Orbits and numerical ranges of zero diagonal matrices with block patterns

Question

Let $A \in \mathbb{R}^{m \times m}$ be a nonsymmetric zero diagonal matrix with a zero/non-zero pattern which is symmetric and persymmetric (i.e. symmetric in the northeast-to-southwest diagonal).

Edit: The matrices are nonsymmetric while the patterns are symmetric and persymmetric.

If $A$ has any kind of block checkerboard pattern, or an offset of a block checkerboard pattern that conserves symmetry and persymmetry (of the pattern), then $A^k, k=2n+1, n\in\mathbb{N}_0$ has also zero diagonal.

1) Is there a necessary and sufficient condition on the zero/non-zero pattern to get a zero diagonal $A^k, k=2n+1, n\in\mathbb{N}_0$?

If $A^k, k=2n+1, n\in\mathbb{N}_0$ has zero diagonal, then the spectrum of $A$ is symmetric with respect to the imaginary axis (proof).

2) Is the numerical range (i.e. field of values, $W(A)=\left\{ \frac{v^* A v}{v^* v}, v \in \mathbb{C}^m, v\ne 0 \right\}$) always symmetric with respect to the imaginary axis as well?

Examples: \begin{align} \pmatrix{0 &2 &0 &-4\\ 1 &0 &2 &0 \\ 0 &-1 &0 &8 \\ -4 &0 &7 &0} \end{align}

\begin{align} \pmatrix{0 &0 &3 &-4\\ 0 &0 &-1 &3 \\ 3 &-1 &0 &0 \\ -4 &3 &0 &0} \end{align}

\begin{align} \pmatrix{0 &-9 &1 &0\\ -9 &0 &0 &-1 \\ -1 &0 &0 &1 \\ 0 &5 &7 &0} \end{align}

Edit: This article: www.math.technion.ac.il/iic/ela/ela-articles/articles/vol26_pp591-603.pdf claims that traceless matrices with n-fold symmetry of the spectrum have the same n-fold symmetry of the numerical range if any product of $A$ and $A^*$ where the number of occurrences of $A$ and $A^*$ is different and nonzero.

Now my matrices have zero diagonal on top of traceless. If the theorem can be specialized for any product where the total number of occurrences is odd, then the proof is done.

Answer 1

Long story short: while there is a nice way to generate the kind of patterns you've been producing (i.e. there is a sufficient condition), there is no necessary/sufficient condition that I'm aware of for more general zero/non-zero patterns.

If $A$ is symmetric and its spectrum is "symmetric with respect to the imaginary axis", then $W(A)$ will also be "symmetric with respect to the imaginary axis".

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Saying that $A$ is symmetric and persymmetric is equivalent to saying that $A$ is symmetric and satisfies $AJ = JA$, where $J$ is the exchange matrix. Notably, this means that

First of all, note that with all your examples, all you've done is taken a matrix of the form $$ M = \pmatrix{0&A\\A^T&0} $$ and applied a permutation similarity. That is: for a suitable permutation matrix $P$, $PMP^T$ will be a "checkerboard matrix". If $M$ is symmetric, then $PMP^T$ will also be symmetric, since we'd find that $$ [PMP^T]^T = P^{TT}M^TP^T = PMP^T $$ If $M$ is also persymmetric and if $PJ = JP$ (which implies that $P^TJ = JP^T$), then $PMP^T$ will also be persymmetric, since we'd find that $$ [PMP^T]J = PMP^TJ = PMJP^T = PJMP^T = JPMP^T = J[PMP^T] $$ We can also show that for $M$ as above, $M^k$ will be traceless for odd $k$ (from which it follows that $[PMP^T]^k$ is traceless). In particular: note that $$ M^2 = \pmatrix{AA^T & 0\\0&A^TA} $$ Thus, $$ M^{2n+1} = \pmatrix{AA^T & 0\\0&A^TA}^{2n} \pmatrix{0 & A\\A^T&0} = \pmatrix{0 & (AA^T)^{2n}\\(A^TA)^{2n}A^T & 0} $$ which is traceless since it is zero on the diagonal.

If $A$ is symmetric and its spectrum is symmetric with respect to the imaginary axis, then $W(A)$ will also be symmetric with respect to the imaginary axis. To see this, it suffices to note that the numerical range is simply the convex hull of the spectrum of $A$ whenever $A$ is symmetric.

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It can also be directly shown that any matrix of the form $$ M = \pmatrix{0&A\\A^T&0} $$ will be "symmetric with respect to the imaginary axis". In particular, if $A = U\Sigma V^T$ is a singular value decomposition, then we write $$ \pmatrix{&A\\A^T} = \pmatrix{U \\ & V} \pmatrix{ & \Sigma\\ \Sigma} \pmatrix{U \\ & V}^T $$ And since $\Sigma$ is diagonal, we can easily determine the spectrum of the matrix $$ \pmatrix{ & \Sigma\\ \Sigma} = \pmatrix{0&1\\1&0} \otimes \Sigma $$ where $\otimes$ denotes the Kronecker product.