5
$\begingroup$

I am trying to understand the intuition behind eigenvalues/eigenvectors through the lens of repeated matrix multiplication:

Given a $2\times2$ matrix $M$ and $2D$ vector $v$, multiplying $v$ repeatedly with $M$ causes the result ($M^n v$) to gravitate towards one of the eigenspaces of $M$ because:

$$M^n v = M^n(\alpha x_1 + \beta x_2) = (\alpha \lambda_1^n x_1 + \beta \lambda_2^n x_2)$$

where $x_1$ and $x_2$ are eigenvectors of $M$ and $\lambda_1$ and $\lambda_2$ the corresponding eigenvalues. As $n$ gets larger $M^n v$ will gravitate towards either $\alpha \lambda_1^n x_1$ or $\beta \lambda_2^n x_2$, whichever has the dominant eigenvalue.

assuming: $v = \alpha x_1 + \beta x_2$

So the above is a way to connect the abstract concept of eigenvalue/eigenvector to something concrete: what happens when you apply a matrix over and over to a vector.

However, the intuition breaks down for me with complex eigenvectors. I know repeated multiplication by a matrix with complex eigenvectors causes the result to either spiral outwards or inwards.

Is there simple math such as above to see why?

Edit: I know similar questions have been asked before, but I ask in the context of repeated matrix multiplication

$\endgroup$
  • $\begingroup$ I don't know why complex eigenvalues give rotation, but the fact that rotation give complex eigenvalues is plain enough to see: if there is rotation, then there are no real eigenvectors, but the characteristic polynomial must have roots. That means that the roots are complex. $\endgroup$ – Arthur Nov 20 '15 at 22:27
  • $\begingroup$ I guess it's not plain to me why if there is rotation, there are no real eigenvectors? What is the characteristic polynomial of a matrix? $\endgroup$ – applecider Nov 20 '15 at 22:31
  • $\begingroup$ The characteristic polynomial is what you use to calculate eigenvalues. It is given by the determinant of the matrix $A-xI$ where $x$ is the unknown and $I$ is the identity matrix. Also, an eigenvector cannot be rotated by the matrix (other than $0^\circ$ or $180^\circ$), because then it is not an eigenvector. $\endgroup$ – Arthur Nov 20 '15 at 22:41
  • $\begingroup$ @applecider : A rotation of a $3$-dimensional space does have a real eigenvalue, namely $1$, but a rotation of the plane through an angle other than $180^\circ$ or $0^\circ$ cannot have real eigenvalues, because if you rotate a vector then you're not multiplying it by a scalar multiple of itself. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 20 '15 at 22:56
3
$\begingroup$

Every square matrix is similar to a matrix in what's called Jordan Canonical Form. This has various properties, but most important here is that it is upper triangular, and the eigenvalues (of both the new and original matrix) are on the diagonal of the resulting matrix.

The way to think about this process is that we change bases, and in that new basis the matrix becomes diagonal. This will help the existing intuition given in the OP about iterating the matrix, because iterating a triangular matrix will simply exponentiate the diagonal entries (and do predictable but somewhat messy stuff to the part above the diagonal).

Now, if an eigenvalue is complex, all the above still holds. However if the original matrix had real entries, then that eigenvalue must be paired with its complex conjugate -- both $a+bi$ and $a-bi$ must be eigenvalues. Then, we can rearrange the standard JCF matrix, as described above, into what is called Real Jordan Canonical Form. Now, instead of the matrix being entirely upper triangular, there will be some $2\times 2$ blocks along the diagonal, a rearrangement of what was a complex eigenvalue-plus-its-conjugate pair in the original JCF. The entries of these $2\times 2$ blocks are exactly the real and imaginary parts of these two complex eigenvalues.

Each of these $2\times 2$ blocks performs a rotation in the two-dimensional space spanned by those two basis elements. Hence, a $6\times 6$ matrix with six complex eigenvalues might be doing three different rotations in three different two-dimensional directions at once.

In the original $2\times 2$ case, the reason that a complex eigenvalue leads to a rotation is that it must appear with its complex conjugate, assuming the original matrix has real entries. As a pair they give a rotation. If the original matrix did not have real entries, then the matrix need not represent a pure rotation, because the two eigenvalues need not be related.

$\endgroup$
2
$\begingroup$

Well, one way to look at it is considering the identity $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$ which tells us that the function $e^{i\theta}$ traces out the unit circle of the complex plane. One might note that if we have an eigenvalue of the form $e^{i\theta}$ then powers of it are of the form $e^{ni\theta}$ - so they are just rotating around the unit circle when they get exponentiated. For instance, the expression $i^n$ will just cycle through the pattern $1,\,i,\,-1,\,i$ making quarter turns around the unit circle and $(-i)^n$ does the same, but in reverse. This sort of tells us why these might be the eigenvalues of a rotation of the plane by a quarter turn. Similar things hold true of any rotation of the plane.

We might think of eigenvalues on the unit circle in the complex plane as telling us that the matrix is acting more or less periodically on the corresponding eigenvector. More generally, if we consider eigenvectors of the form $re^{i\theta}$ as having a scaling component $r$ as well as a rotating/periodic component $\theta$. One may also visualize this by noting directly that the map $z\mapsto re^{i\theta}$ is simple a rotation of the complex plane by $\theta$ along with a dilation by $r$.

$\endgroup$
  • $\begingroup$ Real Jordan block for pair of complex eigenvalues will do the job too :) $\endgroup$ – Evgeny Nov 20 '15 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.