# Invariant Subspaces Examples 1

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 examples regarding invariant subspaces.

## Example 1

**Let $T \in \mathcal L(V)$ and let $U$ be a subspace of $V$. Show that if $U \subseteq \mathrm{null} (T)$ then $U$ is invariant under $T$.**

We want to show that $u \in U$ implies that $T(u) \in U$.

Suppose that $U \subseteq \mathrm{null} (T)$. Then for every vector $u \in U$ we have that $T(u) = 0$. However, $0 \in U$ since $U$ is a subspace (and hence must contain the zero vector) $]] so $0 = T(u) \in U$. Therefore $u \in U$ implies that $T(u) \in U$, so $U$ is invariant under $T$.

## Example 2

**Let $T \in \mathcal L(V)$ and let $U$ be a subspace of $V$. Show that if $\mathrm{range} (T) \subseteq U$ then $U$ is invariant under $T$.**

Once again, we want to show that $u \in U$ implies that $T(u) \in U$.

Suppose that $\mathrm{range} (T) \subseteq U$. For any vector $u \in U$ we must have that $T(u) \in \mathrm{range} (T)$. However, $\mathrm{range} (T) \subseteq U$ so then $T(u) \in U$ as well. Therefore $u \in U$ implies that $T(u) \in U$ so $U$ is invariant under $T$.

## Example 3

**Let $T \in \mathcal L(V)$ and let $U$ and $W$ be subspaces of $V$ that are invariant under $T$. Show that $U \cap W$ is invariant under $T$.**

We want to show that for every vector $v \in U \cap W$ we have that $T(v) \in U \cap W$.

Let $v \in U \cap W$. Then $v \in U$ and $v \in W$. Since $U$ and $W$ are both invariant under $T$ then we have that $T(v) \in U$ and $T(v) \in W$. Therefore $T(v) \in U \cap W$.

Therefore $U \cap W$ is invariant under $T$.

## Example 4

**Let $T \in \mathcal L(V)$ and let $U_1$, $U_2$, …, $U_n$ be any collection of subspaces of $V$ that are invariant under $T$. Show that $\bigcap_{i=1}^{n} U_i$ is invariant under $T$.**

We want to show that for every vector $v \in \bigcap_{i=1}^{n} U_i$ we have that $T(v) \in \bigcap_{i=1}^{n} U_i$.

Let $v \in \bigcap_{i=1}^{n} U_i$. Then $v \in U_1$, $v \in U_2$, …, $v \in U_n$. Since $U_1$, $U_2$, …, $U_n$ are each invariant under $T$ we have that $T(v) \in U_1$, $T(v) \in U_2$, …, $T(v) \in U_n$. Therefore $T(v) \in \bigcap_{i=1}^{n} U_i$ and so $\bigcap_{i=1}^{n} U_i$ is invariant under $T$.

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