Eigenvalues Problem: Show Inverse of Diagonalizable Matrix A

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Discussion Overview

The discussion revolves around demonstrating that the inverse of a diagonalizable matrix \( A \) can be expressed as a polynomial \( q(A) \) of degree less than \( n \), where \( n \) is the degree of the characteristic polynomial of \( A \). The context includes theoretical exploration and mathematical reasoning related to eigenvalues and the Cayley-Hamilton theorem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant states that if \( A \) is a diagonalizable matrix with eigenvalues \( x_1, x_2, \ldots, x_n \), the characteristic polynomial can be expressed as \( p(x) = a_1 x^n + a_2 x^{n-1} + \ldots + a_{n+1} \).
  • Another participant references the Cayley-Hamilton theorem, suggesting that \( p(A) = 0 \) can be utilized in the proof.
  • A further contribution explains that from \( p(A) = 0 \), one can rearrange the equation to isolate the identity matrix and subsequently manipulate it to express the inverse of \( A \) in terms of a polynomial.
  • It is noted that the assumption is made that none of the eigenvalues of \( A \) is zero, which is crucial for the existence of the inverse.

Areas of Agreement / Disagreement

Participants generally agree on the application of the Cayley-Hamilton theorem and the approach to show that the inverse can be expressed as a polynomial. However, the discussion does not reach a consensus on the specifics of the polynomial form or the implications of the assumptions made.

Contextual Notes

The discussion assumes the diagonalizability of matrix \( A \) and the non-zero nature of its eigenvalues, which are critical for the validity of the arguments presented. There are also unresolved steps regarding the manipulation of the characteristic polynomial and the exact form of the polynomial \( q(A) \).

tc
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let A be a diagonalizable matrix with eignvalues = x1, x2, ..., xn
the characteristic polynomial of A is
p (x) = a1 (x)^n + a2 (x)^n-1 + ...+an+1
show that inverse A = q (A) for some polynomial q of degree less than n
 
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By cayley-hamilton p(A)=0, now what can you do with that?
 
Yeah, you use cayley-hamilton theorem
so, you have p(A)=0...
That implies a*A^n + b*A^(n-1) + c*A^(n-2)...+ I= 0
(i've used a,b,c as coefficients)...then take the identity matrix to the other side. Multiply both sides by inverse of A. Then RHS becomes -A^(-1) the LHS shows that the characteristic polynomial is of degree < n.
 
of course, we assuming that none of the eigenvalues is zero, too.
 
o..thx matt and mansi
 

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