Linear Operators

We have looked a little bit into linear Maps, and now we will look at a specific class of linear maps known as linear operators.

Definition: If $V$ is a vector space over the field $\mathbb{F}$, then a Linear Operator of $V$ is a linear map $T : V \to V$. The set of all linear operators of a vector space $V$ is denoted $\mathcal L (V)$.

One remark that we should make is that if $T$ is a linear operator, then $T$ is also a linear transformation. In fact, the notation $T \in \mathcal (V, V)$ is equivalent to that of $T \in \mathcal L (V)$, though we will use the latter notation for convenience.

We will now look at a few examples of linear operators. The simplest example is the linear operator that is the identity map $T : V \to V$ defined by $T(v) = v$ for all $v \in V$. A more complex example would be the linear operator $T : \wp (\mathbb{R}) \to \wp(\mathbb{R})$ defined by $T(p(x)) = xp'(x)$ for all $p(x) \in \wp (\mathbb{R})$.

We will now show a very important theorem regarding operators on finite dimensional vector spaces. We must emphasize that $V$, must be a finite dimensional and $T$ must be a linear operator from $V$ into itself.

Theorem 1: Let $V$ be a finite dimensional vector space and let $T \in \mathcal (V)$ be a linear operator of $B$. Then the following statements are equivalent. a) $T$ is an invertible linear operator. b) $T$ is an injective linear operator. c) $T$ is a surjective linear operator.

• Proof: $a) \implies b)$ Suppose that $T$ is an invertible linear operator. In an earlier theorem from the Invertibility of a Linear Map page, we saw that if $T$ is invertible that $T$ is bijective, and so $T$ in specific is injective.

• $b) \implies c)$. Suppose that $T$ is an injective linear operator. Then $\mathrm{null} (T) = \{ 0 \}$ and so $\mathrm{dim} (\mathrm{null} (T)) = 0$. Therefore we have that:

(1)

\begin{align} \mathrm{dim} V = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) \\ \mathrm{dim} V = \mathrm{dim} (\mathrm{range} (T)) \end{align}

• Now since $V$ is finite dimensional, $\mathrm{range} (T) \subseteq V$ and $\mathrm{dim} (\mathrm{range} (T)) = \mathrm{dim} V$, then we have that $\mathrm{range} (T) = V$ (since any basis of $\mathrm{range} (T)$ is linearly independent in $V$ and has the same length as any basis in $V$, so this basis also spans $V$ and therefore $V = \mathrm{range} (T)$). Thus, $V$ is surjective.

• $c) \implies a)$ Suppose that $T$ is a surjective linear operator. Then $\mathrm{range} (T) = V$, and so we have that $\mathrm{dim} (\mathrm{range} (T)) = \mathrm{dim} V$ and so:

(2)

\begin{align} \mathrm{dim} V = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) \\ \mathrm{dim} V - \mathrm{dim} (\mathrm{range} (T)) = \mathrm{dim} (\mathrm{null} (T)) \\ 0 = \mathrm{dim} (\mathrm{null} (T)) \end{align}

• So $\mathrm{dim} (\mathrm{null} (T)) = 0$, so $\mathrm{null} (T) = \{ 0 \}$, so $T$ is injective. Since $T$ is both injective and surjective, this implies that $T$ is invertible. $\blacksquare$