Eigenvalues and eigenvectors

• Given a linear map from to , we want to find the directions where the inputs do not change
  • Say that we have some subspace , such as plane or line. If , this means that all can do is scale at most
  • This subspace is called an invariant subspace with regards to

Definition: Eigenvalues Let be a linear transformation. Let , and is the span of . If is invariant under , then that means for some , that . is called an eigenvalue of and is an associated eigenvector.

• Essentially, for all of the eigenvectors, the transform can be simplified to multiplication by scalar
• Here is an alternate definition for eigenvalues and eigenvectors, with respect to matrices

Definition: Let . is called an eigenvalue of if there exists such that , and is called the associated eigenvector. is called an eigenvalue of if . That is, if solves

Finding eigenvalues

• Now, how do we actually end up finding eigenvalues?
  • You have to start with setting the transformation and scalar multiplication

We know that the definition of an eigenvalues is that for some linear transformation , that Rearranging the equation, we can write Hence, is not an isomorphism (since it is not one to one). The matrix associated: is not invertible, meaning that This gives us the characteristic polynomial for . The solutions to this polynomial are the eigenvalues of .

Definition: Characteristic polynomial. For some transformation , the characteristic polynomial is:

The values of that solve this equation are the eigenvalues () of .

Example: Find the eigenvalues of Eigenvalues are zeros of The solutions to this equation are the eigenvalues of . In this case, there are no real solutions, so the eigenvalues are complex.

Remark: Let be an eigenvalue of and is an eigenvector associated with . Then,

• This remark shows that any eigenvalue will have an infinite number of associated eigenvectors
  • This is because due to linearity, we can multiply the vector by any real number

If is an eigenvector of , then is also an eigenvector of , for all Suppose is linearly independent of eigenvectors of . Then all elements of is an eigenvector of . Proof: TBA

Eigenvalues and eigenvector for 3x3 matrices

• When you have an upper triangular matrix, the eigen values are simply the values along the diagonal

Example: Find the eigenvalues of Notice that when , the entire column is equal to zero, and therefore the determinant is equal to zero. This means that is one of the eigenvalues of this matrix. The same follows for .

• To find the eigenvector for a given eigenvalue , we have to solve for the system of equations where

Example: Find the eigenvectors for in the previous problem. Our eigenvector is ,where , as . This implies that . We have to solve . This gives us that , , is a free variable, so we can say that is an example of an eigenvector of .