# 1 Eigenvalue and 2 Eigenvectors

1. Apr 10, 2010

### 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!

2. Apr 10, 2010

### 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.

3. Apr 10, 2010

### jakey

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?

4. Apr 10, 2010

### Redbelly98

Staff Emeritus
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.

5. Apr 18, 2010

### 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.