Matrix invertibility question

1. Dec 6, 2011

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!

2. Dec 6, 2011

micromass

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

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

3. Dec 6, 2011

jonnyc1003

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

4. Dec 6, 2011

micromass

Have you heard about minimal/characteristic polynomials??

5. Dec 6, 2011

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?

6. Dec 6, 2011

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??

7. Dec 6, 2011

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?

Last edited: Dec 6, 2011
8. Dec 6, 2011

AlephZero

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

9. Dec 6, 2011

micromass

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. Dec 6, 2011

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.

11. Dec 6, 2011

AlephZero

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...."