Recent content by ImAnEngineer

(ATA)-1ATA = I That's it, thanks :)

It's not true that Ker(A)= empty set, because 0 \in \textrm{ker}(A) . And why do you think it implies that it's square?

That statement does hold indeed, and helped me to get: rank(A) + dim(ker(A)) = rank(A) = n rank(At) + dim(ker(At))= rank(A) + dim(ker(At)) = n + dim(ker(At)) = m Hence: m\geq n But it doesn't get me any further.

This is simply not true. I have a better source ("Linear algebra done wrong"), but I'll use wikipedia to quote: I underlined what I'm trying to proof. I know the theorems but I don't see how they are applicable to a non-square matrix.

Why and why? Note that A is not (necessarily) a square matrix

I found on wikipedia that the following statements are equivalant: 1. Matrix A is left invertible 2. Ax=0 => x=0 I couldn't find the proof so I try to do it myself. From 1. to 2. is easy. Assume A is left invertible. If Ax=0, then x=Ix=A-1Ax=A-10 = 0 . I can't figure out how to do 2=>1. Any...

Ah of course, the triangle inequality in one dimension :) (not really a triangle) |a-c|\leq |a-b|+|b-c| If I choose: a=1, b=-1, c=x |1-x|\leq|1+1|+|-1-x|=2+|1+x|<d So surely: |1+x|<d+2 Thank you!

Homework Statement Show the following for every d>0: For every real number x with |x-1|<d it follows that |1+x|<2+d 2. The attempt at a solution If x-1>0, then |x-1|=x-1<d. Hence x+2 = |x+2| < 2+d. If x-1<0, then |x-1|=-(x-1)<d. Hence x-1>-d => x+1 > 2-d ...?? Is this really possible to do...

Indeed, I get ellipses or hyperbolae, or parbolae depending on the chosen initial conditions. I'll ask my instructor what this is about... kinda strange.

Thank you, that formula makes a lot of sense :) . I was allowed to treat the planets as point particles. I used the formule in my mathematica file. However I get a smaller amount of mechanical energy at t=50 than at t=0 (decrease of 10.6%). I assume I did nothing wrong, because in the...

A bullet at 0.00000001 times the speed of light can kill you, so surely a bullet at the speed of 0.99... c will be able to kill you. You're messing up a couple of things... let me make it clear. According to the observer, the time in the bullet's inertial frame is slower than his own. So if...

Suppose we have a universe consisting of three planets. Their velocities and positions at t=0, and the gravitational constant are known. One can calculate, using differential equations, the positions of the planets at any time. Assuming that the only force acting on the planets is gravity and...

Thanks a lot, I get it now :) ! I get the following system of equations: 2x-ly = 0 2y-lx = 0 z(2+2l) = 0 z²-xy = 1 And when I solve this I get (x,y,z)=(0,0,+/-1).

Homework Statement Find the points on the surface: z²-xy=1 nearest to the origin Homework Equations grad(f)= lambda grad(g) (?) f(x,y,z)=z²-xy (?) g=(0,0,0) (?) The Attempt at a Solution grad(f)=(-y, -x, 2z)= lambda (0,0,0) = (0,0,0) -y=0 => y=0 -x=0 => x=0 2z=0 => z=0 But...

Then it is easy: 0<\left|\frac{xyz}{x^2+y^2+z^2}-0\right|\leq\left|\frac{xyz}{y^2+z^2}\right|\leq\left|\frac{xyz}{2yz}\right|=\left|\frac{x}{2}\right|<\frac{\delta}{2}=\epsilon How did you come up with: yz<=1/2*(y^2+z^2) ? (Why is it true at all?)