Matrix Inversion and the Associative Property

[negation]

Homework Statement

Is it true that:

(ABA^-1)^8 = AB^8A^-1 for all n x n matrix and not just for invertible matrix?

My attempt:

(ABA^-1)^8 =A^8B^8A^-8

 

[DrClaude]

(ABA^-1)^8 =A^8B^8A^-8

That's not correct. You can't "distribute" the exponent like that for matrices.

Start with a simpler problem: what is $(A B A^{-1})^{2}$?

 

[negation]

That's not correct. You can't "distribute" the exponent like that for matrices.

Start with a simpler problem: what is $(A B A^{-1})^{2}$?

I can't see what is written. But I presume you wrote (ABA^-1)^2 based on my intuition of the code.

(ABA^-1)^2 = ((AB)A^-1)^2 = (AB)^2 (A^-1)^2 = (B^2A^2)A^-2

 

[DrClaude]

I can't see what is written. But I presume you wrote (ABA^-1)^2 based on my intuition of the code.

You'll have to find a way to solve that problem, otherwise you'll have difficulty reading most posts in PF!

((AB)A^-1)^2 = (AB)^2 (A^-1)^2

Again, you can't distribute the exponent like that, because multiplication of matrices is not commutative.

Go back to first principles: A² = AA

 

[negation]

You'll have to find a way to solve that problem, otherwise you'll have difficulty reading most posts in PF!


Again, you can't distribute the exponent like that, because multiplication of matrices is not commutative.

Go back to first principles: A² = AA

I haven't been able to troubleshoot it because it occurs at certain chances on the same computer. But usually it works fine.

((AB)A^-1)^2 = (ABA^-1) (ABA^-1)

 

[DrClaude]

((AB)A^-1)^2 = (ABA^-1) (ABA^-1)

Good. Now simplify that expression.

 

[negation]

Good. Now simplify that expression.

AABBA^-1A^-1

Is this a valid operation?

 

[DrClaude]

AABBA^-1A^-1

Is this a valid operation?

Absolutely not! Again, matrix multiplication is not commutative: AB ≠ BA. You can't move around the different matrices, like you would do for ordinary numbers.

What you have is

(ABA^-1)(ABA^-1) = A B A^-1 A B A^-1

Is there anything in there you can simplify?

 

[negation]

absolutely not! Again, matrix multiplication is not commutative: Ab ≠ ba. You can't move around the different matrices, like you would do for ordinary numbers.

What you have is

(aba^-1)(aba^-1) = a b a^-1 a b a^-1

is there anything in there you can simplify?

aa^-1a^-1abb

Should I try to get it simplified to identity matrix?

 

[DrClaude]

aa^-1a^-1abb

Please, you are not allowed to move things around. You have to take every matrix in the order it appears.

What I want you do to is to find two consecutive matrices that you can simplify into something else.

 

[negation]

Please, you are not allowed to move things around. You have to take every matrix in the order it appears.

What I want you do to is to find two consecutive matrices that you can simplify into something else.

Am I allowed to introduce matrices?

 

[DrClaude]

Am I allowed to introduce matrices?

Yes, if they do not change the result, either because you do the same on both sides of an equality (meaning you can only multiply on the extrement left or extreme right), or by introducing the identity matrix I (which is neutral in terms of multiplication).

But, in your case, you do not need to do that. What do you find in the middle of the product?

 

[negation]

Yes, if they do not change the result, either because you do the same on both sides of an equality (meaning you can only multiply on the extrement left or extreme right), or by introducing the identity matrix I (which is neutral in terms of multiplication).

But, in your case, you do not need to do that. What do you find in the middle of the product?

There's A^-1A which can be simplified to an identity I.

If this is valid, I have ABBA^-1, and base on def of first principle

BB = B^2

this gives

AB^2A^-1

 

[negation]

Yes, if they do not change the result, either because you do the same on both sides of an equality (meaning you can only multiply on the extrement left or extreme right), or by introducing the identity matrix I (which is neutral in terms of multiplication).

But, in your case, you do not need to do that. What do you find in the middle of the product?

AB^2A^-1 is correct. Very positive.
Would it also be valid to say that (ABC)^-1 =/= (A^-1B^-1C^-1)?

If the exponent is more than 2, is there an abstract proof I can perform without having to flood the paper with Ai=1 .. . . .. . .Ai=n?

 

[DrClaude]

AB^2A^-1 is correct. Very positive.

If the exponent is more than 2, is there an abstract proof I can perform without having to flood the paper with Ai=1 .. . . .. . .Ai=n?

For an exponent 2ⁿ, you have
(A B A^-1)²ⁿ = ((A B A^-1)²)^2n-1 = (A B² A^-1)^2n-1 which you can use to start the proof.

You can also show the result for an exponent 2, then 4, then get the general formula by induction.

 

[negation]

For an exponent 2ⁿ, you have
(A B A^-1)²ⁿ = ((A B A^-1)²)^2n-1 = (A B² A^-1)^2n-1 which you can use to start the proof.

You can also show the result for an exponent 2, then 4, then get the general formula by induction.

On a related note,

Would it also be valid to say that (ABC)^-1 == (A^-1B^-1C^-1)?
Since operation between inverse are different?

 

[DrClaude]

On a related note,

Would it also be valid to say that (ABC)^-1 == (A^-1B^-1C^-1)?
Since operation between inverse are different?

You have to reverse the order of the matrices:

(AB)^-1 = B^-1A^-1

 

[negation]

You have to reverse the order of the matrices:

(AB)^-1 = B^-1A^-1

Yes I understand I have to. The conudrum comes in when there are more than 2 matrix.

But I'd give it a try:

(ABC)^-1 = C^-1 B^-1 A^-1

 

[DrClaude]

Yes I understand I have to. The conudrum comes in when there are more than 2 matrix.

But I'd give it a try:

(ABC)^-1 = C^-1 B^-1 A^-1

Exactly. Just take them two at a time:

(ABC)^-1 = ((AB)C)^-1= C^-1(AB)^-1 = C^-1B^-1A^-1

 

[negation]

Just bear with me for a moment.

A^-1 = BC then C^-1 = AB is true.

(A^-1)^-1 = (BC)^-1 = C^-1 B^-1

(Inverse of an inverse is the original matrix)

A = C^-1 B^-1

I'm a little confused here. I know I have to multiply both sides but B but nebulous as regard the order.

But from what I at least know:

AB = C^-1 B^-1 B

AB = C^-1

 

[negation]

Exactly. Just take them two at a time:

(ABC)^-1 = ((AB)C)^-1= C^-1(AB)^-1 = C^-1B^-1A^-1

Can I then generalized the form:

(Ai. . . An-2, An-1, An)^-1 = (An)^-1, (An-1) ^-1, (An-2)^-1. . . (Ai)^-1 ? or would it be a cardinal sin?

 

[DrClaude]

Just bear with me for a moment.

A^-1 = BC then C^-1 = AB is true.

(A^-1)^-1 = (BC)^-1 = C^-1 B^-1

(Inverse of an inverse is the original matrix)

A = C^-1 B^-1

I'm a little confused here. I know I have to multiply both sides but B but nebulous as regard the order.

But from what I at least know:

AB = C^-1 B^-1 B

AB = C^-1

What you wrote is correct. You have indeed to multiply B from the right on both sides.

 

[DrClaude]

Can I then generalized the form:

(Ai. . . An-2, An-1, An)^-1 = (An)^-1, (An-1) ^-1, (An-2)^-1. . . (Ai)^-1 ? or would it be a cardinal sin?

Apart from the commas, yes, you can make that generalization.

 

[DrClaude]

Additional note: all we've been writing is only valid if each matrix is individually invertible.

 

[negation]

Apart from the commas, yes, you can make that generalization.

woops. It should be a '.' notation.

What about (-4A)^-1? There are no other matrice(s) for me to swap with.

 

[negation]

Additional note: all we've been writing is only valid if each matrix is individually invertible.

Yes I'm aware.

 

[DrClaude]

What about (-4A)^-1? There are no other matrice(s) for me to swap with.

By definition of the inverse, you have that (cA)^-1 = A^-1/c (where c is a scalar).

 

[Mark44]

Yes I understand I have to. The conudrum comes in when there are more than 2 matrix.

But I'd give it a try:

(ABC)^-1 = C^-1 B^-1 A^-1

Yes, ABC is the inverse of C^-1B^-1A^-1. To verify this, notice that (ABC) * (C^-1B^-1A^-1) = (AB)CC^-1(B^-1A^-1) = (AB)I(B^-1A^-1) = (AB)(B^-1A^-1) = (A)BB^-1(A^-1 = AIA^-1 = AA^-1 = I.

If the product of two matrices is I, the two matrices are inverses of one another. Since (ABC)(C^-1B^-1A^-1) = I, ABC and C^-1B^-1A^-1 are inverses.

Although matrix multiplication is not generally commutative (i.e., AB ≠ BA, in general), matrix multiplication is associative, a fact that I used multiple times in my work above.

Edit: I didn't notice that there was a 2nd page of posts in this thread...

 

[Ray Vickson]

Yes I understand I have to. The conudrum comes in when there are more than 2 matrix.

But I'd give it a try:

(ABC)^-1 = C^-1 B^-1 A^-1

There is no conundrum. Just write ABC = AD, where D = BC; this uses the "associative" property of matrix multiplication: A(BC) = (AB)C, which is why it makes sense to write ABC without parentheses. Anyway, (ABC)^(-1) = (AD)^(-1) = D^(-1) A ^(-1), and D^(-1) = C^(-1) B^(-1).