Given an square matrix \(\mathbf{M}\)

- an
*eigenvector*\(\mathbf{v}\) is a non-zero vector whose direction doesn't change when multiplied by \(\mathbf{M}\), note that \(\mathbf{M}\) has an eigenvector then there are an infinite number of eigenvectors (vectors parallel to \(\mathbf{v}\)) - an
*eigvenvalue*\(\lambda\) is the scale factor associated with some*eigenvector*\(\mathbf{v}\) of \(\mathbf{M}\) has after the multiplication with \(\mathbf{M}\)

\[ \begin{equation} \label{eigenvector} \mathbf{Mv} = \lambda \mathbf{v} \end{equation} \]

Assuming that \(\mathbf{M}\) has at least one eigenvector \(\mathbf{v}\) we can do standard matrix multiplications to find it, first let's manipulate the right side of \eqref{eigenvector} so that it also features a matrix multiplication

\[ \mathbf{Mv} = \lambda \mathbf{Iv} \]

Where \(\mathbf{I}\) is the identity matrix, next we can rewrite the last equation as

\[ \mathbf{Mv} - \lambda \mathbf{Iv} = \mathbf{0} \]

Because matrix multiplication is distributive

\[ \begin{equation} \label{eigenvector-0} (\mathbf{M} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \end{equation} \]

The quantity \(\mathbf{M} - \lambda \mathbf{I}\) must not be invertible, if it had an inverse we could premultiply both sides by \((\mathbf{M} - \lambda \mathbf{I})^{-1}\) which would yield

\[ \begin{align*} (\mathbf{M} - \lambda \mathbf{I})^{-1}(\mathbf{M} - \lambda \mathbf{I})\mathbf{v} &= (\mathbf{M} - \lambda \mathbf{I})^{-1} \; \mathbf{0} \\ \mathbf{v} &= \mathbf{0} \end{align*} \]

The vector \(\mathbf{v = 0}\) fulfills \eqref{eigenvector} however we'll try to find a vector \(\mathbf{v} \not = \mathbf{0}\), if such a condition is added then the matrix \(\mathbf{M} - \lambda \mathbf{I}\) must not have an inverse which also means that its determinant is 0

\[ \left | \mathbf{M} - \lambda \mathbf{I} \right | = 0 \]

If \(\mathbf{M}\) is a \(2 \times 2\) matrix then

\[ \begin{equation} \label{lambda} \left | \mathbf{M} - \lambda \mathbf{I} \right | = \begin{vmatrix} m_{11} - \lambda & m_{12} \\ m_{21} & m_{22} - \lambda \end{vmatrix} = \lambda^2 - (m_{11}+m_{22})\lambda + (m_{11}m_{22} - m_{12}m_{21})= 0 \end{equation} \]

From \eqref{lambda} we can find two values for \(\lambda\) which may be unique/imaginary, a similar manipulation for a \(n \times n\) matrix will yield an \(n\)th degree polynomial, for \(n \leq 4\) we can compute the solutions by analytical methods, for \(n > 4\) only numeric methods are used

The associated eigenvector can be found by solving \eqref{eigenvector-0}

\[ \begin{bmatrix} m_{11} - \lambda & m_{12} \\ m_{21} & m_{22} - \lambda \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

### Applications

List of applications

- if \(\mathbf{M}\) is a transformation matrix then \(\mathbf{v}\) is a vector that
**isn't affected by the rotation part of \(\mathbf{M}\)**, therefore \(\mathbf{v}\) is the rotation axis of \(\mathbf{M}\)

- Shirley, P. and Ashikhmin, M. (2005). Fundamentals of computer graphics. Wellesley, Mass.: AK Peters.
- Proof of formula for determining eigenvalues. (2016). [online] Khan Academy. Available at: https://www.khanacademy.org/math/linear-algebra/alternate_bases/eigen_everything/v/linear-algebra-proof-of-formula-for-determining-eigenvalues [Accessed 7 Mar. 2016].