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 $1\le {g}_{\lambda}$, since the existence of an eigenvalue means the existence of a nonzero 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 ${\lambda}_{1}\ne {\lambda}_{2}$.

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 $V$ finitedimensional.
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 ${g}_{\lambda}\le {m}_{\lambda}$.

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
${M}_{T}=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$
From this matrix, it is easy to see that ${p}_{T}(x)={(x1)}^{2}$ 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 onedimensional subspace of ${\mathbb{R}}^{2}$ generated by $(1,0)$. As a result, $T$ is not diagonalizable.
Title  eigenspace 

Canonical name  Eigenspace 
Date of creation  20130322 17:23:07 
Last modified on  20130322 17:23:07 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  9 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 15A18 