Question 2.14: Let R2 be endowed with Cartesian coordinate system N = {e1^→...
Let R² be endowed with Cartesian coordinate system N=\left\{\overrightarrow{e_{1} },\overrightarrow{e_{2} } \right\} as Fig. 2.17(b) indicated. Investigate the geometric mapping properties of the linear transformation
f\left(x_{1},x_{2} \right)=\left(2x_{1}-x_{2},-3x_{1}+4x_{2} \right)
=\overrightarrow{x}A, where \overrightarrow{x}=\left(x_{1},x_{2} \right) and A=\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}.
It is easy to check that f is indeed a linear transformation. Moreover, f is isomorphic. To see this, suppose f\left(x_{1},x_{2} \right)=0 . This is equivalent to say that
2x_{1}-x_{2}=0 , -3x_{1}+4x_{2} =0 ,
whose only solution is \overrightarrow{x}=\left(x_{1},x_{2} \right)=\overrightarrow{0}. Hence f is one-to-one. Next, for any given \overrightarrow{y}=\left(y_{1},y_{2} \right), solve f\left(x_{1},x_{2} \right)=\left(y_{1},y_{2} \right) , i.e.
\left \{ \begin{matrix} 2x_{1}-x_{2}=y_{1} \\ -3x_{1}+4x_{2} =y_{2} \end{matrix} \right.
\Rightarrow \left \{ \begin{matrix} x_{1}=\frac{4}{5}y_{1}+\frac{1}{5}y_{2} \\ \\ x_{2}=\frac{3}{5}y_{1}+\frac{2}{5}y_{2} \end{matrix} \right. , with A^{-1} =\frac{1}{5}\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}.
The resulting vector \overrightarrow{x}=\left(\frac{4}{5}y_{1}+\frac{1}{5}y_{2},\frac{3}{5}y_{1}+\frac{2}{5}y_{2}\right) is then the (unique)
solution to f\left(\overrightarrow{x}\right)=\overrightarrow{y}. Thus f is onto (see (2.7.8)). It is worth to notice that the above algebraic computation can be simplified as the computation of the inverse A^{-1} of the matrix A:\overrightarrow{y} =\overrightarrow{x}A ⇔\overrightarrow{x}=\overrightarrow{y}A^{-1}.
f maps straight lines into straight lines and preserves ratios of lengths of line segments along the same line. The equation
a_{1}x_{1}+a_{2}x_{2} +b=0 \left(*_{1} \right)
of a line can be written in matrix form as
\overrightarrow{x}\left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right] +b=0.
Hence, the image of this line under f has the equation
\overrightarrow{y}A^{-1}\left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right] +b=0, or
\left(4a_{1}+3a_{2}\right)y_{1}+ \left(a_{1}+2a_{2}\right)y_{2}+5b=0, \left(*_{2} \right)
which represents a straight line. The property that f preserves ratios of lengths segments is contained in the definition of f, a linear transformation. Do you see why? Of course, one can use parametric equation \overrightarrow{x}=\overrightarrow{a}+t\overrightarrow{b} for the line (see (2.5.4)) to prove these results.
f preserves the relative positions of two lines (see (2.5.9)). Suppose the
two lines have respective equation \overrightarrow{x}=\overrightarrow{a_{1}}+t\overrightarrow{b_{1}} and \overrightarrow{x}=\overrightarrow{a_{2}}+t\overrightarrow{b_{2}} where t ∈ R . Note that the image of the line \overrightarrow{x}=\overrightarrow{a_{i}}+t\overrightarrow{b_{i}} under f has the equation \overrightarrow{y}=\overrightarrow{a_{i}}A+t\overrightarrow{b_{i}}A for i=1,2. Therefore, for example,
two lines intersect at a point \overrightarrow{x_{0}}.
⇔ \overrightarrow{b_{1}} and \overrightarrow{b_{2}} are linearly independent.
⇔ \overrightarrow{b_{1}}A and \overrightarrow{b_{2}}A are linearly independent (because A is invertible).
⇔ The two image lines intersect at a point \overrightarrow{x_{0}}A.
Again, one should refer to Ex. <B> 4 of Sec. 2.4 or (2.7.8).
Suppose the x_{1} x_{2} -plane and y_{1} y_{2} -plane are deposited on the same plane. We are posed with the problem: Is there any line coincident with its image under f? If yes, how to find it and how many can we find?
For simplicity, suppose firstly the line passes through the origin \overrightarrow{0}=\left(0,0\right). Then in \left(*_{1} \right) and in \left(*_{2} \right) , b=0 should hold and both lines are coincident if and only if
\frac{4a_{1}+3a_{2}}{a_{1}} =\frac{a_{1}+2a_{2}}{a_{2}} .
There are two algebraic ways to treat the above equation.
(1) One is
a_{2}\left(4a_{1}+3a_{2}\right) = a_{1}\left(a_{1}+2a_{2}\right)
\Leftrightarrow a^{2}_{1} -2a_{1}a_{2}-3a^{2}_{2}=\left(a_{1}-3a_{2}\right) \left(a_{1}+a_{2}\right)=0
\Leftrightarrow a_{1}+a_{2}=0 which results in a_{1}:a_{2}=1:-1 , or
a_{1}-3a_{2}=0 which results in a_{1}:a_{2}=3:1.
This means that the line x_{1} -x_{2}=0 and 3x_{1}+x_{2}=0 are kept invariant under the mapping f .
(2) The other is
\frac{4a_{1}+3a_{2}}{a_{1}} =\frac{a_{1}+2a_{2}}{a_{2}}=\alpha
\Leftrightarrow \left \{ \begin{matrix} \left(4-\alpha \right)a_{1}+3a_{2}=0 \\ a_{1}+\left(2-\alpha \right)a_{2}=0 \end{matrix} \right. or \begin{bmatrix} 4-\alpha & 3 \\ 1 & 2-\alpha \end{bmatrix}\left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right] =\left[\begin{matrix} 0 \\ 0 \end{matrix} \right]
⇔(Suppose there does exist \left(a_{1},a_{2}\right) ≠\left(0,0\right), and it solves equations. Refer to Exs. <A> 2 and <B> 4 of Sec. 2.4.)
det\begin{bmatrix} 4-\alpha & 3 \\ 1 & 2-\alpha \end{bmatrix}=\left|\begin{matrix}4-\alpha & 3 \\ 1 & 2-\alpha \end{matrix}\right|
=\left(4-\alpha\right)\left(2-\alpha\right)-3=\alpha ^{2}-6\alpha +5=0
⇔ α = 1, 5.
The case α = 1 will result in a_{1}+a_{2}=0 which, in turn, implies that x_{1}- x_{2}=0 is an invariant line. While α = 5 results in a_{1}-3a_{2}=0 and hence 3x_{1}+ x_{2}=0 is another invariant line. We obtain the same results as in (1).
Note 1 Method (2) above can be replaced by the following process. By the vector forms of \left(*_{1} \right) and \left(*_{2} \right) , we have:
The line \overrightarrow{x} \left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right] =0 coincides with the line \overrightarrow{x}A^{-1} \left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right] =0.
⇔There exists constant μ such that
A^{-1} \left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right]=μ\left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right] .
⇔\left( A^{-1} -μ I_{2}\right)\left[\begin{matrix} a_{1} \\ a_{2} \end{matrix} \right]=\left[\begin{matrix}0\\ 0 \end{matrix} \right]. \left(*_{3} \right)
⇔det \left( A^{-1} -μ I_{2}\right)=det \left[\frac{1}{5}\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} -\mu\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right]
=\frac{1}{25}\left|\begin{matrix} 4-5\mu & 3 \\ 1 & 2-5\mu \end{matrix} \right|=\frac{1}{5}\left(5\mu ^{2}-6\mu+1 \right)=0.
⇒ \mu = 1 and \mu =\frac{1}{5}.If μ = 1, by \left(*_{3} \right), we have a_{1}-3a_{2}=0, and if \mu =\frac{1}{5}, then a_{1}-a_{2}=0.
Note 2 Traditionally, especially in Cartesian coordinate system, we adopt \left(*_{1} \right) as starting point for almost every computation work about straight lines. Why not use parametric equation in vector form as (2.5.4)? If so, then:
The line \overrightarrow{x}=t\overrightarrow{b} coincides with its image \overrightarrow{x}=t\overrightarrow{b}A under f.
⇔ There exists constant λ such that \overrightarrow{b}A=\lambda\overrightarrow{b}.
⇔ \overrightarrow{b} \left(A-\lambda I_{2}\right)=\overrightarrow{0}.
⇔ \left(since \overrightarrow{b}≠\overrightarrow{0}\right)
det \left(A-\lambda I_{2}\right)= \left|\begin{matrix} 2-\lambda & -3 \\ -1 & 4-\lambda \end{matrix} \right|=\lambda ^{2}-6\lambda +5=0.
⇒ λ = 1, 5.
In case λ = 1, to solve \overrightarrow{b} \left(A-\lambda I_{2}\right)=\overrightarrow{b} \left(A- I_{2}\right)= \overrightarrow{0} is equivalent to solve
\left(b_{1} b_{2}\right) \begin{bmatrix} 2-1 & -3 \\ -1 & 4-1 \end{bmatrix}=0 or b_{1}-b_{2}=0, where \overrightarrow{b} =\left(b_{1}, b_{2}\right).
Hence b_{1}: b_{2}=1 : 1 and the line \overrightarrow{x} =t \left(1,1\right) is an invariant line. In case λ = 5, to solve \overrightarrow{b} \left(A-\lambda I_{2}\right)= \overrightarrow{b} \left(A-5I_{2}\right)=0 is to solve
\left(b_{1} b_{2}\right) \begin{bmatrix} -3 & -3 \\ -1 & -1 \end{bmatrix}=0 or 3b_{1}+b_{2}=0.
Hence b_{1}: b_{2}=1 : -3 and thus the line \overrightarrow{x} =t \left(1,-3\right) is another invariant line.
We call, in Note 2 above, \lambda _{1} =1 and \lambda _{2} =5 the eigenvalues of the square matrix A, and \overrightarrow{x _{1}} =t \left(1,1\right) for t ≠0 eigenvectors of A related to \lambda _{1} and \overrightarrow{x _{2}} =t \left(1,-3\right) for t ≠ 0 eigenvectors of A related to \lambda _{2}=5. When comparing
various algebraic methods mentioned above, the method in Note 2 is the simplest one to be generalized.
We will see its advantages over the others as we proceed. Notice that \overrightarrow{x _{1}}A =\overrightarrow{x _{1}} and \overrightarrow{x _{2}}A =5\overrightarrow{x _{2}}. See Fig. 2.71.
An immediate advantage is as follows. Let \overrightarrow{v_{1}}=\left(1,1\right) and \overrightarrow{v_{2}}=\left(1,-3\right).
Since
det\left[\begin{matrix}\overrightarrow{v_{1}} \\ \overrightarrow{v_{2}} \end{matrix} \right] =det\begin{bmatrix} 1 & 1 \\ 1 & -3 \end{bmatrix}=\left|\begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right|=-4\neq 0,
so \overrightarrow{v_{1}} and \overrightarrow{v_{2}} are linearly independent. Thus B =\left\{ \overrightarrow{v_{1}}, \overrightarrow{v_{2}}\right\} is a basis for R².
What does the linear isomorphism f look like in B? Now,
f\left(\overrightarrow{v_{1}}\right)=\overrightarrow{v_{1}}A=\overrightarrow{v_{1}}=\overrightarrow{v_{1}}+0\cdot \overrightarrow{v_{2}}
\Rightarrow \left[f\left(\overrightarrow{v_{1}}\right)\right] _{B} =\left(1,0\right) , and
f\left(\overrightarrow{v_{2}}\right)=\overrightarrow{v_{2}}A=5\overrightarrow{v_{2}}=0\cdot \overrightarrow{v_{1}}+5 \overrightarrow{v_{2}}
\Rightarrow \left[f\left(\overrightarrow{v_{2}}\right)\right] _{B} =\left(0,5\right).
Therefore, the matrix representation of f with respect to B is
\left[f\right] _{B}=\left[\begin{matrix} \left[f\left(\overrightarrow{v_{1}}\right)\right] _{B} \\ \left[f\left(\overrightarrow{v_{2}}\right)\right] _{B} \end{matrix} \right] =\begin{bmatrix} 1 & 0 \\ 0 & 5 \end{bmatrix}=PAP^{-1} , \left(*_{3} \right)
where
P=\left[\begin{matrix} \overrightarrow{v_{1}} \\ \overrightarrow{v_{2}} \end{matrix} \right]=\begin{bmatrix} 1 & 1 \\ 1 & -3 \end{bmatrix}.
Hence, for any \overrightarrow{x}= \left(x_{1},x_{2}\right) ∈ R² , let \left[\overrightarrow{x}\right] _{B}=\left(\alpha _{1},\alpha _{2}\right)=\overrightarrow{x}P^{-1}, then
\left[f\left(\overrightarrow{x}\right)\right] _{B}= \left[\overrightarrow{x}\right] _{B}\left[f\right] _{B}= \left(\alpha _{1} \alpha _{2}\right)\begin{bmatrix} 1 & 0 \\ 0 & 5 \end{bmatrix}= \left(\alpha _{1} , 5\alpha _{2}\right) \left(*_{4} \right)
(see (2.4.2), (2.6.4) or (2.7.23) if necessary). Parallelograms in Fig. 2.72 show how goodness the representation \left[f\right] _{B} over the original \left[f\right] _{N}=A.
A useful and formal reinterpretation of \left(*_{3} \right) and \left(*_{4} \right) is as follows. Since
PAP^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 5 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} +\begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} +5 \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}
⇒ A = P^{-1} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}+5 P^{-1} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} P = A _{1}+5A _{2},
where
A _{1} = P^{-1} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} P = \frac{1}{4} \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -3 \end{bmatrix}=\begin{bmatrix} \frac{3}{4} & \frac{3}{4} \\ \\ \frac{1}{4} & \frac{1}{4}\end{bmatrix},
A _{2} = P^{-1} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} P = \begin{bmatrix} \frac{1}{4} & \frac{-3}{4} \\ \\ \frac{-1}{4} & \frac{3}{4}\end{bmatrix},
the claimed advantage will come to surface once we can handle the geometric mapping properties of both A _{1} and A _{2}. Note that
\overrightarrow{x} =\left(x_{1},x_{2} \right)\in R^{2} using N=\left\{\overrightarrow{e_{1} },\overrightarrow{e_{2}} \right\}\\\ \downarrow \\\ \overrightarrow{x}P^{-1}=\left[\overrightarrow{x}\right] _{B}=\left(\alpha _{1},\alpha _{2} \right)\in R^{2} using B=\left\{\overrightarrow{v_{1} },\overrightarrow{v_{2}} \right\} \\\ \downarrow \\\ \overrightarrow{x}P^{-1}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} =\left(\alpha _{1},0 \right), the projection of \left(\alpha _{1},\alpha _{2} \right) onto \left(\alpha _{1},0 \right) in B \\\ \downarrow \\\ \overrightarrow{x}P^{-1}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}P =\left(\alpha _{1},0 \right) \left[\begin{matrix}\overrightarrow{v_{1} } \\ \overrightarrow{v_{2} } \end{matrix} \right]= \alpha _{1}\overrightarrow{v_{1} }=\overrightarrow{x } A_{1}.Hence A_{1} defines a linear transformation \overrightarrow{x } ∈ R² → \overrightarrow{x } A_{1} ∈ \ll \overrightarrow{v_{1} } \gg , called the eigenspace of A related to the eigenvalue \lambda_{1}. Also,
\overrightarrow{x } ∈\ll \overrightarrow{v_{1} } \gg ⇔\overrightarrow{x }A_{1}= \overrightarrow{x }
and thus A^{2}_{1}=A_{1} holds. A_{1} is called the projection of R² onto \ll \overrightarrow{v_{1} } \gg along \ll \overrightarrow{v_{2} }\gg . Similarly, A_{2} defined by \overrightarrow{x } ∈R^{2} → \overrightarrow{x }A_{2} ∈\ll \overrightarrow{v_{2} } \gg is called the projection of R² onto \ll \overrightarrow{v_{2} } \gg along \ll \overrightarrow{v_{1} } \gg.For details, see (2.7.51).
We summarize as an abstract result.
Let us return to the original mapping f in Example 2.14. Try to find the image of the square with vertices at \left(1,1\right) , \left(-1,1\right) , \left(-1,-1\right) and \left(1,-1\right) under the mapping f. Direct computation shows that
f\left(1,1\right) =\left(1 1\right) \begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}=\left(1,1\right),
f\left(-1,1\right) =\left(-1 1\right) \begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}=\left(-3,7\right),
f\left(-1,-1\right) =\left(-1 -1\right) \begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}=\left(-1,-1\right)=-f\left(1,1\right),
f\left(1,-1\right) =\left(1 -1\right) \begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}=\left(3,-7\right)=-f\left(-1,1\right).
Connect the image points (1, 1), (−3, 7), (−1,−1) and (3,−7) by consecutive line segments and the resulting parallelogram is the required one. See Fig. 2.72. Try to determine this parallelogram by using the diagonal canonical decomposition of A. By the way, the original square has area equal to 4 units. Do you know what is the area of the resulting parallelogram? It is 20 units. Why?
In general, a linear operator on R² may not be one-to-one, i.e. not an isomorphism.
