Matrix Invertibility and the Minimal Polynomial

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Discussion Overview

The discussion revolves around the question of whether a matrix \( A \) satisfying the equation \( A^2 - A + I = 0 \) is invertible. Participants explore various approaches to demonstrate the invertibility of \( A \), including the use of determinants, eigenvalues, and polynomial properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that since \( \det(A^2 + I) \) will never be zero, \( \det(A) \) must be non-zero, implying \( A \) is invertible.
  • Another participant questions the assertion that \( \det(A^2 + I) \) cannot be zero and proposes calculating eigenvalues instead.
  • A participant notes the challenge of calculating eigenvalues without additional information about \( A \) being arbitrary.
  • Discussion includes the Cayley-Hamilton theorem, with one participant suggesting that \( \det(A) = 1 \) leads to invertibility.
  • Another participant argues that the equation \( A^2 - A + I = 0 \) can be rearranged to show \( A \) is invertible by expressing \( I \) in terms of \( A \).
  • Concerns are raised about the relevance of the characteristic polynomial and minimal polynomial in this context, with some participants debating their necessity.
  • One participant emphasizes that the minimal polynomial divides \( A^2 - A + I \), which relates to the eigenvalues of \( A \).
  • There is acknowledgment that without knowing the size of \( A \), implications about eigenvalues remain uncertain.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of using the minimal polynomial and characteristic polynomial. There is no consensus on the best approach to demonstrate the invertibility of \( A \), and the discussion remains unresolved regarding the implications of the matrix's size on the eigenvalues.

Contextual Notes

Participants note limitations in their arguments based on the lack of information about the size of matrix \( A \) and the assumptions regarding its properties.

jonnyc1003
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I just finished a final in a linear algebra course and was unsure about one of the questions. The question was:

If A^2 - A + I = 0 , show that A is invertible.

My approach was that det(A^2 + I) = det(A)

det(A^2 + I) will never be zero, so det(A) is non-zero and therefore A is invertible.

Is this the right way of doing this problem?

Thanks!
 
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Wh will det(A^2+I) never be 0??

Perhaps the easiest way to approach this is to calculate the eigenvalues...
 
I cannot calculate the eigenvalues as A is an arbitrary matrix. The information in my original post is all that was given.
 
jonnyc1003 said:
I cannot calculate the eigenvalues as A is an arbitrary matrix. The information in my original post is all that was given.

Have you heard about minimal/characteristic polynomials??
 
Oh I see now.

A2 - A + I = 0

A2 - A + det(A)*I = 0 <-- from Cayley-Hamilton

So det(A)=1 and therefore A is invertible

Is this where you were leading me?

I can't recall now whether we were told A is 2x2. Does that matter?
 
Yes, but it needs some tweaking.

For example, the characteristic polynomial doesn't need to be x^2-x+1. But can we say something about the minimal polynomial??
 
Sorry to butt in, but I can't see why you need to use the minimal polynomial or Cayley-Hamilton Theorem here..
You have:
A^2 - A + I = 0

I = A - A^2
I = A(I-A)

So inv(A) = (I - A) and A must be invertible..?
Sorry if I've made a mistake but this is sound as far as I can tell..

Also: this may not be the most obvious way of doing it- you can use minimal polynomials but Jonny I have no idea how you're arguing that the Cayley-Hamilton Theorem implies A2 - A + det(A)*I = 0 ..
The question tells you that A^2 - A + I = 0, what does this imply about the minimal polynomial? And from there what can you deduce about the characteristic polynomial, and hence about the eigenvalues?
 
Last edited:
Zoe-b said:
I = A(I-A)

So inv(A) = (I - A) and A must be invertible..?

I'll buy that.

I don't see where the characteristic polynomial would lead to. You know a factor of it, but if the order of A is > 2, you don't know anything about the other factors.

... unless Micromass has seen something that I haven't
 
AlephZero said:
I'll buy that.

I don't see where the characteristic polynomial would lead to. You know a factor of it, but if the order of A is > 2, you don't know anything about the other factors.

... unless Micromass has seen something that I haven't

Of course the method Zoe posted is superior. But the method with the characteristic polynomial works too. The point is that the minimal polynomail is the least P(X) such that P(A)=0. So if a polynomial is 0 in A then the minimal polynomial divides it.

In our case, we know that the minimal polynomial will divide A^2-A+I.

The crucial part now is that the eigenvalues of our matrix are exactly the roots of the minimal polynomial. So seeing A^2-A+I=0 immediately gives us information about the eigenvalues.
 
  • #10
I believe you've been foiled micromass. As I said, I didn't think the problem mentioned that A was 2x2, which if I'm wrong, and I may well be, you really can't imply anything about the eigenvalues from the given equation.

Mind you, this is a 600 level linear algebra course, and also my first course in linear algebra, so I'm obviously naive on the topic. It appears however, that Zoe-b's solution is THE solution in this case.

Feel free to give your whole spiel now micromass, as I'm satisfied on the topic with the given solution.
 
  • #11
micromass said:
In our case, we know that the minimal polynomial will divide A^2-A+I.

The crucial part now is that the eigenvalues of our matrix are exactly the roots of the minimal polynomial. So seeing A^2-A+I=0 immediately gives us information about the eigenvalues.

OK. After thinking about the special case where A is diagonal, I get it.

Before, I was thinking "yeah, but the other N-2 eigenvalues might be anything..."
 

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