Q18E IfT(x鈫�)=Ax鈫抜s an invertible ... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 1: Q18E (page 1)

If $T (\overset{鈫�}{x}) = A \overset{鈫�}{x}$ is an invertible linear transformation from ${鈩�}^{2}$ to ${鈩�}^{2}$ , then the image $T (惟)$ of the unit circle $惟$ is an ellipse. See Exercise 2.2.54.
a. Sketch this ellipse when, $A = [\begin{matrix} p & 0 \\ 0 & q \end{matrix}]$ where pand qare positive. What is its area?
b. For an arbitrary invertible transformation $T (\overset{鈫�}{x}) = A \overset{鈫�}{x}$ , denote the lengths of the semi major and semi minor axes of $T (惟)$ by aand b, respectively. What is the relationship among a,b, and $d e t (A)$ ?.
c. For the transformation $T (\overset{鈫�}{x}) = [\begin{matrix} 3 & 1 \\ 1 & 3 \end{matrix}]$ , sketch this ellipse and determine its axes. Hint: Consider $T [\begin{matrix} 1 \\ 2 \end{matrix}]$ and $T [\begin{matrix} 1 \\ - 1 \end{matrix}]$

Short Answer

Expert verified

Therefore,

a) Area is $p q 蟺$ .

b) $|d e t A| = \frac{a b 蟺}{蟺} = a b$ .

c) The axes of $T (惟)$ are $\frac{1}{\sqrt{2}} [\begin{matrix} 4 \\ 4 \end{matrix}]$ and $\frac{1}{\sqrt{2}} [\begin{matrix} 4 \\ - 2 \end{matrix}]$ .

Step by step solution

To find area.

a) The unit circle centered in the origin is

$惟 = \{[\begin{matrix} \cos t \\ \sin t \end{matrix}] : t 鈭� [0, 2 蟺)\}$

Therefore,

$T (惟) = \{T [\begin{matrix} \cos t \\ \sin t \end{matrix}] : t 鈭� [0, 2 蟺)\} T (惟) = \{[\begin{matrix} p & 0 \\ 0 & q \end{matrix}] [\begin{matrix} \cos t \\ \sin t \end{matrix}] : t 鈭� [0, 2 蟺)\} T (惟) = \{[\begin{matrix} p \cos t \\ q \sin t \end{matrix}] : t 鈭� [0, 2 蟺)\}$

So, it's an ellipse whose semi major and semi minor axes are pand q, respectively. Then it鈥檚 area is $p q 蟺$ .

To find det(A).

b) For an arbitrary linear transformation $T : {鈩�}^{2} 鈫� {鈩�}^{2}$ , let and be the semi major and the semi minor axis of the ellipse that is $T (惟)$ , respectively.

The area of this ellipse is $a b 蟺$ , while the area of the unit circle is $蟺$ .

Therefore,

$|d e t A| = \frac{a b 蟺}{蟺} = a b$ .

To find axes.

c) The unit circle can be written as

$惟 = \{\cos t . \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \end{matrix}] + \sin t . \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ - 1 \end{matrix}] : t 鈭� [0, 2 蟺)\}$ .

Therefore,

$惟 = \{\cos t . \frac{1}{\sqrt{2}} [\begin{matrix} 4 \\ 4 \end{matrix}] + \sin t . \frac{1}{\sqrt{2}} [\begin{matrix} 4 \\ - 2 \end{matrix}] : t 鈭� [0, 2 蟺)\} .$

So, the axes of $T (惟)$ are $\frac{1}{\sqrt{2}} [\begin{matrix} 4 \\ 4 \end{matrix}]$ and $\frac{1}{\sqrt{2}} [\begin{matrix} 4 \\ - 2 \end{matrix}]$ .