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...
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...