The linear operator A=[0 1 b a], where ab ≠ 0 has the following properties:(1) a2 + 4b 0. Let λ1 =a+√a2+4b/2 and λ2 = a-√a2+4b/2 and vi^→ = (b, λi) for i = 1, 2.1. vi^→A = λi v i^→, i = 1, 2. Thus, λ1 and λ2 are eigenvalues of A with corresponding eigenvectors v1^→ and v2^→.2. and

Question

The linear operator

[latex]A=\begin{bmatrix} 0 & 1 \ b & a \end{bmatrix} ,[/latex] where ab ≠ 0

has the following properties:

(1) a² + 4b > 0. Let [latex]\lambda_{1}=\frac{a+\sqrt{a^{2}+4b} }{2}[/latex] and [latex]\lambda_{2}=\frac{a-\sqrt{a^{2}+4b} }{2}[/latex] and [latex]\overrightarrow{v_{i}} =\left(b,\lambda_{i} ight)[/latex] for i = 1, 2.

[latex]\overrightarrow{v_{i}} A=\lambda_{i}\overrightarrow{v_{i}} ,[/latex] i = 1, 2. Thus, [latex]\lambda_{1} and \lambda_{2}[/latex] are eigenvalues of A with corresponding eigenvectors [latex]\overrightarrow{v_{1}} and \overrightarrow{v_{2}}.[/latex]
[latex]\ll \overrightarrow{v_{1}} \gg and \ll \overrightarrow{v_{2}} \gg[/latex] are invariant lines (subspaces) of R² under A.

In the basis [latex]B = \left\{\overrightarrow{v_{1}},\overrightarrow{v_{2}} ight\},[/latex] A can be represented as

[latex]\left[A ight]_{B}=PAP^{-1}= \begin{bmatrix} \lambda_{1} & 0 \ 0 & \lambda_{2} \end{bmatrix} ,[/latex] where [latex]P =\left[\begin{matrix} \overrightarrow{v_{1}} \ \overrightarrow{v_{2}} \end{matrix} ight]=\begin{bmatrix} b & \lambda_{1} \ b & \lambda_{2} \end{bmatrix}.[/latex]

(see Exs. 4, 5 of Sec. 2.4 and Sec. 2.7.3.) See Fig. 2.53.

(2) a² + 4b = 0. Let λ = [latex]\frac{a}{2} ,\frac{a}{2} and \overrightarrow{v}=\left(-a,2 ight).[/latex]

[latex]\overrightarrow{v} A= \lambda \overrightarrow{v}.[/latex]λ is an eigenvalue of multiplicity 2 of A with corresponding eigenvector [latex]\overrightarrow{v}. [/latex]

[latex]\ll \overrightarrow{v} \gg[/latex] is an invariant line (subspace) of R² under A.

In the basis [latex]B =\left\{\overrightarrow{v},\overrightarrow{e_{1}} ight\},[/latex] A can be represented as

[latex]\left[A ight]_{B}=PAP^{-1}= \begin{bmatrix} \lambda & 0 \ \frac{1}{2} & \lambda \end{bmatrix}= \lambda \begin{bmatrix} 1 & 0 \ \frac{1}{2\lambda} & 1 \end{bmatrix}=\lambda I_{2}+\begin{bmatrix} 0 & 0 \ \frac{1}{2} & 0 \end{bmatrix},[/latex]

where [latex]P =\left[\begin{matrix} \overrightarrow{v} \ \overrightarrow{e_{1}} \end{matrix} ight]=\begin{bmatrix} -a & 2\ 1 & 0 \end{bmatrix}.[/latex]

This is the composite map of a shearing followed by an enlargement with scale λ. See Figs. 2.52 and 2.54.

(3) a² + 4b < 0. A does not have any invariant line. Notice that

[latex]\begin{bmatrix} 0 & 1 \ b & a \end{bmatrix}=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}\begin{bmatrix} b & 0 \ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & 0 \ 0 & a \end{bmatrix}.[/latex]

Hence, the geometric mapping property of A can be described as follows in [latex]N =\left\{\overrightarrow{e_{1}},\overrightarrow{e_{2}} ight\}.[/latex]

See Fig. 2.55.

The operator

[latex]\begin{bmatrix} a & 1 \ b & 0 \end{bmatrix}, where ab ≠ 0[/latex]

is of the same type.

Accepted Answer

Is there any more basic operator than these mentioned from Example 2.3 to Example 2.9 ? No more! It will turn eventually out that these operators are sufficiently enough to describe any operators on R² in various ways, both algebraically and geometrically. Please refer to Secs. 2.7.5 to 2.7.8.

Before we are able to do so, we have to realize, as we learned from these
examples, that the natural Cartesian coordinate system [latex]N =\left\{\overrightarrow{e_{1}},\overrightarrow{e_{2}} ight\}[/latex] is not always the best way to describe all possible linear operators whose definitionsare independent of any particular choice of basis for R². A suitable choice of a basis [latex]B = \left\{\overrightarrow{x_{1}},\overrightarrow{x_{2}} ight\}[/latex] for R², according to the features of a given linear operator or a matrix (acts, as a linear operator in (2.7.7))

[latex]A= \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} , [/latex]

will reduce as many entries [latex]a_{ij} [/latex] to zero as possible. That is, after a change of coordinate from N to B, and in the eyes of B,A becomes

[latex]\left[A ight]_{B}=PAP^{-1}= \begin{bmatrix} \lambda_{1} & 0 \ 0 & \lambda_{2} \end{bmatrix} or \begin{bmatrix} \lambda & 0 \ 1 & \lambda \end{bmatrix} or \begin{bmatrix} 0 & 1 \ b & a \end{bmatrix}. [/latex] (2.7.10)

It is [latex]\left[A ight]_{B} [/latex] that makes algebraic computations easier and geometric mapping properties clearer and our life happier.

Through these examples, for a given real 2 × 2 matrix [latex]A = \left[a_{ij} ight],[/latex] its main features are of two aspects:

1. the rank r(A),
2. the existence of an invariant line (subspace), if any.

We treated these concepts in examples via classical algebraic methods. Now, we reformulate them formally in the language, commonly used in linear algebra, which will be adopted throughout the book.

Definition Let [latex]A = \left[a_{ij} ight]_{2 imes 2}[/latex] be a real matrix and considered as a linear operator as in (2.7.7). If there exist a scalar λ ∈ R and a nonzero vector [latex]\overrightarrow{x}[/latex] ∈ R² such that

[latex]\overrightarrow{x}A= \lambda \overrightarrow{x},[/latex]

then λ is called an eigenvalue (or characteristic root) of A and [latex]\overrightarrow{x}[/latex] an associated or corresponding eigenvector (or characteristic vector). (2.7.11)

In case λ≠0 and [latex]\overrightarrow{x} eq \overrightarrow{0}[/latex] is an associated eigenvector, then this geometrically means that the line [latex]\ll \overrightarrow{x}\gg[/latex] is an invariant line (subspace) of R² under A, and if λ = 1, A keeps each vector along [latex]\ll \overrightarrow{x}\gg ( or point on \ll \overrightarrow{x}\gg )[/latex] fixed, which is called a line of invariant points. If λ = −1, reverse it.

How to determine if A has eigenvalues and how to find them if they do exist? Suppose there does exist a nonzero vector [latex]\overrightarrow{x} [/latex] such that

[latex]\overrightarrow{x}A= \lambda \overrightarrow{x} [/latex] for some scalar λ ∈ R

[latex]\Leftrightarrow \overrightarrow{x} \left(A-\lambda I_{2} ight)=\overrightarrow{0}[/latex]

⇔ (by (3) in (2.7.8))

[latex]det \left(A-\lambda I_{2} ight)= det \begin{bmatrix} a_{11}-\lambda & a_{12} \ a_{21} & a_{22}-\lambda \end{bmatrix} = \left|\begin{matrix} a_{11}-\lambda & a_{12} \ a_{21} & a_{22}-\lambda \end{matrix} ight| [/latex]

[latex]= \lambda^{2} - \left(a_{11}+a_{22} ight)\lambda+\left(a_{11}a_{22}-a_{12}a_{21} ight)=0.[/latex]

This means that any eigenvalue of A, if exists, is a root of the quadratic equation [latex]\lambda^{2} - \left(a_{11}+a_{22} ight)\lambda +a_{11}a_{22}-a_{12}a_{21} =0.[/latex] Solve this equation for real root λ and reverse the processes mentioned above to obtain the corresponding eigenvectors [latex]\overrightarrow{x}[/latex].

Question 2.9: The linear operator A=[0 1 b a], where ab ≠ 0 has the follow...

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