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I came a cross a problem I could not solve today:

If matrices $A$ and $B$ satisfy $A^4 = 0$ and $B=I-A$, prove that $$B^{-1}=I+A+A^2+A^3$$

I doubt they expect me to start calculating 4 matrices that would have a,b,c,d values in them etc

I'd love to know how to work with a powered matrix :)

thanks in advance

Rodrigo de Azevedo
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Lostdawn
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4 Answers4

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They simply expect you to compute$$(\operatorname{Id}-A)\times\left(\operatorname{Id}+A+A^2+A^3\right)$$and to see that it is equal to $\operatorname{Id}$.

José Carlos Santos
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  • Lovely, did not think about it like that.. but if may I ask - is there a method to work with powered matrices ? or its just a blunt calculation – Lostdawn Dec 28 '19 at 09:04
  • I don't know what is it that you mean by “a method to work with powered matrices”. If you do the computation that I mentioned, you will get$$\operatorname{Id}-A+A-A^2+A^2-A^3+A^3-A^4=\operatorname{Id},$$since $A^4=0$. – José Carlos Santos Dec 28 '19 at 09:14
  • @Lostdawn All you need here is $A^nA^m=A^{n+m}$ -- and the problem works just as well if $A$ with $A^4=0$ and $B=1-A$ are taken from any ring – Hagen von Eitzen Dec 28 '19 at 09:15
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You can check by expanding

$$(I-A)(I+A+A^2+A^3)=(I+A+A^2+A^3)(I-A)=I-A^4.$$

You can also consider the generalization of the entire series

$$\frac1{1-a}=1+a+a^2+a^3+a^4+\cdots$$ which converges for $|a|<1$.

Here,

$$(I-A)^{-1}=1+A+A^2+A^3+A^4+\cdots$$ which converges for an obvious reason.

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You could notice that:

$$B=\sum_{i=0}^{3} A^i$$

That is a ""geometric series"" so we can apply the known formula(more formally you can because matrix product is associative and matrix sum is associative and commutative so you can prove the analogous formula for matrixes):

$$B=(I-A^4)(I-A)^{-1}$$

By hyphotesys $A^4=0$:

$$B=(I-A)^{-1}$$ $$B^{-1}=I-A$$

This is why it works :)

Kandinskij
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Since $A^4=0$, $A^n=0$ for $n\geqslant 4$. Then trivially the Neumann series $\sum_{k=0}^\infty = \sum_{k=0}^3 T^k$ converges in the operator norm, and hence $$ \mathrm({Id}-A)^{-1} = \sum_{k=0}^4 T_k = 1+T+T^2+T^3, $$ as was to be shown.

Math1000
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