$n$-th power of a matrix

Question

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

Answer 1

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 $\operatorname{Id}$.

Answer 2

You can check by expanding

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

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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.

Answer 3

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 :)

Answer 4

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.