Eigenspace

Invariant subspaces of linear maps: Eigenvalues and eigenvectors

Eigenspace

We now focus on finding eigenvectors for a given eigenvalue.

Let $V$ be a vector space and $L :V\rightarrow V$ a linear map. For every number $\lambda$ ,

$E_\lambda=\ker{ L -\lambda {I_V}}$ is a linear subspace of $V$ .

The subspace $E_\lambda$ is called the eigenspace of $L$ corresponding to $\lambda$ . This space consists of the zero vector and all eigenvectors with eigenvalue $\lambda$ .

The subset $E_\lambda$ of $V$ is the null space of the linear map $L -\lambda \,I_V$ , and so, by theorem Image space and kernel, a linear subspace of $V$ .

Let $\vec{v}$ be a vector in $V$ . Then $\vec{v}$ belongs to $E_\lambda$ if and only if $( L -\lambda \, I_V)\vec{v}=\vec{0}$ , so if and only if $L( \vec{v})-\lambda\vec{v}=\vec{0}$ , so if and only if $L( \vec{v})=\lambda\vec{v}$ . Therefore, the eigenspace $E_\lambda$ consists of the zero vector and all eigenvectors with eigenvalue $\lambda$ .

A special eigenspace is $E_0$ . It consists of all vectors that are mapped onto $0$ times themselves, that is, onto $\vec{0}$ . Therefore, the eigenspace $E_0$ is $\ker{L}$ , the null space (or kernel) of $L$ .

For most values of $\lambda$ , the space $E_\lambda$ only consists of $\{\vec{0}\}$ . The number $\lambda$ is an eigenvalue if and only if $E_\lambda$ contains a vector $\vec{v}\neq\vec{0}$ , that is, if and only if $\dim {E_\lambda }\gt0$ . Some authors speak only of an eigenspace if $\lambda$ is an eigenvalue.

For a given eigenvalue $\lambda$ the determination of the corresponding eigenvectors is a matter of finding the kernel of a linear map, which is equivalent to solving a system of linear equations.

The process of finding eigenvectors consists of first determining the values of $\lambda$ such that $\dim {E_\lambda }\gt0$ . Then $\lambda$ is an eigenvalue and the vectors in $E_\lambda$ distinct from $\vec{0}$ are the eigenvectors corresponding to the eigenvalue.

Let $L :V\rightarrow V$ be a linear map. We recall that the equation $\det (L_\alpha-\lambda I)=0$ is the characteristic equation of $L$ and that the left side of this equation, $\det (L_\alpha-\lambda \,I)$ , is the characteristic polynomial of $L$ .

Let $L:V\to V$ be a linear map, where $V$ is a vector space of finite dimension $n$ .

A number $\lambda$ is an eigenvalue of $L$ if and only if $\det (L-\lambda \cdot I_V)=0$ .
The eigenvectors corresponding to the eigenvalue $\lambda$ are the solutions $\vec{v}$ distinct from the zero vector of the linear equation $L(\vec{v}) = \lambda\, \vec{v}$ .

Let $\alpha$ be a basis of $V$ . Then the $\alpha$ -coordinates $v_1,\ldots,v_n$ of the eigenvectors of $L$ corresponding to the eigenvalue $\lambda$ may be calculated as the solutions distinct from the zero vector of the system of linear equations

$\left(\,\begin{array}{cccc} a_{11}-\lambda & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22}-\lambda & & \vdots\\ \vdots & & \ddots & \vdots \\ a_{n1} & \ldots & \ldots & a_{nn}-\lambda \end{array}\,\right)\ \ \left(\,\begin{array}{c} v_1\\ v_2\\ \vdots\\ v_n \end{array}\,\right)\ =\ \left(\,\begin{array}{r} 0 \\ 0\\ \vdots\\ 0 \end{array}\,\right)$ where $a_{ij}$ is the $(i,j)$ -entry of $L_\alpha$ .

Suppose that $\lambda$ is an eigenvalue of $L$ . Then there is a vector $\vec{v}\ne\vec{0}$ such that $L(\vec{v}) = \lambda\,\vec{v}$ . The vector $\vec{v}$ then belongs to the kernel of the linear map $L-\lambda\cdot I_V$ . This means that $\det( L-\lambda\cdot I_V) = 0$ .

Conversely: If $\lambda$ is a root of the characteristic polynomial of $L$ , then the linear map $\det( L-\lambda\cdot I_V)$ is not invertible, so its kernel contains a vector $\vec{v}\ne\vec{0}$ . Then $L(\vec{v}) = \lambda\,\vec{v}$ , which means that $\vec{v}$ is an eigenvector of $L$ corresponding to the eigenvalue $\lambda$ .

The matrix

$L_\alpha-\lambda\, I_n=\left(\,\begin{array}{cccc} a_{11}-\lambda & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22}-\lambda & & \vdots\\ \vdots & & \ddots & \vdots \\ a_{n1} & \ldots & \ldots & a_{nn}-\lambda \end{array}\,\right)$
is the matrix of $L -\lambda \cdot I_V$ with respect to the basis $\alpha$ .

Statement 1 can be strengthened to: $\lambda$ is an eigenvalue of $L$ if and only if it is a root of the minimal polynomial of $L$ .

Here is the proof:

If $\lambda$ is a root of the minimal polynomial, then it is also a root of the characteristic polynomial (which is after all a multiple of the minimal polynomial), and therefore, according to statement 1, an eigenvalue of $L$ .
Conversely, if $\lambda$ is an eigenvalue of $L$ corresponding to the eigenvector $\vec{v}$ , then the minimal polynomial $m_L(x)$ satisfies: $\begin{array}{rcl}m_L(\lambda)\,\vec{v}&=&m_L(L)\vec{v} \\&&\phantom{xx}\color{blue}{\lambda\,\vec{v} = L(\vec{v})}\\ &=&\vec{0}\\&&\phantom{xx}\color{blue}{m_L(x)\text{ is the minimal polynomial of }L}\end{array}$

Since the characteristic polynomial always has a complex root, each linear map from a complex vector space of finite dimension to itself has eigenvectors. This is not the case if the dimension of the vector space is infinite. Here is an example: Let $P$ be the vector space of all polynomials in $x$ and let $L_x:P\to P$ be multiplication by $x$ . Suppose that $f(x)$ is a polynomial of $P$ that is an eigenvector of $L_x$ corresponding to eigenvalue $\lambda$ . Then we have $x\cdot f(x) = L_x(f(x)) = \lambda\cdot f(x)$ . Because $f(x)$ , being an eigenvector, is not the null polynomial, the degree of the left-hand side is one bigger than the degree of $f(x)$ , the degree of the right-hand side. This is a contradiction. Therefore, there are no eigenvectors of $L_x$ in $P$ . This example shows that, in case of an infinite dimensional vector space (even over the complex numbers), it may happen that a linear transformation has no eigenvectors.

Let $L : \mathbb R^2\to\mathbb R^2$ be the linear map induced by reflection in the line $x + 3 y = 0$ . What are the eigenvalues of $L$ ?

$\{-1,1\}$

Vectors on the line $x + 3 y = 0$ are mapped onto themselves, so $1$ is an eigenvalue of $L$ . Furthermore, since the normal vector to this line gets mapped to its negative, $-1$ is an eigenvalue as well. A map from $\mathbb R^2$ to $\mathbb R^2$ has at most two different eigenvalues, so the answer is $\{-1,1\}$ .

New example