4
$\begingroup$

We say that a real matrix is rotatable iff after turning it clockwise on $90^{\circ}$ it doesn't change.

I'm interesting about eigenvalues and eigenvectors (belonging to non-zero eigenvalues) of such type of matrices. For example, it is not hard to show that for every tuple of real values $\lambda_1,\ldots,\lambda_{k}$ there exists $n\in\mathbb{N}$ and a rotatable $n\times n$ matrix $A$ such that all $\lambda_i$ are eigenvalues of $A$.

Indeed, let us consider a matrix $$ A_1 = \begin{pmatrix} a & a \\ a & a\\ \end{pmatrix}. $$ Then of course $A_1$ is rotatable and its characteristic polynomial $\chi_{A_1}(x) = x^2-2ax = x(x-2a)$ and of course for every $\lambda$ we can choose $a$ (for example $\lambda/2$).

Now let us show how we can construct a rotatable matrix $A_2$ with prescribed eigenvalues $\lambda_1,\lambda_2$. For example, we can consider a matrix of the form $$ A_2 = \begin{pmatrix} b&0&0&b\\ 0&a & a&0 \\ 0&a & a&0\\ b&0&0&b \end{pmatrix}. $$ Then $\chi_{A_2}(x) =x^2(x-2a)(x-2b)$. And we are done for $a=\lambda_1/2, b = \lambda_2/2$. Of course, using this method we can construct the required rotatable matrix for every $k$ of size $2k$.

Also my experiments show that all eigenvectors $v = (v_1,\ldots,v_n)^T$ belonging to non-zero eigenvalues of roratable matrix are symmetric, that is $v_i = v_{n-i+1}$. It is simple to prove this in the case, where $n=2$. Also I tried to prove it by induction on $n$, but my attempts failed.

My question.

  1. For a given tuple $\lambda_1,\ldots, \lambda_k$ can we construct a rotatable matrix $A$ of size $n\times n$, where $n<2k$, such that all $\lambda_i$ are eigenvalues of $A$.

  2. Is it always true that all eigenvectors of non-zero eigenvalues of rotatable matrix are symmetric? And if the answer is "yes", how to prove this.

$\endgroup$
5
  • 1
    $\begingroup$ Is the (2,4) element of $A_2$ supposed to be 0 rather than $b$? If not, how is it rotatable? $\endgroup$ May 15, 2018 at 11:23
  • $\begingroup$ Doesn't every rotatable matrix become of the form $\begin{bmatrix}M & M \\ M & M\end{bmatrix}$, i.e., $\begin{bmatrix}1 & 1\\ 1 & 1\end{bmatrix} \otimes M$, if you conjugate it by a suitable permutation? That looks like it would simplify the analysis a lot. $\endgroup$ May 15, 2018 at 21:04
  • $\begingroup$ @FedericoPoloni, does this hold for $3\times 3$ matrices? I don't sure. $\endgroup$ May 16, 2018 at 10:30
  • $\begingroup$ @MikhailGoltvanitsa No, only in even dimension. $\endgroup$ May 16, 2018 at 14:28
  • $\begingroup$ In particular your matrices are centrosymmetric (apply the rotation twice will also yield the same matrix) en.m.wikipedia.org/wiki/Centrosymmetric_matrix $\endgroup$
    – user35593
    May 18, 2018 at 15:34

1 Answer 1

4
$\begingroup$

Consider the matrix $$P= \begin{pmatrix} 0 & \ldots & 1 \\ \vdots & 1 & \vdots \\ 1 & \ldots & 0 \end{pmatrix}$$ with $1$s along the "other" main diagonal and $0$s elsewhere. Then $(PA)^t$ is a rotation of the matrix $A$ by $90^\circ$ (you can check this on the basis $E_{i,j}$ of matrices with a 1 in the $(i,j)$ slot and 0s elsewhere). So $A$ is rotatable if and only if $(PA)^t = A$, i.e. $PA=A^t$. Note that $P^2=I$ and $A^tP = A$, so $A^t = AP$ and hence $A$ and $P$ must commute. (For what follows below we assume that $A$ is a real matrix. Otherwise I think all you can say is that $A$ preserves the splitting $\mathbb{C}^n=E_+ \oplus E_-$ as explained below. In the complex case we get symmetric eigenvectors of the $v+Pv$ as well as the skew symmetric vectors $v-Pv$.)

It then follows that $A$ and $A^t$ commute and so, by the Spectral Theorem, $A$ has a unitary basis. Now, notice that $P$ has eigenvalues $1$ and $-1$ with eigenvectors $e_i + Pe_i$ and $e_i-Pe_i$ for $1 \leq i \leq \frac{n+1}{2} $, respectively. Let $E_+ = \ker (P-I)$ and $E_- = \ker (P+I)$. Then $A(E_\pm) \subseteq E_\pm $, so $A$ must preserve the splitting $\mathbb{R}^n = E_+ \oplus E_- $. Note that $\dim E_+ = \left\lfloor \frac{n+1}{2}\right\rfloor$.

Now, the answer to question $2$ is yes if we require the corresponding eigenvalue to be a nonzero real number. If $\lambda$ is an eigenvalue with eigenvector $v \in E_\pm$ and $(\:\: ,\:\: )$ is the standard hermitian inner product on $\mathbb{C}^n$, we have $$ \lambda(v,v)=(v,A^tv) = (v,APv) = \pm \bar{\lambda}(v,v).$$

So $\lambda \in \mathbb{R}$ if and only if $v \in E_+$ and $\lambda \in i \mathbb{R}$ if and only if $v \in E_-$. Note that $v \in E_+$ if and only if $Pv=v$, which is the same as requiring that $v$ is symmetric. Furthermore, note that the eigenvectors in $E_- \setminus \{0\}$ have complex coefficients as they satisfy $Av=i \mu v$. So it is precisely the real eigenvectors of nonzero eigenvalues that are symmetric. The others will be "skew symmetric".

Question $1$ is a bit trickier, and the answer is no in general. Let us treat the even and odd dimensional cases separately.

Let $n=2m$. If $A$ is rotatable it has the form $$A= \begin{pmatrix} B & B^t P \\ PB^t & PBP \end{pmatrix}$$ for some matrix $B$ (here $P$ and $B$ are matrices of size $m\times m$). Now, $\lambda$ is a real eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ Pv \end{pmatrix}$ if and only if $$Bv + B^t v = \lambda v .$$ So $\lambda = 2 \operatorname{Re}(\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B+B^t$. Similarly, $\lambda= i \mu$ is a purely imaginary eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ -Pv \end{pmatrix}$ if and only if $Bv - B^t v = i \mu v$, so $\mu = 2 \operatorname{Im} (\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B-B^t$.

So, if you choose more than $n$ real numbers, or choose any complex numbers $\lambda$ that do not satisfy $\bar{\lambda} = \pm \lambda$, or you choose more than $n$ purely imaginary numbers, then they cannot be eigenvalues of $A$.

Now, in the odd dimensional case $n=2m+1$, $A$ takes the form $$A= \begin{pmatrix} B & u & B^t P \\ u^t & a & (Pu)^t \\ PB^t& Pu & PBP \end{pmatrix}$$ for some matrix $B$, vector $u$ and real number $a$. For $\lambda$ a real eigenvalue of $A$, we have the eigenvector $\begin{pmatrix} v \\ b \\Pv \end{pmatrix}$, so $$(B+B^t)v + bu = \lambda v$$ and $$2(u,v)+ab=\lambda b.$$ Choosing $u=0$ gives $$(B+B^t)v=\lambda v$$ and $ab=\lambda b.$ So we can pick $m$ eigenvectors for $B+B^t$ as above, and also the vector given by $v=0$ and $b=1$, which has eigenvalue $a$. The imaginary eigenvalues $i \mu$ have eigenvectors $\begin{pmatrix} v\\0\\-Pv \end{pmatrix}$, and so $$(B-B^t)v=i\mu v,$$ with the other equation being $0=0$. So as before $\mu$ is twice the imaginary part of an eigenvalue of $B$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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