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 $\mathbf{M}$

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 $\text{(1)}$ so that it also features a matrix multiplication

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

Because matrix multiplication is distributive

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

The vector $\mathbf{v}\mathbf{=}\mathbf{0}$ fulfills $\text{(1)}$ however we’ll try to find a vector $\mathbf{v}\ne \mathbf{0}$, 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

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

From $\text{(???)}$ 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 $n\le 4$ we can compute the solutions by analytical methods, for $n>4$ only numeric methods are used

The associated eigenvector can be found by solving $\text{(2)}$

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 $\mathbf{M}$