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.
Remark Jordan canonical form of a matrix
Let A be as in Example 3.4.
Try to choose a nonzero vector \overrightarrow{v_{3} } so that
\overrightarrow{v_{3} }\left(A-\lambda I_{3} \right) ^{3}=\overrightarrow{0},
\overrightarrow{v_{2} }=\overrightarrow{v_{3} }\left(A-\lambda I_{3} \right) \neq \overrightarrow{0}, and
\overrightarrow{v_{1} }=\overrightarrow{v_{2} }\left(A-\lambda I_{3} \right)=\overrightarrow{v_{3} }\left(A-\lambda I_{3} \right) ^{2} is an eigenvector of A .
Take \overrightarrow{v_{3} }=\overrightarrow{e_{3} }, then \overrightarrow{v_{2} }=\left(0,c,0 \right)=c \overrightarrow{e_{2} } and \overrightarrow{v_{1} }=\left(bc,0,0 \right)=bc \overrightarrow{e_{1} }. Now B=\left\{\overrightarrow{v_{1} },\overrightarrow{v_{2} },\overrightarrow{v_{3} }\right\} is a basis for R³. Thus
\overrightarrow{v_{1} }A=bc\overrightarrow{e_{1} }A=bc\left(\lambda,0,0 \right)= \lambda\overrightarrow{v_{1} }+0\cdot \overrightarrow{v_{2} }+0\cdot \overrightarrow{v_{3} },
\overrightarrow{v_{2} }A=c\overrightarrow{e_{2} }A= c\left(b,\lambda,0 \right)= bc\overrightarrow{e_{1} }+\lambda c \overrightarrow{e_{2} }=1\cdot\overrightarrow{v_{1}} +\lambda\cdot\overrightarrow{v_{2}} +0 \cdot \overrightarrow{v_{3} },
\overrightarrow{v_{3} }A=\overrightarrow{e_{3} }A=\left(0,c,\lambda \right)=c\overrightarrow{e_{2} }+\lambda \overrightarrow{e_{3} }=0\cdot\overrightarrow{v_{1} }+1\cdot\overrightarrow{v_{2} }+\lambda\cdot\overrightarrow{v_{3} }
\Rightarrow \left[A\right] _{B}=PAP^{-1} =\left[\begin{matrix} \lambda & 0 & 0 \\ 1 & \lambda & 0 \\ 0 & 1 & \lambda \end{matrix} \right], where P=\left[\begin{matrix} bc \overrightarrow{e_{1} } \\ c \overrightarrow{e_{2} } \\ \overrightarrow{e_{3} } \end{matrix} \right]. (3.7.25)
\left[A\right] _{B} is called the Jordan canonical form of A (for details, see Sec. 3.7.7). See Fig. 3.35 for λ = 2.