Proof of exchangeability in matrix multiplications with Identity matrix I

In summary, we proved that for a square matrix A where I-A is nonsingular, A * (I-a)^-1 = (I-A)^-1 * A holds true. This is done by manipulating the equations (A*B)^-1 = B^-1 * A^-1 and A^-1 * A = A * A^-1 = I, and using the fact that the identity matrix always commutes.
  • #1
Crution
10
0

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?
 
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  • #2
this may be a place start... you'll need to assume A is non-singular as well...
[tex] (A (I-A)^{-1})^{-1} = .. [/tex]
 
  • #3
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?
 
  • #4
Crution said:
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

[tex] (B^{-1} A^{-1})(AB) [/tex]

Crution said:
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
[tex] (A (I-A)^{-1})^{-1} = (I−A)A^-1 [/tex]

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

so you have
[tex] (A (I-A)^{-1})^{-1} = A^{-1}(I-A)[/tex]

Any ideas where to go from here, think about gettingt back to the LHS in the original question?
 
  • #5
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
starting from
[tex](A (I-A)^{-1})^{-1} = A^{-1}(I-A) [/tex]

then I was thinking take the inverse of both sides
[tex]((A (I-A)^{-1})^{-1})^{-1} = (A^{-1}(I-A))^{-1} [/tex]
[tex]A (I-A)^{-1} = (I-A)^{-1}A [/tex]

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
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
I can't quite follow your working, isn't your second expression what you are trying to prove?
 
  • #9
As a start though, why not expand the expression below to show its true
(I-A)A=A(I-A)
 
  • #10
this is similar to noting that the identity matrix always commutes...
 
  • #11
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
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
Crution said:
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
[tex] (I-A)A = A(I-A)[/tex]

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

can you take it from here?
 
Last edited:
  • #14
oooh ok, yeah nevermind. thanks for tppinting that out ^^
i just didnt get what you where going for
 
  • #15
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
 
  • #16
yeah ok, so that look good

for (I-A)A=A(I-A) just try multying the expressions out
 
  • #17
(I-A)A = A.I-A.A

then use the fact that the identity commutes
 
  • #18
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!
 

FAQ: Proof of exchangeability in matrix multiplications with Identity matrix I

1. What is the concept of exchangeability in matrix multiplications with Identity matrix I?

Exchangeability refers to the property of matrices where the order of multiplication does not affect the final result. This means that when multiplying a matrix with an Identity matrix I, the resulting product will be the same regardless of which matrix is multiplied first.

2. How is proof of exchangeability demonstrated in matrix multiplications with Identity matrix I?

The proof of exchangeability can be demonstrated mathematically by showing that the order of multiplication does not affect the final result. This can be done by using the properties of matrix multiplication and the definition of the Identity matrix I.

3. Why is exchangeability important in matrix multiplications with Identity matrix I?

Exchangeability is important because it allows for easier and more efficient calculations, as the order of multiplication does not need to be considered. It also simplifies the process of solving systems of equations involving matrix multiplications.

4. Can exchangeability be applied to all types of matrices in matrix multiplications with Identity matrix I?

Exchangeability can only be applied to square matrices, as the Identity matrix I is only defined for square matrices. For non-square matrices, the concept of exchangeability does not apply.

5. Are there any exceptions to the proof of exchangeability in matrix multiplications with Identity matrix I?

Yes, there are some exceptions to the proof of exchangeability, such as when one of the matrices is singular or when the matrices do not have the same dimensions. In these cases, the order of multiplication will affect the final result.

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