Proving Similar Matrices Have the Same Characteristic Polynomial

In summary, we have two nxn matrices A and B, where A = QBQ' for some invertible matrix Q. We need to prove that A and B have the same characteristic polynomials. We can prove that they have the same determinant, but we are not sure where to go from there. We can use the identity det(ABC) = det(A)det(B)det(C) and the fact that AI = IA to show that det(A - kI) = det(B - kI), where k is a scalar. Therefore, A and B have the same characteristic polynomials.
  • #1
EvLer
458
0
So, I have stared at this for a while:
Notation: Q' - inverse of Q, != stands for "not equal";

Suppose A and B are nxn matrices such that A = QBQ' for some invertible matrix Q. Prove that A and B have the same characteristic polynomials

I can prove that they have the same determinant, but that is about it. I know that charact. polyn. looks like so:

det(A - I*lambda) = det(B - I*lambda)
det(QBQ' - I*lamda) = det(B - I*lambda)

It would be equal if Q and Q' cancel out, but isn't it true that det(A + B) != det(A) + det(B).
I am not sure where to go from here. Is it correct to multiply expressions inside parenthesis by something on both sides? even if it's in determinant :rolleyes:
I am studying for the final, so any help is appreciated more than ever.
 
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  • #2
det(QBQ' - kI) = det(QBQ' - Q(kI)Q'), where I'm using k instead of [itex]\lambda[/itex] because it's easier to type. Now, do you see why this equality holds?

Q(kI)Q' = kQIQ' = k(QI)Q' = k(Q)Q' = kQQ' = k(I) = kI, so the equality does hold. Use the distributive property of matrix multiplaction twice, then use the fact that det(ABC) = det(A)det(B)det(C), and you should be done.
 
  • #3
Ohhh, OK. So you also used this identity: AI = IA.
Thanks :smile:
 

Related to Proving Similar Matrices Have the Same Characteristic Polynomial

1. What is a characteristic polynomial?

A characteristic polynomial is a polynomial equation that is derived from a matrix. It is used to find the eigenvalues of the matrix, which represent the scaling factors of the matrix's eigenvectors.

2. How is a characteristic polynomial calculated?

To calculate a characteristic polynomial, you first need to find the determinant of the matrix. Then, you replace the main diagonal elements of the matrix with the variable lambda (λ) and subtract the determinant from it. This resulting polynomial is the characteristic polynomial.

3. What is the significance of the characteristic polynomial?

The characteristic polynomial is significant because it helps to determine important properties of a matrix, such as its eigenvalues and eigenvectors. It is also used in various applications, such as solving differential equations and determining stability of dynamical systems.

4. Can a matrix have more than one characteristic polynomial?

No, a matrix can only have one characteristic polynomial. This is because the characteristic polynomial is uniquely determined by the matrix's properties, such as its size and elements.

5. How is the characteristic polynomial related to the determinant of a matrix?

The characteristic polynomial and the determinant of a matrix are closely related. The determinant is the constant term of the characteristic polynomial, and the coefficients of the other terms are determined by the matrix's elements. In fact, the characteristic polynomial is just the determinant of the matrix with the main diagonal elements replaced by the variable lambda (λ).

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