Linear algebra: eigenvalue & character polynomials proof

[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 I am stuck

 

[AlephZero]

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...

 

[ISuckAtMath]

Homework Statement

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

Homework Equations

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

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

 

[Dick]

Homework Statement

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

Homework Equations

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

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

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

 

[ISuckAtMath]

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

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?

 

[Dick]

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?

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

 

[Rockoz]

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?

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.

 

[micromass]

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...

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.

 

[ISuckAtMath]

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

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

 

[Deveno]

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

 

[Dick]

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

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.

 

[AlephZero]

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.

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.

 

[taxidriverhk]

we are given B = CAC^-1

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

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...