= Solution
Take an incident field along $\widehat{\mathbf x}$, without loss of generality because this component has the same circular <eigenmodes> for every direction of input <linear polarization>. Resolve its <Jones vector> into equal circular components:
$$
E_0\widehat{\mathbf x}=\frac{E_0}{\sqrt2}(\mathbf u_++\mathbf u_-),\qquad
\mathbf E_{\rm out}=\frac{E_0}{\sqrt2}(\lambda_+\mathbf u_++\lambda_-\mathbf u_-)
=E_0\begin{pmatrix}a\\-ib\end{pmatrix}.
$$
Using the <complex amplitude> convention $\operatorname{Re}[\mathbf E e^{-i\omega t}]$, the real circular contributions are
$$
\mathbf E_+(t)=\frac{E_0\lambda_+}{2}(\widehat{\mathbf x}\cos\omega t-\widehat{\mathbf y}\sin\omega t),\qquad
\mathbf E_-(t)=\frac{E_0\lambda_-}{2}(\widehat{\mathbf x}\cos\omega t+\widehat{\mathbf y}\sin\omega t).
$$
They counter-rotate and add to
$$
\boxed{\mathbf E_{\rm out}(t)=E_0(a\widehat{\mathbf x}\cos\omega t-b\widehat{\mathbf y}\sin\omega t).}
$$
In general the tip traces an <ellipse> with semiaxis lengths $|a|E_0$ and $|b|E_0$. Equal transmitted circular <wave amplitudes> produce a line; removal of one circular mode, $\lambda_+=0$ or $\lambda_-=0$, produces a <circle>. The special zero-output case has no defined <polarization of waves>.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2021-b4-circular-eigenvectors.png]
{title=Original vector addition of circular eigenmode fields and the resultant <elliptical polarization>}
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