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Arian.D
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Homework Statement
Hi guys,
I'm trying to prove that matrix inversion is continuous. In other words, I'm trying to show that in a normed vector space the map [itex] \varphi: GL(n,R) \to GL(n,R) [/itex] defined by [itex] \varphi(A) = A^{-1}[/itex] is continuous.
Homework Equations
The norm that we're working in the class is [itex] ||A|| = \sup\{ |AX| : |X| \leq 1 \} [/itex] where |X| refers to the Euclidean length of the vector X. So the topology we define on L(Rn) (the set of all linear transformations from Rn onto itself) is defined by the metric topology this norm induces on it.
The results that I already know are:
1- If A is a linear transformation in GL(n,R) and we have [itex] ||B-A|| . ||A^{-1}|| < 1 [/itex] then B is also in GL(n,R).
2- GL(n,R) is an open set in L(Rn).
The Attempt at a Solution
well, I attempted to prove it by the classical epislon-delta definition of continuity, even though I failed I conclude some things that might be useful for a solution:
What I should prove is:
[itex] \forall \epsilon>0 , \exists \delta>0: ||X-A||< \delta \implies ||X^{-1} - A^{-1}||< \epsilon [/itex]
We know that if X is invertible then the inverse of I-X is given by the series:
[tex](I-X)^{-1} = \sum_{k=0}^\infty X^k [/tex]
By writing [itex] X=I-(I-X)[/itex] it's easy to see that:
[tex] X^{-1} = \sum_{k=0}^\infty (I-X)^k[/tex]
So we have:
[tex] (X^{-1}A)^{-1} = A^{-1}X = \sum_{k=0}^\infty (I-X^{-1}A)^k[/tex]This implies that for any matrix X close to A in GL(n,R) (we have defined a topology on GL(n,R), so I can talk about closeness), I can write down:
[tex] A^{-1} = (\sum_{k=0}^\infty (I-X^{-1}A)^k) X^{-1} = \sum_{k=0}^\infty (X^{-1}(X-A))^k X^{-1} [/tex]
Now we can see that:
[tex] ||X^{-1} - A^{-1}|| = ||X^{-1} - \sum_{k=0}^\infty (X^{-1}(X-A))^k X^{-1}|| = ||\sum_{k=1}^\infty (X^{-1}(X-A))^k X^{-1}|| \leq \sum_{k=1}^\infty (||X^{-1}||.||(X-A)||)^k ||X^{-1}||[/tex]
On the other hand, if [itex] ||X-A|| < \delta [/itex] we can conclude that:
[tex] ||X-A||^k < \delta^k \implies ||X^{-1}||^k ||X-A||^k < ||X^{-1}||^k \delta^k \implies \sum_{k=1}^\infty ||X^{-1}||^k||X-A||^k < \sum_{k=1}^\infty ||X^{-1}||^k \delta^k \implies \sum_{k=1}^\infty ||X^{-1}(X-A)||^k < \sum_{k=1}^\infty ||X^{-1}||^k \delta^k [/tex]
[tex]\implies \sum_{k=1}^\infty (||X^{-1}||.||(X-A)||)^k ||X^{-1}|| < \sum_{k=1}^\infty ||X^{-1}||^k \delta^k ||X^{-1}||[/tex]
so far I've shown that:
[tex] ||X^{-1} - A^{-1}|| < \sum_{k=1}^\infty ||X^{-1}||^k \delta^k ||X^{-1}||[/tex]
Now if for any give epsilon I find delta in this inequality then I'm done:
[tex] \sum_{k=1}^\infty ||X^{-1}||^k \delta^k ||X^{-1}|| = \frac{\delta ||X^{-1}||^2}{1-\delta||X^{-1}||}< \epsilon[/tex]
If I could show that delta could be valuated as a function of epsilon then I was done, but unfortunately I have no idea on how to do that.
Maybe whatever I've done so far is nonsense or maybe I'm making it too hard. Any ideas on how to go further with my proof is appreciated. Also if you know a shorter way to prove that matrix inversion is continuous that would be great. I'm pretty bad with writing epsilon-delta continuity proofs I think.
I have another question too, is matrix multiplication continuous?
Oops, I noticed it just now, I posted it on the wrong section :/ Please move it to the homework section. Sorry for that.
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