Let's say a matrix is rotational if it does not change under rotation by 90 degrees. For example, $\begin{pmatrix} 1&0&1\\ 0&2&0\\ 1&0&1 \end{pmatrix}$ is rotational. Show that any eigenvectors $v=(v_1,...,v_n)$ corresponding real eigenvalues of any real rotational matrix has the following property: $v_i=v_{n-i+1}$.


closed as off-topic by Travis, kingW3, Peter Franek, Arnaud D., Maria Mazur Nov 23 '17 at 10:29

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  • 2
    $\begingroup$ You should expose what you tried. Is this an exercise from your professor/book or from your mind? $\endgroup$ – Von Neumann Nov 22 '17 at 17:08
  • $\begingroup$ It is not a homework. It was from an entry exam to a grad school. Trying to prepare. $\endgroup$ – magzhan Nov 22 '17 at 17:26
  • $\begingroup$ I don't understand the question. $\endgroup$ – copper.hat Nov 22 '17 at 17:43
  • $\begingroup$ Your titre should be "Eigenvectors...." $\endgroup$ – Jean Marie Nov 23 '17 at 7:22
  • $\begingroup$ Thanks, fixed it.. $\endgroup$ – magzhan Nov 23 '17 at 7:28

Before beginning, in order to make things clearer for all those who haven't a clear idea of what means invariance by rotation for a matrix, here is the general case for a $4 \times 4$ matrix :

$$A=\begin{pmatrix} a & b & c & a \\ c & d & d & b \\ b & d & d & c \\ a & c & b & a \end{pmatrix}.$$

A first capital remark. Property


you want to establish is only one of the two cases that can occur.

The other case is


Let us show it on the example of

$$A=\begin{pmatrix} -1 & 1 & 1 & -1\\ \ \ 1 & 1 & 1 & \ \ 1\\ \ \ 1 & 1 & 1 & \ \ 1\\ -1 & 1 & 1 & -1 \end{pmatrix}$$

It can be diagonalized under the form: $A=P D P^{-1}$ where $D=diag(-2 \sqrt{2},0,0, 2 \sqrt{2})$ and

$$P=\begin{pmatrix} 0.6533 & 0.6344 & 0.3123 & -0.2706\\ -0.2706 & -0.3123 & 0.6344 & -0.6533\\ -0.2706 & 0.3123 & -0.6344 & -0.6533\\ 0.6533 & -0.6344 & -0.3123 & -0.2706 \end{pmatrix}$$

The columns $P_1,P_2,P_3,P_4$ of $P$ are eigenvectors associated with the given eigenvalues in the same order. We see that the structure of the last column vector $P_4$ verifies (*), but this is not the case for the columns $P_1, P_2, P_3$ that verify the other property (**).

Now for the proof. Let us establish that we are in one of the two cases (*) or (**).

Let $J$ be the $n \times n$ matrix with ones on the second diagonal and zeros elsewhere.

The rotational matrices you consider are a subset (in fact a vector subspace) of the so-called centrosymmetric matrices (see Wikipedia article) which have in particular the following property:


Let us assume that $V$ is an eigenvector of $A$ associated with eigenvalue $\lambda$, i.e.

$$\tag{1}AV=\lambda V.$$

Besides, the property we want to establish, grouping (*) and (**) can be written under the form:

$$\tag{2}JV = V \ \ \text{or} \ \ JV=-V$$

In other words, we want to prove that (1) $\implies$ (2).

Let us left-multiply (1) by $J$, giving $(JA)V=\lambda JV$.

Using relationship (0), this can be written under the form:

$$\tag{3}(AJ)V=\lambda JV$$

which can be read in the following way: $A(JV)=\lambda (JV)$ meaning that $JV$ is an eigenvector associated with the same eigenvalue $\lambda$ as $V$. If the eigenspace associated with $\lambda$ is one-dimensional, then $JV=kV$ for a certain $k$. But, $J$ being an isometry, we have $\|JV\|=\|V\|$. An immediate consequence is that $k=\pm 1$, which is nothing else than property (2) we wanted to establish.

Remark: note the restriction ; we have established the property conditionnaly on the fact that the eigenspace associated with $\lambda$ is 1D....

Some references on centrosymmetric matrices:

An interesting question and answer in (Eigenvalues of centrosymmetric matrix).

An interesting answer in (Eigenvalues of a rotationally symmetric matrix).

A wide view in (http://www.math.ualberta.ca/ami/CAMQ/pdf_files/vol_10/10_4/10_4a.pdf)

  • $\begingroup$ A courageous anonymous downvoting ... surely not on scientific grounds because my answer 1) fixes a bug in the question 2) proves the question 3) makes the connection with centrosymmetric matrices. Personal hatred ? $\endgroup$ – Jean Marie Nov 22 '17 at 22:49
  • $\begingroup$ Thank you very much for the answer and the links. Don't worry on such things :) $\endgroup$ – magzhan Nov 23 '17 at 7:26
  • $\begingroup$ How do we establish the fact that the eigenspace is 1D? $\endgroup$ – magzhan Nov 23 '17 at 8:14
  • $\begingroup$ One must look at the properties of the characteristic polynomial of these matrices. The case of double roots looks very unlikely unless we are in very particular cases (where the double root is in fact $0$), but I have no proof at the moment. $\endgroup$ – Jean Marie Nov 23 '17 at 10:07

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