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Let $M = \begin{bmatrix} 5 & 3 \\ 3 & -6 \end{bmatrix}$ and let $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be a linear transformation defined by $T(x) = Mx$. Let $\mathcal{C}$ be the unit circle in $\mathbb{R}^{2}$ and $T(\mathcal{C}$) be the resulting figure when $T$ is applied to $\mathcal{C}$. What is the area of $T(\mathcal{C})$?

This is what I did:

$$M^{-1} = -\frac{1}{39} \begin{bmatrix} -6 & -3 \\ -3 & 5 \end{bmatrix},$$

so $M$ maps the unit circle to

$$\left\{\begin{bmatrix} u \\v \end{bmatrix} = M\left(\begin{bmatrix} x \\ y \end{bmatrix} \right) \mid x^2 + y^2 = 1\right\}.$$

The equation of this set in the $u, v$ space is

$$\left|\left|M^{-1} \begin{bmatrix}u\\v \end{bmatrix} \right|\right|^{2} = 1. $$

I don't really know how to go on from here. I plugged in $M^{-1}$ and got nowhere to finding the area. Maybe I'm doing the wrong approach.

  • The matrix is symmetric. Find the eigenvalues. Their eigenvectors are going to be perpendicular. Therefore, this map is contracting and/or expanding two perpendicular directions. Therefore, the image is an ellipse. The eigenvalues tell your the lengths of its axes, and therefore the area. Given that the are of the ellipse is $\pi ab$, where $a,b$ are the axes, then you don't even need to compute the eigenvalues. Their product is the determinant. – user647486 Apr 05 '19 at 20:25

1 Answers1

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The area of the image is the area of the original object, multiplied by $|\det(M)|$, so in your case, that would be $|\det(M)|$.

See this answer for a detailed proof

mrtaurho
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