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Say I have a matrix $B$ that is skew symmetric ($B^T=-B$). I want to show that for $A=(I+B)(I-B)^{-1}$ that $A^T=A^{-1}$ is true. All we know is that $B$ is square and that $(I-B)$ is non singular.

Here is how I have approached this:

It suffices to find $A^T$ and verify that $A A^T = I$.

$$A^T=[(I+B)(I-B)^{-1}]^T = [(I-B)^{-1}]^{T}(I+B)^T = (I^T-B^T)^{-1}(I^T+B^T)$$ $$=(I+B)^{-1}(I-B)$$

For this part I have to assume $B$ is non singular, which I think makes this approach inaccurate. Now I verify that $AA^T=I$

$$AA^T = (I+B)(I-B)^{-1}(I+B)^{-1}(I-B) =(I+B)(I-B^{-1})(I+B^{-1})(I-B) $$ $$=(I-B^{-1}+B-BB^{-1})(I-B+B^{-1}-B^{-1}B) = (-B^{-1}+B)(-B+B^{-1})$$ $$= I-B^{-1}B^{-1}+-BB+I$$ which clearly does not equal $I$

Is my transpose incorrect? I have found $A^{-1}$ explicitly to be $A^{-1} = (I-B)(I+B)^{-1}$. So I imagine that the transpose should equal that, but I cannot see what I am doing wrong.

Edit: So using Owen's info I was able to figure this out. The transpose was correct but if I expand it I get $$ A^T = (I+B)^{-1}(I-B) = I+B^{-1}-B-B^{-1}B = I-B+B^{-1}-BB^{-1}$$ $$=(I-B)I+(I-B)B^{-1}=(I-B)(I+B^{-1}) = (I-B)(I+B)^{-1} = A^{-1}$$

Edit2: So apparently I have been making an assumption that is not true. I have been saying that $(I-B)^{-1}=(I-B^{-1})$ which is not true (Here is a related question). To prove this $$(I-B)^{-1}=(I-B^{-1}) \Rightarrow (I-B)(I-B)^{-1}=(I-B)(I-B^{-1})$$ $$I\neq I-B-B^{-1}+BB^{-1} $$

So a new approach would be to $$A^T = (I+B)^{-1}(I-B) = (I+B)^{-1}-(I+B)^{-1}B$$ If I can show that $(I+B)^{-1}B = B(I+B)^{-1}$ then I have what I need to prove the claim, but I am not sure how to accomplish this.

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rioneye
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1 Answers1

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The key observation is that $(I+B), (I-B), (I+B)^{-1}, (I-B)^{-1}$ are all in the algebra generated by $B$. So in particular they commute. Then you can rearrange things to get them to multiply together to equal $I$.

Edit: Let me explain further. Let $B: V \rightarrow V$ be a linear transformation. Let $(B)'=\{T: V\rightarrow V, s. t. BT=TB\}$. Note that $IB=B=BI$ so $I\in (B)'$. Further $B^{-1}B=I=BB^{-1}$ so $B^{-1}\in(B)'$. If $S, T\in (B)'$ then $(S+T)B=SB+TB=BS+BT=B(S+T)$. So $(B)'$ is closed under addition. Let $S\in(B)'$. Let $T=S^{-1}$. Let $v\in V$.

Consider $STB(v)=B(v)$ Also since $S\in(B)'$ we have $SBT(v)=BST(v)=B(v)$. So $STB(v)=SBT(v)\Rightarrow TB(v)=BT(v)$ this is true for all $v\in V$ so $TB=BT$. Thus $(B)'$ is closed under taking inverses. This is enough. You can actually show that everything in $(B)'$ is of the form $\sum_{i=o}^n a_iB^i$ for some constants $a_i$ but you don't need this.

Owen Sizemore
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