Given a 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 if $\mathbf{M}$ has an eigenvector, then there are an infinite number of eigenvectors (vectors parallel to $\mathbf{v}$).
  • An eigenvalue $\lambda$ is the scale factor associated with an eigenvector $\mathbf{v}$ of $\mathbf{M}$ after multiplication by $\mathbf{M}$.
$$ \begin{equation} \label{eigenvector} \mathbf{Mv} = \lambda \mathbf{v} \end{equation} $$

Assuming that $\mathbf{M}$ has at least one eigenvector $\mathbf{v}$, we can perform 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

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

From \eqref{lambda}, we can find two values for $\lambda$, which may be unique or imaginary. A similar manipulation for an $n \times n$ 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 $\mathbf{M}$.