Eigenvector

Certain subspaces of a linear transformation remain mostly unaffected. For the linear transformation $T:V\to V$ which maps a vector space onto itself, some subspaces of $V$, such as $W$ can be invariant, where $T(W)\subseteq W$.

The vectors that are in $W$, which only scale with a linear transformation are called eigenvectors.

Definition §

Eigenvectors & Eigenvalues (Linear Transformation) ^formula1

Let $T:V\to V$ be a linear transformation mapping the vector space $V_{\mathbb{F}}$ (fixed under field $\mathbb{F}$) onto itself. Then an eigenvector of $T$ is given by:

$$T(\vec{v})=\lambda \vec{v}\lambda \in \mathbb{F},\vec{v}\in V,v\ne \vec{0}$$

$\vec{v}$ is an eigenvector, with it’s corresponding eigenvalue being $\lambda$.

Since every transformation can be represented as a matrix, we can solve:

$$[T{]}_{\mathcal{B},\mathcal{B}}\cdot [\vec{v}{]}_{\mathcal{B},\mathcal{B}}=\lambda \cdot [\vec{v}{]}_{\mathcal{B},\mathcal{B}},\vec{v}\ne \vec{0}$$

This leads to a matrix definition of eigenvectors:

Eigenvectors & Eigenvalues (Matrices) ^formula2

Let $\mathbf{A}$ be an $n\times n$ matrix with entries in $\mathbb{C}$ (or $\mathbb{R}$). A scalar, $\lambda \in \mathbb{C}$, is an eigenvalue if there exists a non-zero column matrix, $\vec{v}\in {\mathbb{C}}^n$ such that:

$$\mathbf{A}\vec{v}=\lambda \vec{v}$$

$$(\mathbf{A}-\lambda \mathbf{I})\vec{v}=0$$

Then $\vec{v}$ is an eigenvector of the matrix $\mathbf{A}$

Obtaining eigenvectors §

Obtaining the Eigenvectors from a Matrix

Linear Independence §

Linear Independence of Eigenvectors

If a collection of eigenvectors uniquely map to a set of eigenvalues, then these eigenvectors are linearly independent. I.e. if a set of distinct eigenvectors* $\{{\vec{v}}_{{\lambda }_1},{\vec{v}}_{{\lambda }_2},\dots ,{\vec{v}}_{{\lambda }_i}\}$ corresponds to the set of distinct eigenvalues* $\{{\lambda }_1,{\lambda }_2,\dots {\lambda }_i\}$ such that:

$$v_{{\lambda }_1}\to {\lambda }_1,v_{{\lambda }_2}\to {\lambda }_2,\dots ,v_{{\lambda }_i}\to {\lambda }_i$$

Then the set of eigenvectors is linearly independant.