Question 2.8: The linear operator A = [λ1 0 1 λ2] , where λ1 λ2 ≠ 0 has th...

The linear operator

A=\left[\begin{matrix} \lambda _{1} & 0 \\ 1 & \lambda _{2} \end{matrix} \right] ,   where \lambda _{1}\lambda _{2} ≠ 0

has the following properties:

(1) In case \lambda _{1}≠\lambda _{2} . Let \overrightarrow{v_{1} } = \overrightarrow{e_{1} } ,\overrightarrow{v_{2} } = (1, \lambda _{2} −\lambda _{1}).

1. \overrightarrow{v_{1} }A= \lambda _{1}\overrightarrow{v_{1} } and \overrightarrow{v_{2} }A = \lambda _{2}\overrightarrow{v_{2} }. Thus, \lambda _{1} and \lambda _{2} are eigenvalues of A with corresponding eigenvectors \overrightarrow{v_{1} } and \overrightarrow{v_{2} }.

2. \ll\overrightarrow{v_{1} }\gg and \ll\overrightarrow{v_{2} }\gg are invariant lines of A.

In fact, B = {\overrightarrow{v_{1} },\overrightarrow{v_{2} }} is a basis for R² and

\left[A\right] _{B}=PAP^{-1} =\left[\begin{matrix} \lambda _{1} & 0 \\ 0 & \lambda _{2} \end{matrix} \right] ,   where P=\left[\begin{matrix} \overrightarrow{v_{1} } \\ \overrightarrow{v_{2} } \end{matrix} \right] =\left[\begin{matrix}1 & 0 \\ 1 & \lambda _{1}-\lambda _{2} \end{matrix} \right]

is the matrix representation of A with respect to B (see Exs. <B> 4, 5 of Sec. 2.4 and Sec. 2.7.3). See Fig. 2.50.

(2) In case \lambda _{1}= \lambda _{2} = \lambda _{}. Then

\overrightarrow{e_{1} }A= λ\overrightarrow{e_{1} }

and \ll\overrightarrow{e_{1} }\gg is the only invariant subspace of A. Also

A=\left[\begin{matrix} \lambda & 0 \\ 1 & \lambda \end{matrix} \right] =\lambda I_{2}+\left[\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right]=\lambda \left[\begin{matrix} 1 & 0 \\ \frac{1}{\lambda } & 1 \end{matrix} \right]

shows that A has the mapping behavior as

\begin{matrix} \overrightarrow{x}=(x_1,x_2)\underset{\text{enlargement }\lambda I_2 }{\longrightarrow} (\lambda x_1,\lambda x_2) \\\quad \quad\quad \quad\quad \quad A \searrow \quad \quad \quad\quad \quad\quad \quad \downarrow ^{\text{translation a long}(x_2,0) }\\\overrightarrow{x}A=(\lambda x_1+x_2,\lambda x_2)\quad \quad \end{matrix}

See Fig. 2.51.

A special operator of this type is

S=\left[\begin{matrix} 1 & 0 \\ a & 1 \end{matrix} \right] ,  where a ≠ 0,

which is called a shearing. S maps each point \overrightarrow{x}=\left(x_{1},x_{2} \right) to the point \left(x_{1}+\alpha x_{2},x_{2} \right) along the line passing \overrightarrow{x} and parallel to the x_{1}-axis, to the right if a > 0 with a distance a x_{2} , proportional to the x_{2}-coordinate x_{2} of \overrightarrow{x} by a fixed constant a, and to the left if a < 0 by the same manner. Therefore,

1. S keeps every point on the x_{1}-axis fixed which is the only invariant subspace of it.

2. S moves every point \left(x_{1},x_{2} \right) with x_{2}≠0 along the line parallel to the x_{1}-axis , through a distance with a constant proportion a to its distance to the x_{1}-axis , to the point \left(x_{1}+\alpha x_{2},x_{2} \right) . Thus, each line parallel to x_{1}-axis is an invariant line.

See Fig. 2.52.

Note that linear operators

\left[\begin{matrix} \lambda _{1} & 1 \\ 0 & \lambda _{2} \end{matrix}\right],   where \lambda _{1}\lambda _{2} ≠ 0; \left[\begin{matrix} 1 & a \\ 0 & 1 \end{matrix} \right] , where a ≠ 0,

can be investigated in a similar manner.