4
$\begingroup$

It was mentioned in one MSE answer that eigenvalues of products of square matrices are equal (see the answer of user1551 for Eigenvalues of Matrices and Eigenvalue of product of Matrices)

Let's denote this fact: $ \ \ \ \ $ $\text{eig}(AB)=\text{eig}(BA)$.

  • However .. how can this be explained in the case where matrices don't commute?
  • Does some kind of geometrical interpretation of this statement exist - at least in the case of 3D orthogonal matrices where it is known that they usually don't commute ?
  • Can the statement be extended for a case of product of more number of matrices, for example: $\text{eig}(A_1{A_2} ... A_n)=\text{eig}(A_n{A_{n-1}} ... A_1)= \text{eig}(A_{n-1}{A_{n-2}} ... A_n)=$ etc... ?
$\endgroup$
  • $\begingroup$ You will find some excellent proofs for you first question here. $\endgroup$ – Andreas Caranti Feb 27 '17 at 11:10
  • $\begingroup$ @AndreasCaranti Thank you, they are very interesting. Interestingly also that after the first answer the next ones appeared after .. 4 years :) $\endgroup$ – Widawensen Feb 27 '17 at 11:23
4
$\begingroup$

Hint For invertible $B$, the characteristic polynomial of $AB$ is $$\det(\lambda I - AB) = \det[(\lambda B^{-1} - A) B] = \det (\lambda B^{-1} - A)\det B.$$ Rewrite this to show that $$\det(\lambda I - AB) = \det(\lambda I - BA).$$

Additional hint Now, both sides are evidently continuous functions of $B$, and the set of invertible $n \times n$ matrices is dense in the space of all $n \times n$ matrices.

By applying the above argument inductively, we can also conclude that the eigenvalues of $A_1 \cdots A_r$ coincide with those of any product $A_s \cdots A_r A_1 \cdots A_{s-1}$ given by permuting the factors cyclically. On the other hand, this is not true for general permutations, that is, in general the eigenvalues of $A_{\sigma(1)} \cdots A_{\sigma(r)}$ are not those of $A_1 \cdots A_r$. For a minimal example, take $$A_1 = \pmatrix{1&0\\0&0}, \qquad A_2 = \pmatrix{0&1\\1&0}, \qquad A_3 = \pmatrix{0&1\\0&0} .$$ Then, $A_1 A_2 A_3 = 0$ but $A_1 A_3 A_2 = \pmatrix{1&0\\0&0}$.

You can also show using these ingredients that the trace of an $r$-fold product of square matrices is invariant under cyclic permutation, but we can again see from the above example that the same is not in general true for general permutations.

$\endgroup$
  • $\begingroup$ so it is true for cyclic permutations and it's not true in general for other permutations... but ..your example seems to be quite specific... for example if we added conditions that all matrices have full rank and their product also has full rank it would change something? $\endgroup$ – Widawensen Feb 27 '17 at 11:31
  • $\begingroup$ and thank you for the extension this for a trace... however I have doubts about full rank matrices... $\endgroup$ – Widawensen Feb 27 '17 at 11:40
  • $\begingroup$ You have been right. I've checked for some randomly chosen full rank matrices... cyclic permutation preserves eigenvalues, non-cyclic not.. that's really interesting I didn't expect such distinction... $\endgroup$ – Widawensen Feb 27 '17 at 11:47
  • $\begingroup$ your HINT is excellent - so this cyclicity stems from the elementary cyclicity of $B^{-1}B=BB^{-1}=I$, however I don't really know how to interpret your additional hint...what continuity means here? $\endgroup$ – Widawensen Feb 27 '17 at 12:24
  • 1
    $\begingroup$ I'm glad you found the hint helpful. As for continuity, the hint proves that $\det(\lambda I - AB) = \det(\lambda I - BA)$ (and hence coincidence of the eigenvalues of $AB$, $BA$) for invertible $B$, so it remains to show that equality holds for degenerate $B$. Now, $\det(\lambda I - AB)$ and $\det(\lambda I - BA)$ are both continuous functions of $B$, as they are polynomials in the entries $b_{ij}$ of $B$. Continuous functions that agree on a dense subset must agree everywhere. $\endgroup$ – Travis Feb 27 '17 at 12:35
4
$\begingroup$

In general $AB$ and $BA$ are similar matrices, as $AB=B^{-1}(BA)B$. Hence they represent the same linear transformation but with respect to different bases. Hence they have the same characteristic polynomial. Here "in general" means "provided not both $A$ and $B$ are singular".

The same sort of argument shows that in a group $ab$ and $ba$ have the same order.

So now you have an explanation "in general".

How can one "understand" the result in the special cases? If the matrices have entries from $\mathbb{C}$ we can use a continuity argument -- this is often a useful trick. Just note that the non-singular matrices are dense in the matrix ring, and then choose non-singular $A_n, B_n$ converging to $A,B$. Since the coefficients of the characteristic polynomial are polynomial functions of the matrix elements, they are continuous, so that as $n\to\infty$, $c_{A_nB_n}\to c_{AB}$, and $c_{A_nB_n}=c_{B_nA_n}\to c_{BA}$.

The result is still true over finite fields, but I don't know of a good intuitive reason, you've just got to hack it out.

$\endgroup$
  • $\begingroup$ ok, that's also interesting approach.. could it be used for greater number than $2$ of multiplied matrices ? $\endgroup$ – Widawensen Feb 27 '17 at 12:55
  • $\begingroup$ I have found .... $B^{-1}A^{-1}ABCAB=CAB$ so similarity also exists for cyclic permutations as in the Travis' answer... it seems that cyclicly permutated products are always similar to each other... $\endgroup$ – Widawensen Feb 27 '17 at 13:26
  • $\begingroup$ Yes but of course you only to get the cyclic permutations of the initial product. There's no reason to think the other products share many properties with it. $\endgroup$ – ancientmathematician Feb 27 '17 at 13:27
  • $\begingroup$ Hmm, there is no any common properties ? I'm especially interested in different products of 3 orthogonal matrices and relationships between their eigenvalues.. $\endgroup$ – Widawensen Feb 27 '17 at 13:30
  • 1
    $\begingroup$ $ABC=C^2$ is a rotation through $2\pi/3$; but $ACB=I$. OK? $\endgroup$ – ancientmathematician Feb 27 '17 at 14:06
1
$\begingroup$

On the spectra of cyclic permutation of matrix products

You can show that cyclic permutations of matrix produxts have the same spectra rather easily. Consider $$ A_1 \ldots A_n \vec v = \lambda \vec{v}.$$ Now multiply by $A_n$ on both sides. $$ A_n A_1 \ldots A_{n-1} (A_n \vec{v}) = \lambda (A_n \vec{v}).$$ So $ A_n A_1 \ldots A_{n-1}$ has the same eigenvalues as $ A_1 \ldots A_n$ with corresponding eigenvectors given by the above expressions. You can do this procedure repeatedly to show that all cyclic permutations have the same spectrum. The two matrix case is a special case of this.

Geometric interpretation

Sorry for the hand drawn picture. This gives an explanation for the case where the eigenvalue is $1$ or $-1$. The loci of vectors turned by the same amount due to a rotation matrix form a cone centred at the origin in 3D. When you combine two rotations the eigenvectors corresponding to $1$ lie in the intersection of two such cones, one for each rotation matrix. The picture shows such an intersection. The cones intersect at two different vectors $\vec v$ and $A\vec{v}$ in the picture. When you reverse the order in which the rotation are applied the eigenvector changes from one of these vectors to the other. I cannot think of any geometric interpretation for the case of complex eigenvalues.

enter image description here

$\endgroup$
  • $\begingroup$ Thank you, biryani. You very fine have presented preservation of spectrum under cyclicity. The picture is also very interesting - I have to think it over. If you have some additional ideas I will be very grateful... $\endgroup$ – Widawensen Feb 28 '17 at 7:26
  • 1
    $\begingroup$ I believe the picture extends to the multiple matrix case as well. In that case the eigenvectors would lie on the vertices of a polygon made by arcs of intersecting circles. A cyclic permutation should then take the eigenvector from one vertex of the polygon to the next one. I haven't thought out the consequences of this completely, so take it with a pinch of salt. $\endgroup$ – biryani Feb 28 '17 at 7:32
  • $\begingroup$ Cyclic traveling eigenvectors between vertices of polygon ?... sounds very interesting .. what do you think about the case of exactly three matrices - here we have two orders dividing all 6 cases (as we have seen always 3 in each group can be cyclicly transformed). Do some relation between these two groups an their spectra exist? $\endgroup$ – Widawensen Feb 28 '17 at 7:41
  • $\begingroup$ Your graphical illustration is very interesting for these two rotations. These circles are on a surface of unit sphere,, I suppose. It seems that It is a graphical method of finding the axis of composite rotation. Am I right? $\endgroup$ – Widawensen Feb 28 '17 at 11:26
  • 1
    $\begingroup$ Yes. They are the circles formed by the cones in the answer intersecting the unit sphere. Since the intersection vector is unmoved, it must be the axis of the composite rotation. $\endgroup$ – biryani Feb 28 '17 at 11:41

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.