Matrix Invertibility and the Minimal Polynomial

[jonnyc1003]

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!

 

[micromass]

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

Perhaps the easiest way to approach this is to calculate the eigenvalues...

 

[jonnyc1003]

I cannot calculate the eigenvalues as A is an arbitrary matrix. The information in my original post is all that was given.

 

[micromass]

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??

 

[jonnyc1003]

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?

 

[micromass]

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??

 

[Zoe-b]

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?

 

[AlephZero]

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

 

[micromass]

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.

 

[jonnyc1003]

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.

 

[AlephZero]

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..."