Invariant Subspaces Examples 2

# Invariant Subspaces Examples 2

Recall from the Invariant Subspaces page that if $V$ is a vector space, $U$ is a subspace of $V$, and $T \in \mathcal L (V)$ then the subspace $U$ is said to be invariant under the linear operator $T$ if for every vector $u \in U$ we have that $T(u) \in U$.

We noted that $\{ 0 \}$, $V$, $\mathrm{null} (T)$ and $\mathrm{range} (T)$ were all invariant under $T$.

We also saw that if $V$ is a finite-dimensional vector space and $U$ is a nontrivial subspace of $V$ (that is $U$ is not the zero space $\{ 0 \}$ and not the whole space $V$) then there exists a linear operator $T \in \mathcal L(V)$ such that $U$ is not invariant under $T$.

We will now look at some more examples regarding invariant subspaces.

## Example 1

Let $S$ and $T$ be linear operators on $V$. Suppose that $ST = TS$. Show that then $\mathrm{null} (S)$ is invariant under $T$.

We want to show that $u \in \mathrm{null} (S)$ implies that $T(u) \in \mathrm{null} (S)$.

Let $u \in \mathrm{null} (S)$. Then $S(u) = 0$. Applying the linear operator $T$ to both sides of the equation above and we have that:

(1)
\begin{align} \quad (TS)(u) = 0 \end{align}

However, $ST = TS$. This implies that $(ST)(u) = 0$, that is, $S(T(u)) = 0$. Therefore $T(u) \in \mathrm{null} (S)$. Therefore $\mathrm{null} (S)$ is invariant under $T$.

## Example 2

Let $S$ and $T$ be linear operators on $V$. Suppose that $ST = TS$. Show that then $\mathrm{range} (S)$ is invariant under $T$.

We want to show that $u \in \mathrm{range} (S)$ implies that $T(u) \in \mathrm{range} (S)$.

Let $u \in \mathrm{range} (S)$. Then there exists a vector $w \in V$ such that $S(w) = u$. Applying the linear operator $T$ to both sides of the equation above and we have that:

(2)
\begin{align} \quad (TS)(w) = T(S(w)) = T(u) \end{align}

However, $ST = TS$. This implies that $S(T(w)) = T(u)$. This equation implies that $T(u) \in \mathrm{range} (S)$. Therefore $\mathrm{range} (S)$ is invariant under $T$.