Can You Solve the Nonsingular Matrix Inverse Problem?

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SUMMARY

The discussion centers on solving the nonsingular matrix inverse problem, specifically for an $n \times n$ matrix $A$ with a characteristic polynomial defined as \(p(\lambda)=\lambda^n+a_1\lambda^{n-1}+\ldots+ a_{n-1}\lambda+ a_n\). The solution, provided by participant Sudharaka, demonstrates that the inverse of matrix $A$ can be expressed as \(A^{-1}=-\frac{1}{a_n}\left( A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n\right)\). This formula is crucial for understanding matrix inverses in linear algebra.

PREREQUISITES
  • Understanding of characteristic polynomials in linear algebra
  • Knowledge of matrix operations and properties of nonsingular matrices
  • Familiarity with the concept of matrix inverses
  • Basic proficiency in mathematical notation and manipulation
NEXT STEPS
  • Study the derivation of matrix inverses using characteristic polynomials
  • Explore the implications of nonsingularity in matrix theory
  • Learn about Cayley-Hamilton theorem and its applications
  • Investigate numerical methods for computing matrix inverses
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in advanced matrix theory and applications in computational mathematics will benefit from this discussion.

Chris L T521
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $A$ be an $n\times n$ matrix whose characteristic polynomial is
\[p(\lambda)=\lambda^n+a_1\lambda^{n-1}+\ldots+ a_{n-1}\lambda+ a_n.\]
If $A$ is nonsingular, show that
\[A^{-1}=-\frac{1}{a_n}\left( A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n\right).\]

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Hint:
Use the Cayley-Hamilton theorem.

 
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This week's problem was correctly answered by Sudharaka. You can find his solution below.

Using the Cayley-Hamilton theorem we get,

\[p(A)=A^{n} + a_1A^{n-1} + \ldots + a_{n-1}A+a_{n}I_n=0_n\]

where \(0_n\) is the \(n\times n\) zero matrix.

Since \(A\) is non-singular multiplying by \(A^{-1}\) we get,

\[A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n+a_n A^{-1}=0_n\]

\[\therefore A^{-1}=-\frac{1}{a_n}\left( A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n\right)\]
 

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