# Linear Algebra Matrix Proof

1. Oct 2, 2013

### mreaume

1. The problem statement, all variables and given/known data

Prove or disprove the following statements. I and 0 denote respectively the identity and zero matrix of the same size as A. If A is a square matrix such that A^2 - 3A +2I = 0 then A-cI is invertible whenever c is not equal to 1 and c is not equal to 2.

2. Relevant equations

3. The attempt at a solution

I have factored the function to: (A-2I)(A-I)=0.

However, we can't assume that A=2I and A=I because we are dealing with matrices (i.e. two non zero matrices can produce the zero matrix when multiplied together). I have a feeling that Eigenvectors might be related to this question, but I don't know how to apply that concept in this scenario.

2. Oct 2, 2013

### LCKurtz

Do you have the Cayley-Hamilton theorem available?

3. Oct 2, 2013

### Dick

Yes, think about eigenvalues indirectly. Suppose A-cI is not invertible. That means there is a nonzero vector v such that (A-cI)v=0. So Av=cv. Can you use that to show (A^2-3A+2I)v=0 can only be true if c=2 or c=1?

Last edited: Oct 2, 2013
4. Oct 2, 2013

### mreaume

LCKurtz, I do not have the Cayley-Hamilton theorem available. It has not been presented in class yet.

Dick, I have understood the following:

(A-cI)v=0
So Av=cv
I imagine that we need det(A-cI)=0 in order to prove that A-cI is invertible.

I am stuck here. I don't know how to relate A^2 -3A +2I to the previous equations.

5. Oct 2, 2013

### Office_Shredder

Staff Emeritus
Go with Dick's suggestion. If Av = cv, try to evaluate (A2 - 3A + 2I)v, which you know has to be zero by the statement of the problem.

6. Oct 2, 2013

### mreaume

Ok. Here is what I have:

(A^2 -3A +2I) = 0
A^2v -3Av +2Iv = 0
AAv -3Av + 2v = 0
(c^2)v - 3cv + 2v = 0
(c^2 - 3c + 2)v = 0
((c-2)(c-1))v=0

Therefore (A-cI) is not invertible when c=1 c=2.

I think I understand now. Does everything seem to be in order in my procedure?

7. Oct 2, 2013

### Dick

Yes, that's it. The important point is that (c-2)(c-1)v=0 is ONLY possible if c=1 or c=2, since v is nonzero.

8. Oct 2, 2013

9. Oct 3, 2013

### HallsofIvy

Staff Emeritus
Your last statement is wrong- you are asserting that "(A-cI) is not invertible when c=1 c=2" when what you want to prove is "A- cI is invertible as long as long as $c\ne 1$ and $c\ne 2$", the converse. Did you mean to say "(A-cI) is not invertible only when c=1 c=2"?

10. Oct 3, 2013

### Mandelbroth

OP, you should definitely not take this as excessively pedantic. What you have written, though somewhat correct in the vernacular language, is incorrect in terms of mathematical formalism.

11. Oct 3, 2013

So then how *can* you show that (A-cI) **is** invertible when c /= 1, 2?

12. Oct 3, 2013

### Mandelbroth

It is shown. What he has said is different from that.

13. Oct 3, 2013

Sorry "/= 1,2" is meant as "not equal to 1, 2" which, as HallsofIvy pointed out, is not proven.

14. Oct 3, 2013

### Dick

It is if you the organize the thoughts correctly. We assumed A-cI is NOT invertible and concluded that then there is a nonzero vector such that (c-1)(c-2)v=0. That's a contradiction unless c=1 or c=2. Hence if c is not equal to 1 or 2 then A-cI IS invertible. Treat it as a proof by contradiction.

15. Oct 8, 2013