# Eigenvalue

The eigenvector of a linear transformation is the vector that changes by a scalar factor, referred to as an eigenvalue (typically denoted λ), when the linear transformation is applied to the eigenvector. To clarify, consider the following products of a 2 × 2 matrix and several column vectors:

 1.  2.  3.  4.  Notice that in the 1. and 3., the resulting vector is a multiple of the corresponding column vector on the left-hand side of the equation. These vectors, in this case and , are referred to as eigenvectors. When a transformation is applied to an eigenvector, the eigenvalue is the value by which the eigenvector is scaled to yield the result of the transformation. Referencing 1. and 3.,

 1.  3.  the corresponding eigenvalues are 4 and 2 respectively.

## Formal definition

Let A be an n × n matrix. If there is a non-zero vector, x, and some scalar λ, such that, then λ is the eigenvalue corresponding to eigenvector x. Note that eigenvalues and eigenvectors can be equivalently defined in either the language of matrices or transformations. Here we have defined it in terms of matrices, where A is the matrix representation of some transformation T, and x is the coordinate vector of eigenvector v. In the language of transformations, ## How to find eigenvalues

To find an eigenvalue, set the equation above equal to 0, then factor out the eigenvector:  Then, multiply λ by an n × n identity matrix, I, the size of A. Doing so doesn't change the value of λ and allows us to combine it with A: Since x is an eigenvector, we know that it must have a non-zero value, or the solution to the above expression would be trivial. In cases where the matrix, (λI - A), is invertible, eigenvector x would equal 0; we want to find solutions such that this is not the case. Recall that if the determinant of a matrix is equal to 0, that the matrix is not invertible. Thus, we take the determinant of the above expression and set it equal to 0 in order to solve for the eigenvector of the system. This equation is referred to as the characteristic equation for matrix A. Taking the determinant of A - Iλ produces an nth degree polynomial in terms of λ, the roots of which are the eigenvalues of A such that where ci is a constant and i = 1, 2, 3, ..., n.

Example

Find the eigenvalues for the following:

1. A = 2. A = 1. First, find A - Iλ:    Then, compute the determinant:    Thus, the characteristic polynomial is λ2 - 4λ + 3, and the eigenvalues of A are the roots of this expression:     Thus, λ = 1 or 3 are the eigenvalues of A.

2. First, find A - Iλ:    Then, compute the determinant:     Thus, the characteristic polynomial is -λ3 - 2λ2 + 3λ, and the eigenvalues of A are the roots of this expression:       Thus, λ = -3, 0, 1 are the eigenvalues of A.