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 $\lbrace v\in V\mid Tv=\lambda v\rbrace$ 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 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 $\lambda_1\ne \lambda_2$
5. In fact, if $W_{\lambda}'$ is the sum of eigenspaces corresponding to eigenvalues of $T$ other than $\lambda$ then $W_{\lambda}\cap W_{\lambda}'=0$

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 $g_{\lambda}\le m_{\lambda}$
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