Characteristic Eq for Matrix problem

In summary, the coefficient of x in the expansion of det(x*I-A) can be found by taking the sum from i = 1 to n of the determinant of the Aii minor, where Aii is the submatrix of A formed by deleting row i and column i. This can also be expressed as the sum of the product of eigenvalues divided by each eigenvalue of A. This method can be used for diagonal and diagonalizable matrices.
  • #1
Zoe-b
98
0

Homework Statement


Where A is an n x n matrix and I is the n x n identity
In the expansion of det(x*I-A), show that the coefficient of x is equal to the sum from i = 1 to n of the determinant of the Aii minor. (where Aii = the submatrix of A formed by deleting row i and column i)


Homework Equations


if X1, X2, ... Xn are the eigenvalues of A then the coefficient of x is equal to the sum from i = 1 to n of (product of eigenvalues)/Xi.


The Attempt at a Solution


I have tried quite a few different ways of doing this but getting nowhere..
using the definition of a determinant as a sum of products over permuations of Sym (n) and then using the Leibnix rule for integrals I seem to find that (where a11 is the top left entry of A and so on)
d(det(x*I-A)/dx = sum (from i =1 to n) of det(x*I-A)/(x-aii)

The constant coefficient of this will be equal to the x coefficient of det(x*I-A) but the RHS only cancels directly to what I want if aii is an eigenvalue, which obviously isn't always the case. I tried to use row permutations but this obviously changes the Aii minors and so doesn't seem to work. Any ideas? I was trying to follow the proof on

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TY9-4H5MYJD-1&_user=126524&_coverDate=06%2F30%2F2006&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_searchStrId=1630570352&_rerunOrigin=google&_acct=C000010360&_version=1&_urlVersion=0&_userid=126524&md5=26cc582e2a0f1358d2f4b36931f4954e&searchtype=a

(sorry for the long link!) but the notation is a bit beyond me. Am I on the right track at all or is there a better way to attempt this? Sorry for not including full working but without the maths language its near impossible to be clear.
Thanks
 
Physics news on Phys.org
  • #2
Can you prove it for when A is diagonal
 
  • #3
Yes I understand for when it is diagonal but altho row/column operations of form (row i -> row i + d*row j) won't change the determinant of the matrix, they will change the determinant of the minors.. I don't see how to relate the diagonal form back to the original matrix.
 
  • #4
How about for a diagonaliable matrix? That is the key here.
 
  • #5
Hmmmn I'm still pretty clueless but I'll give it another go tomorrow! Thanks
 

1. What is a characteristic equation for a matrix?

A characteristic equation for a matrix is an equation that is used to find the eigenvalues of a given matrix. It is represented by det(A - λI) = 0, where A is the matrix, λ is the eigenvalue, and I is the identity matrix.

2. How do I solve a characteristic equation for a matrix?

To solve a characteristic equation for a matrix, you can follow these steps:

  • Calculate the determinant of A - λI.
  • Set the determinant equal to 0.
  • Solve for λ using algebraic methods.
  • The values of λ that satisfy the equation are the eigenvalues of the matrix.

3. Why is the characteristic equation important?

The characteristic equation is important because it allows us to find the eigenvalues of a matrix, which are essential in many applications of linear algebra. Eigenvalues are used to find eigenvectors, diagonalize matrices, and solve differential equations.

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

No, a matrix can only have one characteristic equation. However, different matrices can have the same characteristic equation, which means they have the same eigenvalues.

5. Is there a shortcut to finding the characteristic equation?

There is no shortcut to finding the characteristic equation, as it involves calculating the determinant of a matrix and solving for the eigenvalues. However, there are some techniques that can make the process easier, such as using properties of determinants and factoring techniques.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
4K
  • Calculus and Beyond Homework Help
Replies
6
Views
4K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
12K
  • Linear and Abstract Algebra
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Back
Top