Help Showing d/dt of Char. Polynomial

[oobob]

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 P_A(t) refers to the characteristic polynomial of an n by n matrix A and is expressed in terms of t. Also, P_A_i(t) 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:

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

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

\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)

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

 

[member 553083]

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

 

[wais]

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!