# Major and Minor axis of an ellipse obtained from applying a linear transformation to the unit circle

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 $$2$$.

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

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 $$2$$.
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.