Can (I-A)^{-1} Be Expressed as a Series When A^4 = 0?

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    Polynomial Proof
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The discussion establishes that for a square matrix A where \( A^4 = 0 \), the inverse \( (I - A)^{-1} \) can be expressed as the finite series \( I + A + A^2 + A^3 \). Furthermore, if \( A^{n+1} = 0 \), the inverse can be generalized to \( (I - A)^{-1} = I + A + A^2 + ... + A^n \). The proof utilizes the property of matrix multiplication and the definition of the inverse, confirming the relationship through polynomial expansion.

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delgeezee
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Let A be a square matrix,

a) show that $$(I-A)^{-1}= I + A + A^2 + A^3 if A^4 = 0$$

b) show that $$(I-A)^{-1}= I + A + A^2+...+A^n $$ if $$
A^{n+1}= 0$$
 
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Re: polynomial problem proof?

B is the inverse of A iff AB = BA = I
so
try
(I -A )( I + A + A^2 + A^3) = I-A + A - A^2 + A^2 - A^3 + A^3 - A^4 = I - A^4 = I
 

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