# Proof of exchangeability in matrix multiplications with Identity matrix I

1. Mar 15, 2012

### Crution

1. The problem statement, all variables and given/known data

Let A be a square matrix such that I-A is nonsingular.

Prove that A * (I-a)^-1 = (I-A)^-1 * A

2. Relevant equations

Now I think that

A^-1 * A = A*A^-1 = I
and
I*A = A*I
are relevant for this.

3. 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?

2. Mar 15, 2012

### lanedance

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

3. Mar 15, 2012

### Crution

thank you it looks like that might be the way to go.
however, i dont 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?

4. Mar 15, 2012

### lanedance

try multplying and see if it behaves an inverse should

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

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?

5. Mar 15, 2012

### 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.

6. Mar 16, 2012

### 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.

7. Mar 19, 2012

### 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!

8. Mar 19, 2012

### lanedance

9. Mar 19, 2012

### lanedance

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

10. Mar 19, 2012

### lanedance

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

11. Mar 19, 2012

### 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

12. Mar 19, 2012

### 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

13. Mar 19, 2012

### lanedance

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...

$$(I-A)A = A(I-A)$$

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?

Last edited: Mar 19, 2012
14. Mar 19, 2012

### Crution

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

15. Mar 19, 2012

### Crution

man, now i thought i finally had it, but as i'm trying do write it down I cant 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

16. Mar 19, 2012

### lanedance

yeah ok, so that look good

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

17. Mar 19, 2012

### lanedance

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

then use the fact that the identity commutes

18. Mar 19, 2012

### 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