Polynomials Applied to Linear Operators Examples 2

Recall from the Polynomials Applied to Linear Operators page that we can apply a polynomial to a linear operator $T \in \mathcal L(V)$ by defining $T^m = \underbrace{T \circ T \circ ... \circ T}_{\mathrm{m \: times}}$ where $m$ is a positive integer. Furthermore, if $T$ is an invertible linear operator then we define $T^{-m} = \underbrace{T^{-1} \circ T^{-1} \circ ... \circ T^{-1}}_{\mathrm{m \: times}}$.

We will now look at some more examples regarding polynomials applied to linear operators.

Example 1

Let $T$ be a linear operator over the finite-dimensional vector space $V$. Let $v \in V$ be a nonzero vector, and let $p$ be a nonzero polynomial of the smallest degree for which $p(T)v = 0$. Prove that then every root of $p$ is an eigenvalue of $T$.

Let $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{F}$ be the roots of the polynomial $p$. Then we have that $p(x)$ can be factored as:

(1)

\begin{align} \quad p(x) = c(x - \lambda_1)(x - \lambda_2) ... (x - \lambda_m) \end{align}

Replace $x$ with $T$ and set this equal to $0$ to get that:

(2)

\begin{align} \quad p(T)(v) = c(T - \lambda_1 I)(T - \lambda_2 I)...(T - \lambda_m I)(v) = 0 \end{align}

Consider one of the roots, $\lambda_j$ where $1 ≤ j ≤ m$. Then $(T - \lambda_j)(v) = 0$ implies that $T(v) = \lambda_j v$, so $\lambda_j$ is an eigenvalue of $T$. Since $\lambda_j$ was arbitrary, we have that then every nonzero, $\lambda_1, \lambda_2, ..., \lambda_m$ is an eigenvalue of $T$.