Invariant Subspaces

# Invariant Subspaces

 Definition: Let $V$ be a vector space over the field $\mathbb{F}$, and let $T$ be a linear operator from $V$ to $V$, that is $T \in \mathcal L (V)$. A subspace $U$ of $V$ is said to be Invariant Under $T$ if for all $u \in U$ we have that $T(u) \in U$.

Alternatively we can say that the subspace $U$ is invariant under $T$ if the operator $T$ restricted to the domain subspace $U$ denoted $T \mid_U$ is an operator on $U$. Thus every element $u$ in the subspace $U$ gets mapped to an element $T(u)$ which is also in $U$.

Before we look at some examples of invariant subspaces, we will first acknowledge the following theorem which will provide us the existence of a linear operator for which a subspace $U$ of $V$ is not invariant under provided that $U$ is a nontrivial subspace.

 Theorem 1: If $V$ is a finite-dimensional vector space over the field $\mathbb{F}$ and $U$ is a nontrivial subspace of $V$, then there exists a linear operator $T \in \mathcal L (V)$ such that $U$ is not invariant under $T$.
• Proof: Suppose that $U$ is a nontrivial subspace of $V$. Then $U \neq \{ 0 \}$ and $U \neq V$. We will now construct a linear operator $T$ for which $U$ is not invariant under $T$.
• Let $u \in U$ be such that $u \neq 0$, and let $w \in V \setminus U$. Both of these vectors $u$ and $w$ exist since $U \neq \{ 0 \}$ and $U \neq V$. Now since $V$ is finite-dimensional and $U$ is a subspace of $V$, then $U$ is also finite-dimensional. Extend the linearly independent set $\{ u \}$ to a basis $\{ u, v_1, v_2, ..., v_n \}$ of $V$. If we define $T \in \mathcal L (V)$ by:
(1)
$$T(au) = T(au + a_1v_1 + a_2v_2 + ... + a_nv_n) = aw$$
• Hence $T(u) = w$. However $T(u) = w \not \in U$ and so $U$ is not invariant under $T$. $\blacksquare$

The contrapositive of Theorem 1 tells us that if $U = \{ 0 \}$ or if $U = V$, then $U$ is invariant under every linear operator $T \in \mathcal (V)$.

Let's now look at some examples of invariant subspaces

## Example 1: The Zero and Whole Space as Invariant

Recall that if $V$ is a vector space then both the zero space $\{ 0 \}$ and the whole space $V$ are subspaces of $V$.

Let $T \in \mathcal (V)$. The zero subspace $\{ 0 \}$ is invariant under $T$. This can easily be seen since $T(0) = 0 \in \{ 0 \}$.

Furthermore, the whole space $V$ is also trivially invariant under $T$ since for all elements $v \in V$ we have that $T(v) \in V$.

## Example 2: The Null Space as Invariant

If $T \in \mathcal (V)$, then the null space $\mathrm{null} (T) = \{ v \in V : T(v) = 0 \}$ is invariant under $T$. We note that $0 \in \mathrm{null} (T)$ since $T(0) = 0$. Since every element $v \in V$ is such that $T(v) = 0$, then the image of $\mathrm{null}(T)$ under $T$ will only contain $0$ which, once again, is in $\mathrm{null}(T)$.

## Example 3: The Range Space as Invariant

Recall that if $T \in \mathcal (V)$ then the range $\mathrm{range} (T) = \{ T(v) : v \in V \}$ is invariant under $T$. This can easily be seen since if $u \in \mathrm{range}(T)$, then $T(u) \in \mathrm{range}(T)$ by the definition of the subspace $\mathrm{range}(T)$.

## Example 4: The Differentiation of Polynomials Operators

Let $T \in \mathcal ( \wp_5 (\mathbb{R}), \wp_5 (\mathbb{R}))$ be defined by $T(p(x)) = p'(x)$ for all $p(x) \in \wp_5 (\mathbb{R})$. Consider the subspace $\wp_4 (\mathbb{R})$ of $\wp_5 (\mathbb{R})$. If we have a polynomial $p(x) \in \wp_4 (\mathbb{R})$, then $T(p(x)) = p'(x)$ will be a polynomial of degree less than or equal to $3$, and so $T(p(x)) \in \wp_4 (\mathrm{R})$, so $\wp_4 (\mathbb{R})$ is invariant under $T$.

More generally, if $T \in \mathcal ( \wp_5 (\mathbb{R}), \wp_5 (\mathbb{R}))$ is defined by $T(p(x)) = p'(x)$, then all of the subspaces $\wp_{m-1} (\mathrm{R})$, $\wp_{m-2} (\mathrm{R})$, …, $\wp_{1} (\mathrm{R})$, $\wp_{0} (\mathrm{R})$ are all invariant under $T$ since any polynomial in any of these subsets are mapped to polynomials of lesser degree which is still contained in the chosen subspace.

## Example 5

Let $T \in \mathcal L(v)$ and suppose that $U_1$ and $U_2$ are subspaces of $V$ and are both invariant under $T$. Prove that $U_1 \cap U_2$ is also invariant under $T$.

Suppose that $U_1$ and $U_2$ are both invariant under $T$, and suppose that $u \in U_1 \cap U_2$. Then $u \in U_1$ and $u \in U_2$. Since $U_1$ is invariant under $T$, then $T(u) \in U_1$, and since $U_2$ is invariant under $T$, then $T(u) \in U_2$ and so $T(u) \in U_1 \cap U_2$, so $U_1 \cap U_2$ is invariant under $T$.

## Example 6

Let $T \in \mathcal L(V)$ and suppose that $U_1$, $U_2$, …, $U_m$ are subspaces of $V$, all of which invariant under $T$. Prove that $U_1 + U_2 + ... + U_m$ is also invariant under $T$.

Suppose that $u \in U_1 + U_2 + ... + U_m$. Then $u = u_1 + u_2 + ... + u_m$ where $u_j \in U_j$ for $j = 1, 2, ..., m$. Applying the linear operator $T$ to both sides of the equation above and we have that:

(2)
\begin{align} \quad T(u) = T(u_1 + u_2 + ... + u_m) = T(u_1) + T(u_2) + ... + T(u_m) \end{align}

Since $U_1$, $U_2$, …, $U_m$ are all invariant substances under $T$, then since $u_j \in U_j$, we have that $T(u_j) \in U_j$ for $j = 1, 2, ..., m$. Hence $T(u) \in U_1 + U_2 + ... + U_m$ and so $U_1 + U_2 + ... + U_m$ is invariant under $T$.