Determine the image and area of the unit disk under linear transformation

Question

Determine the image and area of the unit disk D = {($x$, $y$) | $x^2 + y^2 ≤ 1$ } under the linear transformation T: R$^2$ $\rightarrow$ R$^2$ defined by T([, ]) = $ \begin{bmatrix} 1 & -\frac{\sqrt{3}}{2} \\ \sqrt{3} & \frac{1}{2} \end{bmatrix}$$ \begin{bmatrix} x \\ y \end{bmatrix}$

So far I've taken the inverse of A = $ \begin{bmatrix} 1 & -\frac{\sqrt{3}}{2} \\ \sqrt{3} & \frac{1}{2} \end{bmatrix}$ and I've gotten $\frac{1}{2}$$ \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\sqrt{3} & 1 \end{bmatrix}$. Not sure how to proceed with the solution, however.

Taken inverse of the matrix, but not sure where to go from there — prismarine, Sep 22 '20 at 01:28
Always add your attempts to your question ...It's better in order to get answers and hints. — user577215664, Sep 22 '20 at 01:28
Sorry, I am a new user. I didn't know the protocol, I'll make sure I do that next time — prismarine, Sep 22 '20 at 01:29
@JCAA How did you calculate that? I understand that area of unit disk will be , but how do you know the image is an ellipse? — prismarine, Sep 22 '20 at 01:33
@La'QuaviousJackson, substitute $$x=\frac{X}{2}+Y\frac{\sqrt 3}{2}\ y=-X\sqrt 3+Y$$ in the original equation. — SarGe, Sep 22 '20 at 01:36
@SarGe You don't have to distribute the 1/2 when calculating the inverse? — prismarine, Sep 22 '20 at 01:40
@SarGe Or wait...is this just det(A) • area of disk formula? So det(A) = 2 and area of unit disk is , so area(T(D)) is just 2? — prismarine, Sep 22 '20 at 01:42
You do not need to calculate much, just the determinant of the matrix. The rest applies to any invertible linear transformation. — markvs, Sep 22 '20 at 03:25
You got two answers Do you need anything else. If not, accept one of them. — markvs, Sep 22 '20 at 23:19

score 0 · Accepted Answer · 2020-09-22T03:20:17.877

Compose the rotation through $π/3$ with $(x,y)\mapsto(2x,y)$.

The former "is" $\begin{pmatrix}1/2&-\sqrt3/2\\\sqrt3/2&1/2\end{pmatrix}$; the latter $\begin{pmatrix}2&0\\0&1\end{pmatrix}$.

The rotation preserves area and shape. The other one takes the unit circle to the ellipse $ (\dfrac x2)^2+y^2=1$. But that has area $2π$ (by this Show that the Area of image = Area of object $\cdot |\det(T)|$? Where $T$ is a linear transformation from $R^2 \rightarrow R^2$, or a little calculus).

markvs · Answer 2 · 2020-09-22T03:36:06.777

With linear transformations, the area of the image of a figure is the area of the figure times the determinant of the transformation. The area of a unit circle is $\pi$ and the determinant of your matrix is $2$. So the area of the image of a unit circle is $2\pi$:

Show that the Area of image = Area of object $\cdot |\det(T)|$? Where $T$ is a linear transformation from $R^2 \rightarrow R^2$

The image of a quadric under a non-singular linear transformation is a quadric. The image of a bounded figure is a bounded figure. The only bounded quadrics are ellipses (follows from the classification of quadrics). Hence the image of a unit circle is an ellipse.

Determine the image and area of the unit disk under linear transformation

2 Answers2