Proof of exchangeability in matrix multiplications with Identity matrix I

[Crution]

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 respect to I without having an inversion in it. but somehow I can't get it to work.

Any ideas?

 

[lanedance]

this may be a place start... you'll need to assume A is non-singular as well...
$$(A (I-A)^{-1})^{-1} = ..$$

 

[Crution]

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 guessing that I have to transform this further so I can show they are equal by using this formula: A^-1 * A = A * A^-1

But how do I get there?

 

[lanedance]

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

try multplying and see if it behaves an inverse should

$$(B^{-1} A^{-1})(AB)$$

is this correct: (I−A) * A^-1 = A^-1 * (I−A)

2. Where should I go from here?
I'm guessing that I have to transform this further so I can show they are equal by using this formula: A^-1 * A = A * A^-1

But how do I get there?

Ok so you get
$$(A (I-A)^{-1})^{-1} = (I−A)A^-1$$

continuing to multiply through and as you suggest using A^-1 * A = A * A^-1 = I
$$(A (I-A)^{-1})^{-1} = (I−A)A^{-1} = A^{-1}−I = A^{-1}I −A^{-1}A = A^{-1}(I-A)$$

so you have
$$(A (I-A)^{-1})^{-1} = A^{-1}(I-A)$$

Any ideas where to go from here, think about gettingt back to the LHS in the original question?

 

[Dick]

You don't need to assume A is nonsingular. I would start from A=IA=(I-A)^(-1)(I-A)A. Now show (I-A)A=A(I-A). Continue from there.

 

[lanedance]

starting from
$$(A (I-A)^{-1})^{-1} = A^{-1}(I-A)$$

then I was thinking take the inverse of both sides
$$((A (I-A)^{-1})^{-1})^{-1} = (A^{-1}(I-A))^{-1}$$
$$A (I-A)^{-1} = (I-A)^{-1}A$$

Though as mentioned this assumes A is non-singular, which may be a leap we don't require. With this in mind I would try Dick's method.. also because he's usually always right about these things.

 

[Crution]

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:

(I-A)^-1*______A_______ = (I-A)^-1*______________A

leaving me with

(I-A)^-1*A = (I-A)^-1*A

That obviously would be great!

 

[lanedance]

I can't quite follow your working, isn't your second expression what you are trying to prove?

 

[lanedance]

As a start though, why not expand the expression below to show its true
(I-A)A=A(I-A)

 

[lanedance]

this is similar to noting that the identity matrix always commutes...

 

[Crution]

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

 

[Crution]

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

 

[lanedance]

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

still not sure I follow, are you starting by assuming the answer and working back from it? and i can't see how you get from the red and green to the last line, or follow the whole idea...

how about this, you know
$$(I-A)A = A(I-A)$$

now substitute it on the RHS for (I-A)A in the below expression
$$A = (I-A)^{-1}(I-A)A$$

can you take it from here?

 

[Crution]

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

 

[Crution]

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

A*(I-A)^-1

= (I-A)^-1*(I-A)*A*(I-A)^-1

because: (I-A)*A=A*(I-A) i can rewrite as

(I-A)^-1*A*(I-A)*(I-A)^-1

which is again: (I-A)^-1*A*I

= (I-A)^-1*A

 

[lanedance]

yeah ok, so that look good

for (I-A)A=A(I-A) just try multying the expressions out

 

[lanedance]

(I-A)A = A.I-A.A

then use the fact that the identity commutes

 

[Crution]

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!