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'eigenspace'
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| Title of object: |
eigenspace |
| Canonical Name: |
Eigenspace |
| Type: |
Definition |
| Created on: |
2007-07-08 12:30:19 |
| Modified on: |
2007-07-08 14:20:48 |
| Classification: |
msc:15A18 |
Preamble:
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Content:
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 \emph{eigenspace} corresponding to $\lambda$. Let us write this set $W_{\lambda}$.
$W_{\lambda}$ can be viewed as the kernel of the linear transformation $T-\lambda I$. As a result, $W_{\lambda}$ is a subspace of $V$. 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. Also $W_{\lambda}$ is an invariant subspace under $T$ ($T$-invariant). Another simple consequence is that $W_{\lambda_1}\cap W_{\lambda_2}=0$ iff $\lambda_1\ne \lambda_2$.
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:
\begin{enumerate}
\item If $m_{\lambda}$ is the algebraic multiplicity of $\lambda$, then $g_{\lambda}\le m_{\lambda}$.
\item 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 $\lambda\in S_T$.
\item In other words, if $p_T(x)$ splits over $k$, then $T$ is diagonalizable iff $V=W$.
\end{enumerate}
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
\begin{center}$M_T=
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}.$
\end{center}
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. |
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