Question 3.4: The linear operator A= [λ 0 0 b λ 0 0 c λ] , where bc ≠ 0 ha...
The linear operator
A= \left[\begin{matrix} \lambda & 0 & 0 \\ b & \lambda & 0 \\ 0 & c & \lambda \end{matrix} \right], where bc ≠ 0
has the following properties:
1. A satisfies its characteristic polynomial −(t − λ)³, i.e.
\left(A-\lambda I_{3} \right) ^{3}=O_{3\times 3}
but
A-\lambda I_{3} = \left[\begin{matrix} 0& 0 & 0 \\ b & 0 & 0 \\ 0 & c & 0 \end{matrix} \right] \neq O_{3\times 3}, \left(A-\lambda I_{3} \right) ^{2}= \left[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ bc & 0 & 0 \end{matrix} \right] \neq O_{3\times 3}.
2. Hence, A has eigenvalues λ, λ, λ with associated eigenvectors t\overrightarrow{e_{1} },t\in R and t ≠ 0 and A is not diagonalizable.
3. Notice that A can be written as the sum of the following linear operators:
A=\lambda I_{3} + \left(A-\lambda I_{3} \right)
= \lambda \left[\begin{matrix} 1& 0 & 0 \\ \frac{b}{\lambda } & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] + \left[\begin{matrix} 0& 0 & 0 \\ 0 & 0 & 0 \\ 0 & c & 0 \end{matrix} \right].
Note that \ll \overrightarrow{e_{1} } \gg is the only invariant line (subspace). See Fig. 3.35 for some geometric feeling.
