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

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

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