For the linear operator \(T\) of Exercise 3, prove that \[ (T^2 - I)(T - 3I) = 0. \]
Solution
Goal: Show that the operator in Exercise 3 satisfies \((T^2-I)(T-3I)=0\).
1. Read \(T\) from Exercise 3:for \(x=(x_1,x_2,x_3)\in\mathbb R^3\),\[T(x_1,x_2,x_3)=(3x_1,\;x_1-x_2,\;2x_1+x_2+x_3).\]In matrix form on the standard basis,\[M=\begin{pmatrix}3&0&0\\1&-1&0\\2&1&1\end{pmatrix}.\]
2. Compute \(T^2\).
2.1. A direct product gives\[M^2=\begin{pmatrix} 9&0&0\\2&1&0\\9&0&1 \end{pmatrix},\]so\[T^2(x_1,x_2,x_3)=(9x_1,\;2x_1+x_2,\;9x_1+x_3).\]
3. Compute \(T-3I\) and \(T^2-I\).
3.1. Since\[M-3I=\begin{pmatrix}0&0&0\\1&-4&0\\2&1&-2\end{pmatrix},\quad M^2-I=\begin{pmatrix}8&0&0\\2&0&0\\9&0&0\end{pmatrix},\]every column of \(M^2-I\) except the first is zero.
4. Multiply \((M^2-I)(M-3I)\).
4.1. The second and third columns of \(M^2-I\) are zero, so only the first column can contribute.
4.2. The first column of \(M-3I\) is \((0,1,2)^T\), and row arithmetic gives\[\begin{aligned} [8,0,0]\cdot(0,1,2)&=0,\\ [2,0,0]\cdot(0,1,2)&=0,\\ [9,0,0]\cdot(0,1,2)&=0. \end{aligned}\]Hence \((M^2-I)(M-3I)=0\), and therefore
4.3. \((T^2-I)(T-3I)=0\).