# Linear Algebra Matrix Proof

## Homework Statement

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.

## 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.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

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.

## 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.

Do you have the Cayley-Hamilton theorem available?

Dick
Homework Helper
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:
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.

Office_Shredder
Staff Emeritus
Gold Member
2021 Award
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.

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?

Dick
Homework Helper
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?

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.

HallsofIvy
Homework Helper
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.
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"?

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

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"?
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.

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

So then how *can* you show that (A-cI) **is** invertible when c /= 1, 2?
It is shown. What he has said is different from that.

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

Dick