Invariant Subspaces

Defn

We say $W<V$ is $T$-invariant if $T(W)\subset W$ i.e., $\forall v\in W, T(v)\in W$

Let $W$ be an invariant subspace of $V$. Characteristic polynomial $P_{T_W}$ of $T_W$ divides the characteristic polynomial $P_T$ of $T$

Choose a basis $\alpha=\lbrace v_1,...,v_k\rbrace $ of $W$

Extend this to a basis $\beta=\lbrace v_1,...,v_k,v_{k+1},...,v_n\rbrace $ of $V$

Then $[T]_\beta=\begin{pmatrix}|&|& &|\\ T(v_1)&T(v_1)&...&T(v_1)\\|&|& &|\end{pmatrix}=\begin{pmatrix}A&B\\0&C\end{pmatrix}$

$T(v_j)\in W=span(\alpha)$

$\forall j\in \lbrace 1,...,k\rbrace ,T(v_j)=\sum ^k_{i=1} a_{i,j}v_i$

$A=[T_w]_\alpha$

$det([T]_\beta-tI)=det(T-tI)$

$P_T(t)=P_{T_W}(t)q(t)$ thus $P_{T_W}$ divides $P(T)$

Let $W$ be $T$-cyclic subspace of $V$ generated by a non-zero $v\in V$. Let $k=dim(V)$, then:

Since $v\neq0$ so $\lbrace v\rbrace $ is linearly independent

Let $j$ be the largest integer st $\beta=\lbrace v,T(v),T^2(v),...T^{j-1}(v)\rbrace $ is linearly independent

Since $V$ is finite-dimensional; thus, such a $j$ exists

Suppose $span(\beta)$ was $T$-invariant.

Since $v\in w$ and $w$ was T-variant, so $\beta\subset W\implies span(\beta)\subset W$

As $W$ is the smallest $T$-variant subspace containing $v$, so $W\subset span(\beta)$

$\beta$ is a basis of $W$ so $j=k$

To show $span(\beta)$ is $T$-invariant