Proving Linear Algebra Statement: A and B in Mn(R) with B invertible

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SUMMARY

The discussion focuses on proving the linear algebra statement involving matrices A and B in Mn(R), where B is invertible. The goal is to demonstrate that |A - cI| = |B-1AB - cI|. Key concepts include the determinant properties of matrix products, specifically that |AB| = |A||B| and |B-1| = 1/|B|. The user seeks guidance on applying these properties to solve the problem.

PREREQUISITES
  • Understanding of determinants in linear algebra
  • Familiarity with matrix operations and properties
  • Knowledge of invertible matrices and their characteristics
  • Basic concepts of eigenvalues and eigenvectors
NEXT STEPS
  • Study the properties of determinants, including |AB| = |A||B|.
  • Learn about the implications of matrix invertibility on determinants.
  • Explore the relationship between eigenvalues and the determinant of a matrix.
  • Investigate proofs involving matrix similarity and their determinants.
USEFUL FOR

Students studying linear algebra, particularly those tackling matrix theory and determinant properties, as well as educators looking for examples of matrix proofs.

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Homework Statement



If A and B [tex]\in[/tex] Mn(R) and B is invertible

show that

l A-cI l = l B-1AB-cI l

Homework Equations



N/A

The Attempt at a Solution



i've no idea how to prove this. can give me any clue?

l AI-cI l = l ABB-1-cI l

am i in the right way?
 
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recall that for square matrices

[tex] <br /> det(AB) = det(A)det(B)<br /> [/tex]

and for an invertible square matrix,

[tex] <br /> det(B^{-1}) = det(B)^{-1}<br /> [/tex]

How can you apply these?
 

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