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)

\begin{align} \quad T^2(u) = T(T(u)) = T(\lambda u) = \lambda T(u) = \lambda^2 u \\ \quad T^3(u) = T(T^2(u)) = T(\lambda^2 u) = \lambda^2 T(u) = \lambda^3 u \\ \quad \quad \quad \vdots \quad \quad \quad \\ \quad T^n(u) = T(T^{n-1}(u)) = T(\lambda^{n-1} u) = \lambda^{n-1} T(u) = \lambda^n u \end{align}

Substituting this into $T^3 - 6T^2 + 11T - 6I = 0$ and we have that:

(2)

\begin{align} \quad T^3 - 6T^2 + 11T - 6I = 0 \\ \quad \lambda^3 - 6\lambda^2 + 11 \lambda - 6 \lambda = 0 \\ \quad (\lambda - 1)(\lambda - 2)(\lambda - 3) = 0 \end{align}

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)

\begin{align} \quad v = T(v) + v - T(v) \\ \quad v = T(v) + (v - T(v)) \end{align}

We have that $T(v) \in \mathrm{range} (T)$ clearly. Furthermore, we note that:

(4)

\begin{align} \quad T(v - T(v)) = T(v) - T^2(v) = 0 \end{align}

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)

\begin{align} \quad T(u) = 0 \\ \quad T(T(w)) = 0 \\ \quad T^2(w) = 0 \\ \quad T(w) = 0 \end{align}

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)

\begin{align} \quad T^2(u) - 4I(u) = 0 \\ \quad (T^2 - 4I)(u) = 0 \\ \quad (T + 2I)(T - 2I)(u) = 0 \end{align}

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)

\begin{align} \quad T^2(u) = T(T(u)) = T(-2u) = -2T(u) = -2(-2u) = 4u \end{align}

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)

\begin{align} \quad T^2(u) = T(T(u)) = T(2u) = 2T(u) = 2(2u) = 4u \end{align}

Once again we have that $4$ is an eigenvalue of $T^2$.