Help Showing d/dt of Char. Polynomial

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In summary: C_A(t)_{1j} and \det(C_A_i(t))_1j with \det(C_A(t))_1j. Therefore, the equation holds true. In summary, the equation \frac{d}{dt}P_A(t) = \sum_{i=1}^n P_A_i(t) is proven by expanding the left and right sides using the properties of determinants and identifying the corresponding terms.
  • #1
oobob
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I've been working on a problem all day but I can't seem to go anywhere with it. I have to prove the following equation holds, in which [tex]P_A(t)[/tex] refers to the characteristic polynomial of an n by n matrix A and is expressed in terms of t. Also, [tex]P_A_i(t)[/tex] is the characteristic polynomial of the principal submatrix formed by removing row i and column i from A, also expressed in terms of t. The equation is:

[tex]\frac{d}{dt}P_A(t) = \sum_{i=1}^n P_A_i(t) [/tex]

I've been trying to prove it by induction, using the following formula for the left side. In the formula, for a given k, [tex]E_k(A)[/tex] is the sum of the k-by-k principal minors of A.:

[tex]\frac{d}{dt}P_A(t) = nt^n^-^1 - (n-1)E_1(A)t^n^-^2 + (n-2)E_2(A)t^n - ... \pm E_n_-_1(A) [/tex]

I'm not sure how to arrive at the right side of the first equation at all. I've tried rearranging it, but I don't know how to make up for the fact that all of the (n-1) by (n-1) principal submatrices of A are different. My only interpretation of the right hand side is as the sum of the characteristic polynomials of all (n-1) by (n-1) principal submatrices of A, but I can't seem to draw anything from that. Am I on an entirely wrong path, or am I just missing a step somewhere?

Thanks,
Oobob
 
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  • #2
The key to solving this problem is to use the fact that the characteristic polynomial of a matrix can be written as a determinant of its coefficients. This allows us to expand the right side of the equation using the properties of determinants. Specifically, we can use the Laplace expansion along a row or column to expand the determinant.Let P_A(t) = det(C_A(t)), where C_A(t) is an n by n matrix of coefficients of the characteristic polynomial. Then the left side of the equation can be written as:\frac{d}{dt}P_A(t) = \frac{d}{dt}\det(C_A(t)) Using the properties of determinants, we can expand the determinant on the right side of the equation along any row or column. Let's choose row 1 to expand the determinant:\frac{d}{dt}P_A(t) = \sum_{j=1}^n (-1)^{1+j} C_A(t)_{1j} \det(C_A(t))_1j where \det(C_A(t))_1j is the determinant of the matrix formed by removing row 1 and column j from C_A(t). Now, let's look at the right side of the equation. For each i in the summation, P_A_i(t) = det(C_A_i(t)). We can rewrite this as:\sum_{i=1}^n P_A_i(t) = \sum_{i=1}^n \det(C_A_i(t)) Again, using the properties of determinants, we can expand this summation along row 1 to get:\sum_{i=1}^n P_A_i(t) = \sum_{i=1}^n \sum_{j=1}^n (-1)^{1+j} C_A_i(t)_{1j} \det(C_A_i(t))_1j Comparing the two sides of the equation, we see that they are identical if we identify C_A_i
 
  • #3


It seems like you are on the right track with your approach using induction. One suggestion would be to try expanding the right side of the first equation using the formula for the characteristic polynomial of a principal submatrix. This will help you see the relationship between the two equations and how they can be equated. Another approach could be to use the fact that the characteristic polynomial of a matrix is equal to the determinant of the matrix, and then use the product rule for differentiating determinants. This may lead you to a more direct proof of the equation. Keep exploring and trying different approaches, and don't be afraid to seek help from a classmate or your instructor if you are still stuck. Good luck!
 

1. What is a characteristic polynomial?

A characteristic polynomial is a polynomial that is used to find the eigenvalues of a square matrix. It is typically denoted as p(λ) and is defined as det(A - λI), where A is the original matrix and I is the identity matrix.

2. What is the purpose of finding the characteristic polynomial?

The characteristic polynomial helps to determine the eigenvalues of a matrix, which in turn can provide information about the linear transformations that the matrix represents. It is also useful in solving differential equations and in finding the diagonalization of a matrix.

3. How do you find the derivative of a characteristic polynomial?

The derivative of a characteristic polynomial can be found using the chain rule. The derivative of p(λ) is equal to the derivative of det(A - λI), which is equal to det(A - λI) * (-I), where I is the identity matrix. This can be simplified to -p'(λ)I, where p'(λ) is the derivative of p(λ).

4. Why is it important to find the derivative of a characteristic polynomial?

The derivative of a characteristic polynomial is important because it helps to find the eigenvalues of a matrix at a specific point in time. This is useful in many fields, such as physics, engineering, and economics, where understanding the behavior of a system over time is crucial.

5. Are there any applications of the derivative of a characteristic polynomial?

Yes, the derivative of a characteristic polynomial has various applications in different fields. In physics, it is used to solve problems in mechanics, electromagnetism, and quantum mechanics. In engineering, it is used in control systems and signal processing. In economics, it is used to model and analyze economic systems.

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