Symmetric, irreducible, tridiagonal matrix: Eigenvalues

In summary, the conversation discusses the properties of symmetric, irreducible, tridiagonal matrices and upper Hessenberg matrices. It is shown that symmetric, irreducible, tridiagonal matrices cannot have multiple eigenvalues, while upper Hessenberg matrices with non-zero subdiagonal elements can only have one associated eigenvector if they have a multiple eigenvalue. The solution involves using the Jordan canonical form and the rank of A-\lambdaI.
  • #1
Scootertaj
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Homework Statement


A) Let A be a symmetric, irreducible, tridiagonal matrix. Show that A cannot have a multiple eigenvalue.
B) Let A be an upper Hessenberg matrix with all its subdiagonal elements non-zero. Assume A has a multiple eigenvalue. Show that there can only be one eigenvector associated with it.



2. Relevant definitions
Symmetric: A = AT
Irreducible: not a good definition for it, but essentially it's well-connected
Tridiagonal: obvious
Upper Hessenberg: http://en.wikipedia.org/wiki/Hessenberg_matrix Essentially, it's a mix between a tridiagonal and a upper triangular. It's tridiagonal with non-zero elements above its 3rd subdiagonal.



The Attempt at a Solution


A) It seems contradiction would be the best combined with using the Jordan canonical form of A-[itex]\lambda[/itex]I.
Jordan canonical form: http://en.wikipedia.org/wiki/Jordan_normal_form
So, we know that we can get the Jordan canonical form of A which has all [itex]\lambda[/itex]'s in the diagonal.
Thus, is there anyway we can look at the canonical form of A-[itex]\lambda[/itex]I and deduce from it's rank that it has no multiple eigenvalues?
B) The hint says to use a similar strategy as A
 
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  • #2
Just a bump.
 
  • #3
Just another bump. This is a tough problem.
 

1. What is a symmetric, irreducible, tridiagonal matrix?

A symmetric, irreducible, tridiagonal matrix is a special type of square matrix where all the elements are zero except for those on the main diagonal, the diagonal above it, and the diagonal below it. Additionally, the matrix is symmetric, meaning that it is equal to its transpose. It is also irreducible, meaning that it cannot be reduced to a smaller block matrix by permuting rows and columns.

2. What are the eigenvalues of a symmetric, irreducible, tridiagonal matrix?

The eigenvalues of a symmetric, irreducible, tridiagonal matrix are the values that satisfy the equation Ax = λx, where A is the matrix, x is an eigenvector, and λ is the eigenvalue. These eigenvalues can be found by solving a characteristic equation using techniques such as the QR algorithm or the bisection method.

3. How can I calculate the eigenvalues of a symmetric, irreducible, tridiagonal matrix?

There are several methods for calculating the eigenvalues of a symmetric, irreducible, tridiagonal matrix. One common approach is the QR algorithm, which involves iteratively transforming the matrix into a similar tridiagonal matrix with the same eigenvalues. Another method is the bisection method, which involves narrowing down the possible eigenvalues by repeatedly dividing the interval in which they are located.

4. Why are symmetric, irreducible, tridiagonal matrices important in mathematics?

Symmetric, irreducible, tridiagonal matrices have many important applications in mathematics, especially in linear algebra and numerical analysis. They are used in solving systems of linear equations, computing the eigenvalues and eigenvectors of a matrix, and in various other algorithms. They also have connections to other areas of mathematics such as graph theory and combinatorics.

5. Can a symmetric, irreducible, tridiagonal matrix have complex eigenvalues?

Yes, a symmetric, irreducible, tridiagonal matrix can have complex eigenvalues. This is because the eigenvalues of a matrix are determined by its characteristic equation, which can have complex solutions. However, if the matrix is real, then its complex eigenvalues will always occur in complex conjugate pairs.

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