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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)

Widawensen
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    The main difference between imaginary and real eigenvalues is that imaginary eigenvalues are imaginary, whereas real eigenvalues are real. – Gerry Myerson Jul 05 '16 at 13:18
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    I wouldn't worry about the differences too much. The eigenvalues are what they are, and I would focus on the geometric intuition behind their use as opposed to whether they are real or complex. Matrices provide us a nice little way to skirt around several irritating aspects of mappings, but the tradeoff is that the intuition hardly exists, in my experience. – The Count Jul 05 '16 at 13:39
  • @GerryMyerson Are you joking Gerry? Is there no differences in effects on linear operation? – Widawensen Jul 05 '16 at 14:41
  • @GerryMyerson I have from below mathreadler' s remarks that imaginary eigenvalues generate 2x2 blocks in decomposition not just 1x1. – Widawensen Jul 05 '16 at 15:43
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    If you are working in a complex vector space, there is no difference between real and nonreal eigenvalues, in terms of their significance for linear operators. If you are working in a real vector space, there is no such thing as a nonreal eigenvalue. What you may be discussing is a matrix that has a nonreal eigenvalue when viewed over the complex numbers, and you want to know what happens to that matrix when you restrict your attention to a real vector space. Then, yes, there is a normal form where the nonreal eigenvalues give rise to $2\times2$ blocks. Continued... – Gerry Myerson Jul 05 '16 at 23:17
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    ... see, e.g., https://en.wikipedia.org/wiki/Jordan_normal_form#Real_matrices – Gerry Myerson Jul 05 '16 at 23:17
  • @GerryMyerson Yes, my attention is directed towards real vector space, that with geometric interpretation. Complex vector space is hard (for me) to interpret geometrically, I don't know even whether it is possible. The really interesting fact is that we need, somehow mysteriously, complex numbers, as in the case of rotations, to construct evidently real geometric transformations. – Widawensen Jul 06 '16 at 04:18
  • Real transformations are represented by real matrices. It's only when you try to force them into diagonal form that you need nonreal numbers. But when you do that, you are representing them with respect to a nonreal basis of a complex vector space. – Gerry Myerson Jul 06 '16 at 05:37
  • @GerryMyerson So we can probably conclude that nonreal basis vs. real basis give quite different geometric operations, a stretching is very different from a rotation. – Widawensen Jul 06 '16 at 05:50
  • I suppose. Then again, a stretching is very different from a shearing, or a reflecting. – Gerry Myerson Jul 06 '16 at 06:58
  • Rotation is unique for physical objects. Shearing, stretching, reflecting are not so frequent in the real world as rotation. – Widawensen Jul 14 '16 at 05:17

2 Answers2

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You can prove @lftaberas result by

  1. realizing the coefficients to the characteristic polynomial $\det({\bf A}-\lambda{\bf I})$ must all be real

and then

  1. 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.

mathreadler
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  • Interesting second remark, but how to use it? We only return in effect to real coefficients, What is it giving us? – Widawensen Jul 05 '16 at 15:02
  • Yes. Ok, I forgot that part. Here's maybe a better explanation for how a 3D rotation can be represented. If we insist on working with real numbers in our matrices we will need to go to the generalization of block-diagonalizations. – mathreadler Jul 05 '16 at 15:12
  • $(\lambda-a)(\lambda -a^)={\lambda}^2-\lambda(a+a^)+aa^*={\lambda}^2-2Re{a}+\lvert{a}\rvert^2$ – Widawensen Jul 05 '16 at 15:16
  • Yep and if you do the 2x2 matrix block multiplied by it's transpose and using $s^2+c^2=1$ you may find something interesting too. – mathreadler Jul 05 '16 at 15:19
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    I know $RR^T=I$. – Widawensen Jul 05 '16 at 15:21
  • So I understand that with above decomposition we always obtain $2x2$ blocks of a characteristic form for a rotation matrix (and there is no other possibility) since that there is no preservation of vector direction? – Widawensen Jul 05 '16 at 15:32
  • Thank you mathreadler for information about "generalization" of eigenvalue decomposition, it is something new for me. I suppose separate blocks e.g. 4x4 don't exist in this decomposition, only 2x2 ones. – Widawensen Jul 05 '16 at 15:37
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    These blocks have something common with the fact that algebra of complex numbers can be replaced by operations on 2x2 matrices? – Widawensen Jul 05 '16 at 15:40
  • If it is a rotation matrix we should always be able to find such a decomposition, fixing everything not in the plane of rotation to eigenvalue 1 and the plane to such a 2x2 block. If all you want to do is to rotate, 2x2 blocks should be enough, but there exist other kinds of block diagonalizations better suited for other operations where blocks could be larger than 2x2. If you read about groups and abstract algebra you will make friends with types of numbers which benefit from block diagonalizations. The complex numbers are as you say an easy example of numbers we can "fit" into 2x2 matrices. – mathreadler Jul 05 '16 at 15:42
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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.

lftabera
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  • I have just no intuition for this imaginary space. How it can be generated from matrix with only real values ? It is somehow counter intuition. – Widawensen Jul 05 '16 at 14:06
  • Could we also risk a statement that set of real and set of imaginary eigenvalues cuts the whole space of matrix $A$ into two orthogonal subspaces ? ( as in the case of 3D rotations). – Widawensen Jul 05 '16 at 14:11