Table of Contents

Polynomials Applied to Linear Operators Examples 1
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 examples regarding polynomials applied to linear operators.
Example 1
Let $T$ be a linear operator on $V$. Suppose that $T^3  6T^2 + 11T  6I = 0$. Prove that then if $\lambda$ is an eigenvalue of $T$ then $\lambda = 1$ or $\lambda = 2$ or $\lambda = 3$.
Let $\lambda$ be an eigenvalue of $T$. Then for some nonzero vector $u \in V$ we have that $T(u) = \lambda u$. Note that:
(1)Substituting this into $T^3  6T^2 + 11T  6I = 0$ and we have that:
(2)Therefore we have that $\lambda = 1$ or $\lambda = 2$ or $\lambda = 3$.
Example 2
Let $T$ be a linear operator on $V$ such that $T = T^2$. Show that then $V = \mathrm{null} (T) \oplus \mathrm{range} (T)$.
To show that $V = \mathrm{null} (T) \oplus \mathrm{range} (T)$ we must show that $V = \mathrm{null} (T) + \mathrm{range} (T)$ and that $\mathrm{null} (T) \cap \mathrm{range} (T) = \{ 0 \} $$.
Notice that every vector $v \in V$ is such that:
(3)We have that $T(v) \in \mathrm{range} (T)$ clearly. Furthermore, we note that:
(4)Therefore $v  T(v) \in \mathrm{null} (T)$. Hence we have that $V = \mathrm{null} (T) + \mathrm{range} (T)$.
We now want to show that $\mathrm{null} (T) \cap \mathrm{range} (T) = \{ 0 \}$. We will show that $\mathrm{null} (T) \cap \mathrm{range} (T) \subseteq \{ 0 \}$ and that $\{ 0 \} \subseteq \mathrm{null} (T) \cap \mathrm{range} (T)$.
Let $u \in \mathrm{null} (T) \cap \mathrm{range} (T)$. Therefore $u \in \mathrm{null} (T)$ and $u \in \mathrm{range} (T)$. Hence $T(u) = 0$. Furthermore, there exists a vector $w \in V$ such that $T(w) = u$. Hence we have that:
(5)Since $T(w) = u$, we have that the equation above implies that $u = 0$. Therefore $\mathrm{null} (T) \cap \mathrm{range} (T) \subseteq \{ 0 \}$.
Furthermore, if $u \in \{ 0 \}$ then $u = 0$ and clearly $0 \in \mathrm{null} (T)$ (since $T(0) = 0$) and $0 \in \mathrm{range} (T)$ (also since $T(0) = 0$) so $0 \in \mathrm{null} (T) \cap \mathrm{range} (T)$.
Therefore $\mathrm{null} (T) \cap \mathrm{range} (T) = \{ 0 \}$. Thus $V = \mathrm{null} (T) \oplus \mathrm{range} (T)$.
Example 3
Let $T$ be a linear operator over $V$. Prove that $4$ is an eigenvalue of $T^2$ if and only if $2$ or $2$ is an eigenvalue of $T$.
$\Rightarrow$ Suppose that $4$ is an eigenvalue of $T^2$. Then for some nonzero vector $u \in V$ we have that $T^2(u) = 4u$, and so:
(6)From the equation above we have that either $(T + 2I)(u) = 0$ or $(T  2I)(u) = 0$. In the first case we have that $(T + 2I)(u) = 0$ implies that $T(u) + 2u = 0$ so $T(u) = 2u$, and so $2$ is an eigenvalue of $T$. In the second case we have that $(T  2I)(u) = 0$ implies that $T(u)  2u = 0$ so $T(u) = 2u$, and so $2$ is an eigenvalue of $T$.
$\Leftarrow$ Suppose that $2$ is an eigenvalue of $T$. Then for some nonzero vector $u \in V$ we have that $T(u) = 2u$. Apply $T$ to both sides of this equation again to get that:
(7)Therefore $4$ is an eigenvalue of $T^2$. Now suppose that instead $2$ is an eigenvalue of $T$. Then for some nonzero vector $u \in V$ we have that $T(u) = 2u$. Apply $T$ to both sides of this equation again to get that:
(8)Once again we have that $4$ is an eigenvalue of $T^2$.