Eigenvalues and Eigenvectors

Let $A\in {\mathbb{F}}^{n\times n}$ be a square matrix and $\underset{―}{v}\in {\mathbb{F}}^n$ be a non-zero vector. $\underset{―}{v}$ is an eigenvector of $A$ if and only if

$$A\underset{―}{v}=\lambda \underset{―}{v}$$

where $\lambda \in \mathbb{F}$ is called an eigenvalue of $A$ if $\underset{―}{v}$ exists.

The linear transformation $A\underset{―}{v}$ where $\underset{―}{v}$ is an eigenvector of $A$ is equivalent to the scalar transformation $\lambda \underset{―}{v}.$

Eigenvalues satisfy the following properties:

$$\sum _{i=1}^n{\lambda }_i=\mathrm{tr}(A)$$ $$\prod _{i=1}^n{\lambda }_i=|A|$$

The expression $|A-\lambda I_n|$ is an $n^{\mathrm{th}}$-order polynomial in $\lambda$ called the characteristic polynomial of $A.$ The roots $|A-\lambda I_n|=0$ are the eigenvalues of $A.$

The set of all distinct eigenvalues of $A$ is called the eigen-spectrum ${\sigma }_A$ of $A.$

$${\sigma }_A=\{\lambda :|A-\lambda I_n|=0\}$$

The characteristic polynomial of $A$ will admit $m$ distinct complex roots where $1\le m\le n.$ Therefore the cardinality of the eigen-spectrum ${\sigma }_A$ is $m.$

For each eigenvalue ${\lambda }_i\in {\sigma }_A,$ the non-zero solution to

$$(A-{\lambda }_i{I}_n)\underset{―}{v_i}=\underset{―}{0}$$

yields the corresponding eigenvector $\underset{―}{v_i}$ of $A.$

The eigen-space ${\mathcal{E}}_{\lambda }$ of $A$ with respect to eigenvalue $\lambda$ is the vector space that is spanned by all eigenvectors of $A.$

The algebraic multiplicity ${\mu }_A\left(\lambda \right)$ of $\lambda$ with respect to the matrix $A$ is the number of times $\lambda$ appears as a root of the characteristic polynomial of $A.$ The sum of the algebraic multiplicities of all eigenvalues of $A$ is $n.$

The geometric multiplicity ${\gamma }_A\left(\lambda \right)$ of $\lambda$ with respect to the matrix $A$ is the number of linearly independent eigenvectors of $A$ and is equivalent to the dimension of the eigen-space of $A$ with respect to $\lambda .$

The algebraic and geometric multiplicities are related by:

$$1\le {\gamma }_A\left(\lambda \right)\le {\mu }_A\left(\lambda \right)\le n\mathrm{\forall }\lambda \in {\sigma }_A$$

which means that while distinct eigenvalues correspond to linearly independent eigenvectors, repeated eigenvalues do no necessarily correspond to multiple linearly independent eigenvectors.

$A$ will admit $n$ linearly independent eigenvectors if and only if:

$${\gamma }_A\left({\lambda }_i\right)={\mu }_A\left({\lambda }_i\right)\mathrm{\forall }{\lambda }_i\in {\sigma }_A.$$

$A$ will admit fewer than $n$ linearly independent eigenvectors if and only if:

$${\gamma }_A\left({\lambda }_i\right)<{\mu }_A\left({\lambda }_i\right)\mathrm{\exists }{\lambda }_i\in {\sigma }_A$$

and $A$ is referred to as defective.

For a non-defective matrix $A,$ the $n$ linearly independent eigenvectors will form a basis for ${\mathbb{F}}^n$ and is called an eigen-basis.

Let $A$ be a non-defective matrix with $n$ linearly independent eigenvectors ${\underset{―}{v}}_1,\dots ,{\underset{―}{v}}_n$ and corresponding eigenvalues ${\lambda }_1,\dots ,{\lambda }_n.$ The following matrices can be constructed:

$$P=\left[\begin{array}{ccc}\uparrow & & \uparrow \\ {\underset{―}{v}}_1& \dots & {\underset{―}{v}}_n\\ \downarrow & & \downarrow \end{array}\right]$$ $$\mathrm{\Lambda }=\left[\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& \ddots & 0\\ 0& 0& {\lambda }_n\end{array}\right]$$

where $P$ is the eigenvectors in columnar form stacked horizontally to form a square matrix and $\mathrm{\Lambda }$ is a diagonal matrix with eigenvalues along the diagonal.

The diagonalisation of $A$ is the decomposition:

$$A=P\mathrm{\Lambda }{P}^{-1}$$