Eigenvalue

linear-algebra

Definition

Eigenvalue

Let $f:V\to V$ be a linear mapping. A scalar $\lambda \in F$ is called Eigenvalue of $f$ if there exists a vector $\mathbf{x}\in V\setminus \{0\}$ such that:

$$\exists \mathbf{x}\in V\setminus \{0\}:f(\mathbf{x})=\lambda \cdot \mathbf{x}$$

The vectors $\mathbf{x}\in V\setminus \{0\}$ are called Eigenvectors.

Matrix Representation

Matrix Representation

Instead of using a linear mapping, a matrix representation can be used as a criteria.

Let $A\in F^{n\times n}$ be a square matrix. A scalar $\lambda \in F$ is called Eigenvalue if there exists a vectors $\mathbf{x}\in F^n\setminus \{0\}$ such that:

$$\exists \mathbf{x}\in F^n\setminus \{0\}:A\cdot \mathbf{x}=\lambda \cdot \mathbf{x}$$

Another way of writing this for a square matrix can be done using determinants:

$$\det (\lambda \cdot I_n-A)=0$$

where $I_n$ is the identity matrix.

Characteristic Polynomial

Definition

Characteristic Polynomial

The characteristic polynomial of a square matrix $A\in F^{n\times n}$ is the determinant:

$${\chi }_A(\lambda )=\det (\lambda \cdot I_n-A)$$

Note that the degree of this polynomial is $n$ and the leading coefficient is $1$. Thus, eigenvalues are the rots of ${\chi }_A(\lambda )$. Since a polynomial of degree $n$ (over a field $F$) can have at most $n$ distinct roots, an $n\times n$ square matrix has at most $n$ eigenvalues.

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Example

Matrix Example

Matrix Example

$$A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\in {\mathbb{R}}^{2\times 2}$$

First, compute:

$$\lambda \cdot I_2-A=\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]-\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]=\left[\begin{array}{cc}\lambda -1& -2\\ -3& \lambda -2\end{array}\right]$$

Then, compute the determinant:

$$\begin{array}{rl}\det (\lambda \cdot I_2-A)& =\left|\begin{array}{cc}\lambda -1& -2\\ -3& \lambda -2\end{array}\right|\\ & =(\lambda -1)\cdot (\lambda -2)-(-2)\cdot (-3)\\ & ={\lambda }^2-2\cdot \lambda -1\cdot \lambda +2-6\\ & ={\lambda }^2-3\cdot \lambda -4\end{array}$$

This quadratic polynomial has two roots:

$$\{\begin{array}{ll}{\lambda }_1& =4\\ {\lambda }_2& =-1\end{array}$$

Thus, ${\lambda }_1=4$ and ${\lambda }_2=-1$ are eigenvalues of $A$.