eigenspace

Let $V$ be a vector space over a field $k$ . Fix a linear transformation $T$ on $V$ . Suppose $\lambda$ is an eigenvalue of $T$ . The set $\{v\in V\mid Tv=\lambda v\}$ is called the eigenspace (of $T$ ) corresponding to $\lambda$ . Let us write this set $W_{\lambda}$ .

Below are some basic properties of eigenspaces.

1.

$W_{\lambda}$ can be viewed as the kernel of the linear transformation $T-\lambda I$ . As a result, $W_{\lambda}$ is a subspace of $V$ .
2.

The dimension of $W_{\lambda}$ is called the geometric multiplicity of $\lambda$ . Let us denote this by $g_{\lambda}$ . It is easy to see that $1\leq g_{\lambda}$ , since the existence of an eigenvalue means the existence of a non-zero eigenvector corresponding to the eigenvalue.
3.

$W_{\lambda}$ is an invariant subspace under $T$ ( $T$ -invariant).
4.

$W_{\lambda_{1}}\cap W_{\lambda_{2}}=0$ iff $\lambda_{1}\neq\lambda_{2}$ .
5.

In fact, if $W_{\lambda}^{\prime}$ is the sum of eigenspaces corresponding to eigenvalues of $T$ other than $\lambda$ , then $W_{\lambda}\cap W_{\lambda}^{\prime}=0$ .

From now on, we assume $V$ finite-dimensional.

Let $S_{T}$ be the set of all eigenvalues of $T$ and let $W=\oplus_{\lambda\in S}W_{\lambda}$ . We have the following properties:

1.

If $m_{\lambda}$ is the algebraic multiplicity of $\lambda$ , then $g_{\lambda}\leq m_{\lambda}$ .
2.

Suppose the characteristic polynomial $p_{T}(x)$ of $T$ can be factored into linear terms, then $T$ is diagonalizable iff $m_{\lambda}=g_{\lambda}$ for every $\lambda\in S_{T}$ .
3.

In other words, if $p_{T}(x)$ splits over $k$ , then $T$ is diagonalizable iff $V=W$ .

For example, let $T:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be given by $T(x,y)=(x,x+y)$ . Using the standard basis, $T$ is represented by the matrix

$M_{T}=\begin{pmatrix}1&1\\ 0&1\end{pmatrix}.$

From this matrix, it is easy to see that $p_{T}(x)=(x-1)^{2}$ is the characteristic polynomial of $T$ and $1$ is the only eigenvalue of $T$ with $m_{1}=2$ . Also, it is not hard to see that $T(x,y)=(x,y)$ only when $y=0$ . So $W_{1}$ is a one-dimensional subspace of $\mathbb{R}^{2}$ generated by $(1,0)$ . As a result, $T$ is not diagonalizable.

Title	eigenspace
Canonical name	Eigenspace
Date of creation	2013-03-22 17:23:07
Last modified on	2013-03-22 17:23:07
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	9
Author	CWoo (3771)
Entry type	Definition
Classification	msc 15A18