Matrix Algebra Find an inverse for I-A

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SUMMARY

The discussion centers on finding the inverse of the matrix expression I - A, where A is a nilpotent matrix satisfying A^n = 0 for some n > 1. The solution involves recognizing that for |r| < 1, the geometric series formula 1/(1-r) can be applied by substituting r with A. This leads to the conclusion that I - A can be expressed as (I - A)(I + A + A^2 + ... + A^{n-1}), confirming the inverse exists under the given conditions.

PREREQUISITES
  • Understanding of nilpotent matrices and their properties
  • Familiarity with geometric series and convergence
  • Knowledge of matrix algebra and operations
  • Basic concepts of linear transformations
NEXT STEPS
  • Study the properties of nilpotent matrices in linear algebra
  • Learn about geometric series and their applications in matrix theory
  • Explore matrix inversion techniques for different types of matrices
  • Investigate the implications of matrix transformations in higher dimensions
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Students of linear algebra, mathematicians, and anyone interested in advanced matrix theory and its applications in various fields.

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Suppose A^n=0 for some n>1. Find an inverse for I-A.

I don't understand how to do this. It isn't homework, but I'm just studying.
 
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If r were a real number with |r|<1 then 1/(1-r)=1+r+r^2+r^3+... It's a geometric series. Substitute r=A and see what you can conclude. Then prove it's true.
 
Or, essentially the same thing,
I- A^n= (I- A)(I+ A^2+ A^3+ \cdot\cdot\cdot+ A^{n-1})

And, of course, since A^n= 0, I- A^n= ?
 

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