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.
Suppose ab ≠0. Consider the linear operator
f\left(\overrightarrow{x} \right) =f\left(x_{1},x_{2} \right) =\left(bx_{2},x_{1}+ax_{2} \right).
Let \overrightarrow{x} = \left(x_{1},x_{2} \right) ≠ \overrightarrow{0} . Then,
\ll\overrightarrow{x}\gg is an invariant line (subspace).
⇔There exists a constant λ ≠ 0 such that f\left(\overrightarrow{x} \right) =\lambda\overrightarrow{x} , i.e.
b x_{2}=\lambda x_{1}
x_{1}+a x_{2}=\lambda x_{2}.
Since b ≠0, x_{1}≠0 should hold. Now, put x_{2} =\frac{\lambda }{b}x_{1} into the second equation and we get, after eliminating x_{1},
\lambda ^{2}-a\lambda -b=0.
The existence of such a nonzero λ depends on the discriminant a²+4b ≥ 0.
Case 1 a² + 4b > 0. Then \lambda ^{2}-a\lambda -b=0. has two real roots
\lambda _{1} =\frac{a+\sqrt{a^{2}+4b } }{2} , \lambda _{2} =\frac{a-\sqrt{a^{2}+4b } }{2}
Solve f\left(\overrightarrow{x} \right) =\lambda _{i}\overrightarrow{x} ,i=1,2 , and we have the corresponding vector \overrightarrow{v_{i} } =\left(b,\lambda _{i}\right) . It is easy to see that \overrightarrow{v_{1}} and \overrightarrow{v_{2}} are linearly independent. Therefore, f has two different invariant lines \ll \overrightarrow{v_{1}}\gg and \ll \overrightarrow{v_{2}}\gg . See Fig. 2.53.
Case 2 a² + 4b = 0. λ² − aλ − b = 0 has two equal real roots
\lambda =\frac{a}{2} ,\frac{a}{2}.
Solve f\left(\overrightarrow{x} \right) =\lambda \overrightarrow{x}, and the corresponding vector is \overrightarrow{v}=(−a, 2). Then \ll \overrightarrow{v}\gg is the only invariant line. See Fig. 2.54.
Case 3 a² + 4b < 0. f does not have any invariant line. See Fig. 2.55.
Summarize these results in
