Recent content by Crution

thanks that was enough help :-)

thank you! wow I feel stupid. well so (I-A)A=A(I-A) I*A - A2 = A*I - A2 A-A2 = A-A2 alright thanks for all your help!

man, now i thought i finally had it, but as I'm trying do write it down I can't seem to figure out how to proof (I-A)A=A(I-A) anymore. I expanded it to (I-A)*(I-A)^-1*(I-A)*A = (I-A)^-1*(I-A)*A*(I-A). But now what? Ok so overall I will now do this:A=I*A=(I-A)^-1*(I-A)*A subsitute in...

oooh ok, yeah nevermind. thanks for tppinting that out ^^ i just didnt get what you where going for

because that would mean I could do this: A*(I-A)^-1 = (I-A)^-1*A (I-A)^-1*(I-A)*A*(I-A)^-1 = (I-A)^-1*(I-A)^-1*(I-A)*A (I-A)^-1*(I-A)*A*(I-A)^-1*I = (I-A)^-1*(I-A)^-1*(I-A)*A*I leaving (I-A)^-1*A = (I-A)^-1*A 0 = 0

thats what i did: (I-A)*A=A*(I-A) (I-A)*(I-A)^-1*(I-A)*A = (I-A)^-1*(I-A)*A*(I-A) What I was trying to ask is: can i do this (I-A)*(I-A)^-1*(I-A)*A*I = (I-A)^-1*(I-A)*A*(I-A)*I

Homework Statement I'm sorry this doesn't look too nice but it is supposed to be two matricces. Show: |1 a1-b1 a1+b1| |1 a1 b1| |1 a2-b2 a2+b2|=2*|1 a2 b2| |1 a3-b3 a3+b3| |1 a3 b3| without evaluating the determinants. Homework Equations...

Thank you both for your help! So if I get this right then Dick suggested to do this: Since: A = I*A = (I-A)^-1*(I-A)*A I can rewrite:A*(I-A)^-1 = (I-A)^-1*A (I-A)^-1*(I-A)*A*(I-A)^-1 = (I-A)^-1*(I-A)^-1*(I-A)*A question is, am i allowed to cancel in the following way...

thank you it looks like that might be the way to go. however, i don't really know where to go from there: 1. Is taking the inverse an equivalent tranformation? in other words after using (A*B)^-1 = B^-1 * A^-1 is this correct: (I−A) * A^-1 = A^-1 * (I−A) 2. Where should I go from here? I'm...

Homework Statement Let A be a square matrix such that I-A is nonsingular. Prove that A * (I-a)^-1 = (I-A)^-1 * A Homework Equations Now I think that A^-1 * A = A*A^-1 = I and I*A = A*I are relevant for this. The Attempt at a Solution I tried to express (I-a)^-1 in...