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 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 $W_{\lambda }^{\prime }$ 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 $g_{\lambda }\le m_{\lambda }$.

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

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 ${ℝ}^2$ generated by $(1,0)$. As a result, $T$ is not diagonalizable.