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

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The discussion focuses on proving that the inverse of the matrix (I - A) can be expressed as a finite series when A raised to the fourth power equals zero. It is shown that (I - A)^{-1} equals I + A + A^2 + A^3 under the condition A^4 = 0. Additionally, if A raised to the power of (n+1) equals zero, the inverse can be represented as a series I + A + A^2 + ... + A^n. The proof involves verifying that multiplying (I - A) by the series yields the identity matrix. This confirms the validity of the series representation for the inverse in both cases.
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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
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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