# 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.

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