Notes - Linear Algebra MT23, Invariant subspaces
Flashcards
Suppose $T : V \to V$ is a linear transformation. What is a $T$-invariant subspace?
$U \le V$ such that $T(U) \le U$.
Suppose $U \le V$ is $T$-invariant, and $U$ has basis $[e _ 1, \ldots, e _ k]$, extended to $[e _ {k+1}, \ldots, e _ n]$. Then what can be said about
\[Te_j\]
for $1 \le j \le k$?
\[Te_j = \sum^k_{i=1} \alpha_i e_i\]
i.e. no $e _ {k+1}, \ldots, e _ n$ terms.
Suppose $U \le V$ is $T$-invariant, and $U$ has basis $[e _ 1, \ldots, e _ k]$, extended to $[e _ {k+1}, \ldots, e _ n]$. Then how does the matrix of $T$ with respect to this basis look, and what fact does this allow you to deduce about the characteristic polynomial $\chi _ T$ in terms of $T \vert _ U$ and $\overline T$?
\[\left(\begin{array}{c|c}
\left. T \right|\_U & * \\
\hline
0 & \overline{T}
\end{array}\right)\]
and
\[\chi_T(x) = \chi\_{\left. T \right|\_U}(x) \cdot \chi\_{\overline T}(x)\]