## 1. Definition

We have seen that a matrix $A$ multiplying a vector $\vec{x}$ is in fact a linear transformation.

We now are interested in the vector $\vec{x}$ whose image $\vec{y}$ (under the linear transformation) is linearly dependent to itself, namely :

$\lambda$ and $\vec{x}$ satisfying $(1)$ are called respectively **eigenvalues** and **eigenvectors**.

In $(1)$ $\boldsymbol{\vec{0}}$ is **excluded**. Indeed we know anyway that $\underbrace{A\,\vec{0}}_{ \vec{0} }

= \underbrace{ \lambda\, \vec{0} }_{ \vec{0} } \,$ is true for any value of $\lambda$.

## 2. Calculation

From $(1)$ we can obtain :

In $(5)$, the multiplication matrix by vector of $(4)$ has been decomposed. The columns $\vec{C_1}, \dots, \vec{C_n}$ must be linearly dependent since the $x_i$ cannot equal zero because of $(1)$. Therefore their determinant equal zero.

Once the $\lambda$’s (eigenvalues) are found by solving the determinant, we replace them in the matrix in $(4)$ for determining the associated eigenvectors.

#### Example I

Let $h$ : $h \colon \, \mathbb{R}^2 \to \mathbb{R}^2$, $ \vec{x} \mapsto \left( \begin{smallmatrix} 2 & 1 \\ 3 & 4 \end{smallmatrix} \right) \vec{x}$.

Let’s determine its eigenvalues and eigenvectors.

Based on $(4)$, we have :

The columns of the matrix in $(6)$ are linearly dependent, thus :

From $(7)$, we get : $\lambda_1 =1$ and $\lambda_2 =5$

To determine the eigenvector $\vec{x}_1$ associated to $\lambda_1$, we replace $\lambda_1$ in $(6)$ :

$ \iff 1x+1y = 0 \iff x=-y$. Let $x= \beta$.

To determine the eigenvector $\vec{x}_2$ associated to $\lambda_2$, we replace $\lambda_2$ in $(6)$ :

$ \iff -3x+1y = 0 \iff y=3x \iff x=\frac{y}{3}$. Let $x= \beta$.

## 3. But more concretely ?

Figure 10.1 shows a shape before and after the transformation (in dashed line) under the linear transformation of example I.

The eigenvector $\,\color{green}{\vec{x_1}}=\left(\begin{smallmatrix} -1 \\ -3 \end{smallmatrix}\right)$, as well as the ordinary vector $\color{navy}{\vec{x_2}}=\left(\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right)$ are represented too. The **eigenvector** does **not rotate** under the linear transformation.

## Recapitulation

An **eigenvector** is a vector whose image (under the linear transformation) is linearly dependent to itself, namely $\underbrace{ A\vec{x} }_{ \vec{y} } = \lambda \vec{x}$.

First the **eigenvalues** are found by solving the **characteristic polynomial** and after that the associated eigenvectors can be found.

The **characteristic polynomial** is obtained by computing the determinant of the matrix $A -\lambda I\,$.

By definition, an eigenvector does not equal to the null vector.