Polynomials Applied to Linear Operators

Suppose that $T$ is a linear operator from the vector space $V$ to $V$. Then the operator $T \circ T = T^2$ is an operator from $V$ to $V$. In fact, we can compose $T$ with itself multiple times, that is, for $m$ as a positive integer we can define the operator $T^m$ as follows:

(1)

\begin{align} T^m = \underbrace{T \circ T \circ ... \circ T}_{\mathrm{m \: times}} \end{align}

If $m = 0$, then we can define $T^0$ to be the identity operator on $V$.

Furthermore, for a positive integer $m$ we can define $T^{-m} = (T^{-1})^m$ provided that $T$ is an invertible linear operator, that is:

(2)

\begin{align} (T^{-1})^m = \underbrace{T^{-1} \circ T^{-1} \circ ... \circ T^{-1}}_{\mathrm{m \: times}} \end{align}

So now we have a formal definition of a power of a linear operator $T$. Suppose that $p(x) \in \wp (\mathbb{F})$ suh that $p(x) = a_0 + a_1x + a_2x^2 + ... + a_mx^m$. Then we can define the operator of $p$ applied to $T$ as:

(3)

\begin{align} \quad p(T) = a_0I + a_1T + a_2T^2 + ... + a_mT^m \end{align}

Proposition 1: If $T \in \mathcal L(V)$, then the transformation $S : \wp (\mathbb{F}) \to \mathcal L (V)$ defined by $S(p(x)) = p(T)$ is linear.

• Proof: Let $T \in \mathcal L (V)$. To show that $S$ is linear, we must show that $S$ has both the additivity property and the homogeneity property.

• First let $p(x), q(x) \in \wp (\mathbb{F})$. For the additivity property, we have that:

(4)

\begin{align} \quad S(p(x) + q(x)) = p(T) + q(T) = S(p(x)) + S(q(x)) \end{align}

• Now let $p(x) \in \wp (\mathbb{F})$ and let $a \in \mathbb{F}$. Then for the homogeneity property we have that:

(5)

\begin{align} \quad S(ap(x)) = ap(T) = a S(p(x)) \end{align}

• Therefore $S$ is a linear transformation. $\blacksquare$