Let $A$ be a squared matrix, and suppose there exists an $n\in \Bbb N$ in a way that $A^n=0$. Show that $I-A$ is invertible and that $(I-A)^{-1}=I+A+\cdots+A^{n-1}$
I don't have a clue where to start from.
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related question http://math.stackexchange.com/questions/336869/expanding-ia-1/ – Oiale Apr 05 '13 at 18:12
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duplicate question – chango Apr 05 '13 at 18:14
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Why the downvote? – Git Gud Apr 05 '13 at 18:14
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There are several related questions. Just google "nilpotent matrix invertible". (Note that $A^n = O$ for some $n$ is equivalent to $A$ being nilpotent.) – TMM Apr 05 '13 at 18:15
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1Of course this has nothing to do with matrices, it holds in every ring. – Martin Brandenburg Apr 05 '13 at 18:22
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@MartinBrandenburg my proof doesn't use matrices at all and juliens only a bit. – Dominic Michaelis Apr 05 '13 at 18:28
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Set $B:=I+A+\ldots+A^{n-1}$ and compute $(I-A)B$ by expanding the product: $$ (I-A)B=(I-A)(I+A+\ldots+A^{n-1})=I+A+\ldots+A^{n-1}-A(I+A+\ldots+A^{n-1}) $$ $$ =I+A+\ldots+A^{n-1}-A-A^2-\ldots-A^{n-1}-A^n=\;? $$
Observe the cancellations: you are left with $I$. This suffices to show that $I-A$ is invertible with inverse $B$. This is by the rank-nullity theorem: a square matrix is right invertible if and only if it is invertible. Or you can observe that $B$ and $I-A$ commute, so $(I-A)B=I$ yields also $B(I-A)=I$.
Julien
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@DominicMichaelis I actually refrained from doing the quasinilpotent generalization in a unital Banach algebra...That would have been closer to your answer. – Julien Apr 05 '13 at 18:17
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How do you check a matrix $B$ is the inverse of matrix $A$? You multiply them together and see if you get the identity matrix. This is just the definition of inverse matrix.
You have a matrix $I-A$ and you want to check that its inverse is $1+A+A^2+...+A^{n-1}$ given that $A^n=0$.
So multiply together and after doing lots of cancellations etc you get $I - A^n = I$, which is what we were after!
fretty
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2I didn't read any of the answers a priori...I just typed my own. However now I have read his answer I feel that Julian's is a bit more technically phrased (the question does not need to go into the rank-nullity theorem and the abstractness of quasinilpotents in unitary Banach algebras, I doubt the person asking the question wants this either). I decided to take a down to earth approach by first explaining what we were trying to do in a simple manner. Of course the maths is the same but on a question like this surely 99% of answers will be! I guess it is up to the person asking the question... – fretty Apr 05 '13 at 18:43
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I think my answer is very down to earth. I mentioned the rank-nullity theorem just because it is a good opportunity to recall that left invertible is enough with square matrices. (You omitted to mention that the product had to be $I$ in both directions). Then if you read, I mention that the two matrices commute, so it is not needed. And finally, I only mentioned quasinilpotents in comments. This being said, I upvoted your answer. – Julien Apr 05 '13 at 20:20
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Well wenn $A^n=0$ then we konw that every eigenvalue of $A$ is zero, taking jordannormalform one sees that $I-A$ is invertible cause it is similar to a triangular matrix with every value 1 on the diagonal. Afterwards you make something like the geommetric series
$$\frac{1}{1-x}= 1+x+x^2+\dots$$ now we have matrices but it is essentially the same it give syou $$(I-A)^{-1}=1+ A +A^2 + \dots+ A^{n-1} + \underbrace{A^n}_{=0}+\dots$$ But as every $A^k$ for $k\geq n$ is zero it is finite.
As Git mentioned it is unneccessary to go to infty
We can even take the finite geometric series which says that $$\sum_{i=0}^{n-1} x^i = \frac{x^n -1}{x-1}=\frac{1-x^n}{1-x}$$
We have $$(I-A)^{-1}= (I-A^n) \cdot (I-A)^{-1} = \sum_{i=0}^{n-1} A^i = I + A + A^2 + \dots + A^{n-1}$$
Julien
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Dominic Michaelis
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Under which conditions for the matrix $A$ is this formula $(I-A)^{-1} = I + A + A^2 + ...$ valid ? – Curtis74 Jun 17 '20 at 16:33