# 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. 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. 2.
3. 3.

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

4. 4.

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

5. 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. 1.

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

2. 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. 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 Eigenspace 2013-03-22 17:23:07 2013-03-22 17:23:07 CWoo (3771) CWoo (3771) 9 CWoo (3771) Definition msc 15A18