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 Post subject: inverse of matrices
PostPosted: Sat, 29 Oct 2011 19:18:11 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 matrices
PostPosted: Sat, 29 Oct 2011 20:33:14 UTC 
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kakasky2012 wrote:
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 matrices
PostPosted: Sat, 29 Oct 2011 20:52:39 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?


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 Post subject: Re: inverse of matrices
PostPosted: Sat, 29 Oct 2011 21:07:53 UTC 
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kakasky2012 wrote:
I don't understand. How did you figure out that no matrices A and B satisfy the condition stated in the problem?


Assume A+B-1=A-1+B-1, then subtract A+B-1 from both sides and you get 0=-1

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 Post subject: Re: inverse of matrices
PostPosted: Sat, 29 Oct 2011 21:17:41 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 matrices
PostPosted: Sat, 29 Oct 2011 21:27:52 UTC 
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kakasky2012 wrote:
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:

(A+B)(A^{-1}+B^{-1})=I+BA^{-1}+AB^{-1}=0.

Is this enough? Let's see.

If you've assumed that, then clearly you've implicitly assumed that both A and B 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 A+B is invertible yet. But consider

(A+B)(A^{-1}+B^{-1}), by the given, this is I, so A+B has a right inverse. Is it also the left-inverse? I'll leave that to you, but remember, A,B are square matrices, and the existence of a right inverse implies that (A+B) is surjective, hence. . .

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 Post subject: Re: inverse of matrices
PostPosted: Sat, 29 Oct 2011 21:37:34 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 matrices
PostPosted: Sat, 29 Oct 2011 21:40:51 UTC 
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kakasky2012 wrote:
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 matrices
PostPosted: Sun, 30 Oct 2011 10:21:45 UTC 
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Shadow wrote:
kakasky2012 wrote:
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:

(A+B)(A^{-1}+B^{-1})=I+BA^{-1}+AB^{-1}=0.

Is this enough? Let's see.

If you've assumed that, then clearly you've implicitly assumed that both A and B 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 A+B is invertible yet. But consider

(A+B)(A^{-1}+B^{-1}), by the given, this is I, so A+B has a right inverse. Is it also the left-inverse? I'll leave that to you, but remember, A,B are square matrices, and the existence of a right inverse implies that (A+B) is surjective, hence. . .


I think you mean (A+B)(A^{-1}+B^{-1})-I=I+BA^{-1}+AB^{-1}=0.

Or slightly differently, if we are prepared to break the symmetry, we can convert the condition I+BA^{-1}+AB^{-1}=0 into C=BA^{-1} satisfying C^2+C+I=0. That way we only need to implicitly assume A is invertible (and of course we get C has inverse -C-I as a result of C^2+C+I=0, so B is invertible), and continue on the same track from there onwards.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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