2): v1 = (1,2,1)^T , v2 = (3,2,1)^T. The vector v = (24,-8,-4)^T belongs to the subspace V. Find its coordinates (c1,c2)^T = [v]E relative to the ordered basis E = [v1,v2]. Homework...

Alright, I'm going to put this altogether to see if it makes sense: Proof: Let's take B to be singular, then there exists an x that is not 0 such that Bx = 0. Thus: C = AB => Cx = (AB)x = A(Bx) = 0. C would have to be singular, as there exists an x not equal to 0 that makes Cx = 0...

Well, we know that if B is invertible and x is a solution to Bx = y: Bx = y => (B-1B)x = B-1y => x = B-1y. This x would then have to be nonzero as y is nonzero. Then: ABx = A(Bx) = A(BB-1y) = Ay = 0. This can't happen though, since C = AB is invertible (and y is not 0), so A has...

I suppose I am a little lost here. I understand how the proof for B works, however A has me stumped. I suppose it is the fact that (AB)x = (Ax)B is illegal, yet (AB)x = A(Bx) is not. I am also confused as to what exactly showing y in the form of Bx means...is there any other way to state this...

If A and B are invertible, we know their product must also be invertible. Other than that, I'm not too sure unless you are hinting at something involving elementary matrices.

Homework Statement Let A and B be NxN matrices, and assume that their product C = AB is invertible. Without using determinants, prove that A and B must both be invertible. Homework Equations If a NXN matrix A is invertible: Ax = 0 has only the trivial solution 0. The Attempt at...

Alright, thanks. That problem was giving me many problems as well, and I am glad that forums such as these exist!