Question

If$\mathit{T}\mathbf{\left(}\overrightarrow{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$is an invertible linear transformation from${\mathbf{ℝ}}^{\mathbf{2}}$to${\mathbf{ℝ}}^{\mathbf{2}}$, then the image $\mathit{T}\mathbf{\left(}\mathbf{\Omega }\mathbf{\right)}$of the unit circle$\mathit{\Omega }$is an ellipse. See Exercise 2.2.54.
a. Sketch this ellipse when,$\mathbf{A}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{p}& \mathbf{0}\\ \mathbf{0}& \mathbf{q}\end{array}\mathbf{\right]}$where pand qare positive. What is its area?
b. For an arbitrary invertible transformation$\mathit{T}\mathbf{\left(}\overrightarrow{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$, denote the lengths of the semi major and semi minor axes of$\mathit{T}\mathbf{\left(}\mathbf{\Omega }\mathbf{\right)}$by aand b, respectively. What is the relationship among a,b, and$\mathit{d}\mathit{e}\mathit{t}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$?.

c. For the transformation $\mathit{T}\left(\overrightarrow{x}\right)\mathbf{=}\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]$, sketch this ellipse and determine its axes. Hint: Consider $\mathit{T}\left[\begin{array}{c}1\\ 2\end{array}\right]$and $\mathit{T}\left[\begin{array}{c}1\\ -1\end{array}\right]$

Short answer

Therefore,

a) Area is$pq\pi$.

b)$\left|detA\right|=\frac{ab\mathrm{\pi }}{\mathrm{\pi }}=ab$.

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

Step-by-step solution

To find area. 

a) The unit circle centered in the origin is

$\Omega =\left\{\left[\begin{array}{c}\cos t\\ \sin t\end{array}\right]:t\in [0,2\mathrm{\pi })\right\}$

Therefore,

$$\begin{gathered} T\left(\Omega \right)=\left\{T\left[\begin{array}{c}\cos t\\ \sin t\end{array}\right]:t\in [0,2\mathrm{\pi })\right\} \\ T\left(\Omega \right)=\left\{\left[\begin{array}{cc}p& 0\\ 0& q\end{array}\right]\left[\begin{array}{c}\cos t\\ \sin t\end{array}\right]:t\in [0,2\mathrm{\pi })\right\} \\ T\left(\Omega \right)=\left\{\left[\begin{array}{c}p\cos t\\ q\sin t\end{array}\right]:t\in [0,2\mathrm{\pi })\right\} \end{gathered}$$

So, it's an ellipse whose semi major and semi minor axes are pand q, respectively. Then it’s area is $pq\mathrm{\pi }$.

To find det(A).

b) For an arbitrary linear transformation $T:{\mathrm{ℝ}}^2\to {\mathrm{ℝ}}^2$, let and be the semi major and the semi minor axis of the ellipse that is $T\left(\Omega \right)$, respectively.

The area of this ellipse is $ab\mathrm{\pi }$, while the area of the unit circle is$\mathrm{\pi }$.

Therefore,

$\left|detA\right|=\frac{ab\mathrm{\pi }}{\mathrm{\pi }}=ab$.

To find axes. 

c) The unit circle can be written as

$\Omega =\left\{\cos t.\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right]+\sin t.\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]:t\in [0,2\mathrm{\pi })\right\}$.

Therefore,

$\Omega =\left\{\cos t.\frac{1}{\sqrt{2}}\left[\begin{array}{c}4\\ 4\end{array}\right]+\sin t.\frac{1}{\sqrt{2}}\left[\begin{array}{c}4\\ -2\end{array}\right]:t\in [0,2\mathrm{\pi })\right\}.$

So, the axes of $T\left(\Omega \right)$are $\frac{1}{\sqrt{2}}\left[\begin{array}{c}4\\ 4\end{array}\right]$and $\frac{1}{\sqrt{2}}\left[\begin{array}{c}4\\ -2\end{array}\right]$.