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 Post subject: inverse of matricesPosted: Sat, 29 Oct 2011 19:18:11 UTC

Joined: Sun, 16 Oct 2011 20:09:16 UTC
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Problem: Find conditions that the matrices A and B have to satisfy in order for the following to be valid: (A+B)−1=A−1+B−1.

My solution. The first condition is that A and B have to have the same dimensions. Then by expanding the two equalities: (A+B)⋅(A−1+B−1)=I and (A−1+B−1)⋅(A+B)=I, I deduced the second condition: A and B have both to be invertible, and third condition is: \Rightarrow AB^{-1}+BA^{-1} =A^{-1}B+B^{-1}A. Is this answer enough? or there are other conditions that A and b have to satisfy?

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 20:33:14 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Problem: Find conditions that the matrices A and B have to satisfy in order for the following to be valid: (A+B)−1=A−1+B−1.

My solution. The first condition is that A and B have to have the same dimensions. Then by expanding the two equalities: (A+B)⋅(A−1+B−1)=I and (A−1+B−1)⋅(A+B)=I, I deduced the second condition: A and B have both to be invertible, and third condition is: \Rightarrow AB^{-1}+BA^{-1} =A^{-1}B+B^{-1}A. Is this answer enough? or there are other conditions that A and b have to satisfy?

Matrices form a ring, if you have your condition that implies that 1=0, which is impossible, so NO matrices satisfy that.

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 20:52:39 UTC

Joined: Sun, 16 Oct 2011 20:09:16 UTC
Posts: 8
I don't understand. How did you figure out that no matrices A and B satisfy the condition stated in the problem?

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 21:07:53 UTC
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I don't understand. How did you figure out that no matrices A and B satisfy the condition stated in the problem?

Assume , then subtract from both sides and you get

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 21:17:41 UTC

Joined: Sun, 16 Oct 2011 20:09:16 UTC
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Sorry this is what I mean by my condition:

(A+B)^{-1} = A^{-1} + B^{-1} i.e inverse of sum of two matrices is equal to the sum of inverse of A and inverse of B

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 21:27:52 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Sorry this is what I mean by my condition:

(A+B)^{-1} = A^{-1} + B^{-1} i.e inverse of sum of two matrices is equal to the sum of inverse of A and inverse of B

Ah, well let's see then, you did the right first thing:

.

Is this enough? Let's see.

If you've assumed that, then clearly you've implicitly assumed that both and are invertible and have the same dimension, so that is covered. Now does this give you the rest of what you want? Well, you don't know that is invertible yet. But consider

, by the given, this is , so has a right inverse. Is it also the left-inverse? I'll leave that to you, but remember, are square matrices, and the existence of a right inverse implies that is surjective, hence. . .

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 21:37:34 UTC

Joined: Sun, 16 Oct 2011 20:09:16 UTC
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Sorry, I understand what you did but I am still not able to go farther. Can you continue your proof?

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 Post subject: Re: inverse of matricesPosted: Sat, 29 Oct 2011 21:40:51 UTC
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Sorry, I understand what you did but I am still not able to go farther. Can you continue your proof?

What part don't you get?

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 Post subject: Re: inverse of matricesPosted: Sun, 30 Oct 2011 10:21:45 UTC
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Sorry this is what I mean by my condition:

(A+B)^{-1} = A^{-1} + B^{-1} i.e inverse of sum of two matrices is equal to the sum of inverse of A and inverse of B

Ah, well let's see then, you did the right first thing:

.

Is this enough? Let's see.

If you've assumed that, then clearly you've implicitly assumed that both and are invertible and have the same dimension, so that is covered. Now does this give you the rest of what you want? Well, you don't know that is invertible yet. But consider

, by the given, this is , so has a right inverse. Is it also the left-inverse? I'll leave that to you, but remember, are square matrices, and the existence of a right inverse implies that is surjective, hence. . .

I think you mean .

Or slightly differently, if we are prepared to break the symmetry, we can convert the condition into satisfying . That way we only need to implicitly assume is invertible (and of course we get C has inverse as a result of , so B is invertible), and continue on the same track from there onwards.

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