In other words, there must be an orthogonal basis of real eigenvectors and one of the eigenvalues must be 1 and the other 0 . Hence, to check that a given matrix corresponds to a linear polariser, one...In other words, there must be an orthogonal basis of real eigenvectors and one of the eigenvalues must be 1 and the other 0 . Hence, to check that a given matrix corresponds to a linear polariser, one should verify that one eigenvalue is 1 and the other is 0 and furthermore that the eigenvectors are real vectors.
To find the components E_{x^{\prime}}, E_{y^{\prime}} on the \widehat{\mathbf{x}}^{\prime}, \widehat{\mathbf{y}}^{\prime} basis: \[\mathbf{E}=E_{x^{\prime}} \widehat{\mathbf{x}}^{\prime}+E_{y^...To find the components E_{x^{\prime}}, E_{y^{\prime}} on the \widehat{\mathbf{x}}^{\prime}, \widehat{\mathbf{y}}^{\prime} basis: \mathbf{E}=E_{x^{\prime}} \widehat{\mathbf{x}}^{\prime}+E_{y^{\prime}} \widehat{\mathbf{y}}^{\prime} , \nonumber we first write the unit vectors \widehat{\mathbf{x}}^{\prime} and \widehat{\mathbf{y}}^{\prime} in terms of the basis \hat{\mathbf{x}}, \hat{\mathbf{y}} (see Figure \PageIndex{1} ) \[\begin{aligned} &\widehat{\mathbf{x}}^{\prime…
