# Linear algebra: eigenvalue & character polynomials proof

1. Dec 11, 2011

### ISuckAtMath

we are given B = CAC^-1

Prove that A and B have the same characteristic polynomial
given a hint: explain why ƛIn = CƛInC^-1

what I did was:
B = CAC^-1
BC = CA
Det(BC) = Det(CA)
Det(B) Det(C) = Det(C) Det(A)
Now they’re just numbers so I divide both sides by Det(C)
Det(B) = Det(A)

not im stuck

2. Dec 11, 2011

### AlephZero

Re: need help with a eigenvalue determinant proof

The determinant is one term in the characteristic polynomial, but you need more than that to show the characteristic polynomials are the same.

The eigenvalues are the roots of the characteristic polynomial so you need to show that every eigenvalue of B is also eigenvalue of A, and vice versa.

If λ is an eigenvalue of B, then there is a vector x such that Bx = λx.

So you need to find a vector y such that Ay = λy.

You are only given one thing that connects A and B, so just plug and chug...

3. Dec 11, 2011

### ISuckAtMath

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

Suppose that C is an invertible matrix, and you are told that

B = CA(C^-1)

prove that A and B have exactly the same characteristic polynomial

do not assume A and B are triangular or diagonal matrices

2. Relevant equations

given hint: explain why ƛIn = CƛIn(C^-1)

3. The attempt at a solution

B = CA(C^-1)

BC = CA

Det(BC) = Det(CA)

Det(B) Det(C) = Det(C) Det(A)

now that they're just numbers, i divded both sides by Det(C)

Det(B) = Det(A)

i don't know what to do next

4. Dec 11, 2011

### Dick

That's fine so far. But it doesn't help. What expression involving Det gives you the characteristic polynomial?

5. Dec 11, 2011

### ISuckAtMath

well i know det(ƛIn - A)= 0

thus giving the characteristic polynomial (ƛ-ƛ1)(ƛ-ƛ2)....(ƛ-ƛn)

do i set set det(ƛIn - A) = det(ƛIn - B)?

with the given vector v: Av = ƛv and Bv = ƛv

therefore Av = Bv?

6. Dec 11, 2011

### Dick

Ok, so the characteristic polynomial for B=CAC^(-1) is det(CAC^(-1)-lambda*I). Now pay attention to the hint.

7. Dec 11, 2011

### Rockoz

You would want to start with det(λI - A) and conclude that it equals det(λI-B), i.e. their characteristic polynomials are the same. Starting with det(λI-A), think about the relationship between A and B, then the hint, and recall that matrice have a distributive property which allows you to factor.

Last edited: Dec 11, 2011
8. Dec 12, 2011

### micromass

Staff Emeritus
Re: need help with a eigenvalue determinant proof

Note that proving that the eigenvalues are thesame for A and B isn't enough. You also need to prove that the multiplicities are the same.

9. Dec 12, 2011

### ISuckAtMath

is this correct?

B = CAC^(-1)

B-λIn = CAC^(-1) - λIn

using the hint: λIn = CλInC^(-1)

B-λIn = CAC(^-1) - CλInC^(-1)

B-λIn = C[ AC^(-1) - λInC^(-1) ] factored out C

B-λIn = CC^(-1)( A-λIn)

B-λIn = A-λIn, CC^(-1) cancel each other out

therefore det(B-λIn) = det(A-λIn)

so they're the same characteristic polynomial

10. Dec 12, 2011

### Deveno

Re: need help with a eigenvalue determinant proof

what you want to do is compare:

det(xI - A) and det(xI - B) = det(xI - CAC-1).

here is a hint:

CIC-1 = I

11. Dec 12, 2011

### Dick

No! The only way B-lambda*I=A-lambda*I is if A=B!. You can't prove that. You went wrong when you changed AC^(-1) into C^(-1)A in the third line. You can't do that. Matrices don't necessarily commute. Just factor the C^(-1) out on the right and the C on the left. Then take the det.

Last edited: Dec 12, 2011
12. Dec 12, 2011

### AlephZero

Re: need help with a eigenvalue determinant proof

Oops. But that is easily fixed up. Just start the argument by saying

If λ is a diagonal matrix of all the eigenvalues of B, then there is a matrix x such that Bx = λx.

13. Dec 12, 2011

### taxidriverhk

Re: need help with a eigenvalue determinant proof

You are trying to prove that det(λI - B) = det(λI - A)
and there is basically no way to prove that B = A because they are really not equal to each other

You are given a hint that λI = CλIC-1
and B = CAC-1
then you substitute these two into det(λI - B)

you should be able to prove that det(λI - B) = det(λI - A)

*you will need to use the properties of determinants, and also matrix multiplication is associative, be careful when factoring any matrix out...

Last edited: Dec 12, 2011