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We have a linear transformation $T: \Bbb R^2 \to \Bbb R^2$ whose matrix is $$\begin{pmatrix} 2&3\\ 0&2 \end{pmatrix}. $$ Suppose we start with the unit circle in $\Bbb R^2$, we know that its image under the transformation is an ellipse. I am trying to find the major and minor axis of the ellipse.

I tried finding the eigenvalues and eigenvector but there is only one eigenvector $(1,0)^t$ corresponding to the eigenvalue $2$.

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  • $\begingroup$ What about finding the preimage of the canonical basis and then going forward? $\endgroup$ – Tito Eliatron Nov 5 '20 at 11:09
  • $\begingroup$ @TitoEliatron Can you explain a bit more how that will help? $\endgroup$ – User8976 Nov 5 '20 at 11:11
  • $\begingroup$ Sorry.... it was a bad idea $\endgroup$ – Tito Eliatron Nov 5 '20 at 11:21
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Here is one way to do it.

A parametric representation of the ellipse is $$E(\theta)=(2\cos\theta+3\sin\theta,2\sin\theta),$$ where $0\leq\theta\leq 2\pi$. The major axis in the direction which maximises the length of $E(\theta)$, and the minor axis in the direction which minimises the length. The square of the length is $$\|E(\theta)\|^2=4\sin^2\theta+(2\cos\theta+3\sin\theta)^2$$ Elementary calculation shows that derivative of this is $3(4\cos 2\theta+3\sin 2\theta)$, whose four zeros in $[0,2\pi]$ are $$\arctan(2),\arctan(2)+\pi/2,\arctan(2)+\pi,\arctan(2)+3\pi/2$$

The major axis is therefore in the direction of $\arctan(2)$, and its length is $8$, and the minor axis is in the direction of $\arctan(2)+\pi/2$, and its length is $2$.

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If we define $$ A=\pmatrix{2 & 3 \\ 0 & 2} $$

then the cartesian equation of the ellipse is $x^T M x=1$, where: $$ M=(A^{-1})^TA^{-1}=\pmatrix{1/4 & -3/8 \\ -3/8 & 13/16}. $$

Its semi-axes are given by $1/\sqrt{\lambda_i}$, where $\lambda_1=1$ and $\lambda_2=1/4$ are the eigenvalues of $M$, while the eigenvectors are aligned with the axes.

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