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 $\mathbf{M}$
$$$$\label{eigenvector} \mathbf{Mv} = \lambda \mathbf{v}$$$$

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

$$$$\label{eigenvector-0} (\mathbf{M} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}$$$$

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{align*} \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{align*}

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 $\mathbf{M}$

References