1 Eigenvalue and 2 Eigenvectors

[jakey]

Hi all,

Let's say we have a symmetric matrix A with its corresponding diagonal matrix D. If A has only 1 eigenvalue, how do we show that there exists 2 eigenvectors?

thanks!

 

[HallsofIvy]

What you have written makes no sense. If a matrix has an eigenvalue, then there exist an infinite number of eigenvectors. Do you mean "2 independent eigenvectors"? And are you talking about a 2 by 2 matrix?

A matrix is "diagonalizable" if and only if it has a "complete set of eigenvectors"- that is, there is a basis for the vector space consisting of eigenvalues of the matrix. If A is an n by n matrix, then it must have n independent eigenvectors. If it has only one eigenvalue, then there must exist n independent eigenvectors corresponding to that one eigenvalue.

 

[jakey]

What you have written makes no sense. If a matrix has an eigenvalue, then there exist an infinite number of eigenvectors. Do you mean "2 independent eigenvectors"? And are you talking about a 2 by 2 matrix?

A matrix is "diagonalizable" if and only if it has a "complete set of eigenvectors"- that is, there is a basis for the vector space consisting of eigenvalues of the matrix. If A is an n by n matrix, then it must have n independent eigenvectors. If it has only one eigenvalue, then there must exist n independent eigenvectors corresponding to that one eigenvalue.

Hi HallsofIvy,

I'm sorry, i forgot to mention that it's a 2x2 matrix. Yes, is there a general method to find 2 independent eigenvectors?

 

[Redbelly98]

In this case, every vector is an eigenvector. Pick one. Then find another vector orthogonal to it.

If you need to, you can normalize the vectors.

 

[Noxide]

Given:
A is symmetric

Conclusion:
if A is symmetric then that means A equals its transpose and is of size nxn (THIS IS ALWAYS TRUE)
if A is symmetric then it has n independent eigenvectors (THIS IS ALWAYS TRUE)

The matrix factorization of A is SDS^(-1)
The n columns of S are the n independent eigenvectors and for a symmetric matrix those eigenvectors are orthogonal when A is symmetric and can be made orthonormal (which makes finding the factorization of A a lot easier, via Gram-Schmidt)
The diagonal entries of the Diagonal matrix D are the eigenvalues associated with the eigenvectors

So, to find those diagonal entries and those independent eigenvectors the general form is as follows where A is an nxn matrix, x is an n-dimensional vector, and d is a constant, $ is the n-dimensional zero vector and I is the nxn identity matrix.

Ax=dx
Ax-dx=$
(A-dI)x=$
det(A-dI)=0 solve for all values of d
(this is called the characteristic equation which gives the n-degree polynomial which is used to determine the values of d (your eigenvalues that satisfy the equation))Edit: Hope this helps, I'm taking intro linear algebra this semester so if anything is wrong please let me know.