Eigenvectors and Eigenvalues

Definition¶

Let $V$ be a vector space defined over a field $\mathbb{F}$ and $L: V \rightarrow V$ be a linear map. A vector $\bm{v}~\in~V$, where $\bm{v} \neq 0$, is called an eigenvector of the map $L$ if

where $\lambda~\in~\mathbb{F}^{1}$ is the eigenvalue of the eigenvector $\bm{v}$.

The set of all eigenvalues of a linear map $L$ is called the spectrum of $L$, denoted $\textrm{spec}L$.

An vector space is said to be degenerate if different eigenvectors have the same eigenvalue.

Basis of Eigenvectors¶

Let $L: V \rightarrow V$ be linear map, where $V$ is a vector space of dimension $n$. The space $V$ can have a basis that consists of vectors of eigenvectors of $L$, say $\{\bm{v}_1, \bm{v}_2, \ldots, \bm{v}_n \}$ with eigenvalues $\{\lambda_1, \lambda_2, \ldots, \lambda_n \}$ - such that $L(\bm{v}_{i}) = \lambda_i \bm{v}_i$.

If so, then the matrix associated to the linear map $L$, $M_{L}$, can be written with respect to this basis as

This also works the other way around. If there exists a basis, $\{\bm{v}_1, \bm{v}_2, \ldots, \bm{v}_n \}$, such that $M_{L}$ can be written diagonally (as above) then vectors in the basis are eigenvectors of $L$ and the associated value on the diagonal is the eigenvalues.

Conditions on a Basis of Eigenvectors

Calculating Eigenvectors and Eigenvalues¶

Let $L:V \rightarrow V$ be a linear operator, where $V$ is a vector space of $\textrm{dim}V<\infty$ defined over $\mathbb{F}$.

$\lambda~\in~\mathbb{F}^{1}$ is an eigenvector of $L$ if and only if

where $\textrm{Det}$ is the determinant and $M_L$ is the matrix associated to $L$ (note this is independent of the choice of basis for $M_L$).

One can define the characteristic polynomial of $L$ as $P(\lambda)=\textrm{Det}(M_{L} - \lambda \mathbb{I})$, the roots of the characteristic polynomial then give the eigenvalues of $L$.

The corresponding eigenspace to an eigenvalues $\lambda$ is then given by

where $\textrm{Ker}$ is the kernel.

Real Matrices¶

Let $M~\in~\mathbb{M}(\mathbb{R})$ be symetric, $M^{t}=M$, then there exists a matrix $B~\in~\mathbb{M}(\mathbb{R})$ such that

where $\{ \lambda_1, \lambda_2, \ldots, \lambda_n \}$ are eigenvalues of $M$ and $B$ has an orthonormal basis of eigenvectors as its columns such that $B^{t}B=\mathbb{I}$.

Further Properties¶

Further Resource¶

• 3Blue1Brown Video: Eigenvectors and eigenvalues | Chapter 14, Essence of linear algebra