Imaginary eigenvalues
Sophia Terry
I'm still developing my skills in LinearAlgebra and I ponder just what are the main differences between effects of imaginary and real eigenvalues on linear operations specially taking into account their geometric interpretation?
Is there somewhere the list of these differences?
Is it true that if we have imaginary eigenvalues then it is necessary for some subspace of a space generated by a matrix $A$ that we have no preserved directions of vectors in this subspace as for example it is in the case of 2D i 3D rotations?
If so how to apply these imaginary eigenvalues for generating this subspace? (in the case of rotations to generate the plane)
$\endgroup$ 112 Answers
$\begingroup$You can prove @lftaberas result by
- realizing the coefficients to the characteristic polynomial $\det({\bf A}-\lambda{\bf I})$ must all be real
and then
- using the fact that each zero for such a polynomial is either real or part of a pair of complex conjugates which multiplied together build a second degree polynomial of real coefficients.
When comes to practical considerations, complex eigenvalues can be put on block-diagonal form for a real 2x2 block. This is a kind of a "generalization" of eigenvalue decomposition:
$$ {\bf A} = {\bf T}^{-1}{\bf DT}\hspace{0.5cm}\text{where}\hspace{0.5cm}{\bf D} = \left[\begin{array}{r|rr} 1&0&0\\\hline0&c&-s\\0&s&c\end{array}\right]$$
The column corresponding to the 1 is the vector describing the axis of rotation ( left intact ). The 2x2 block corresponds to the complex pair $c\pm is$. ( $c,s$ stand for cos, sin ) and it's corresponding columns in $\bf T$ span the plane being rotated. The lines in the matrix are just there to highlight the 1x1 and 2x2 blocks along the diagonal.
$\endgroup$ 9 $\begingroup$If you have a real matrix $A$ and $a+bi$ is an eigenvalue, then $a-bi$ is also an eigenvalue. Now, if $v$ is an eigenvector associated to $a+bi$, then the conjugate $\overline{v}$ is an eigenvector associated to $a-bi$. Show that the (complex) space $\langle v,\overline{v}\rangle$ in $\mathbb{C}^n$ is the same as $\langle Re(v),Im(v)\rangle$ in $\mathbb{C}^n$ and that the (real) space $\langle Re(v),Im(v)\rangle$ in $\mathbb{R}^n$ has no preserved direction.
$\endgroup$ 2