Finding a matrix for which a polynomial can be associated to a sum of matrices

Question

I want to find (if it exists) a matrix $M$ with integer coefficients such that :

$a\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\,+ b\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0 \end{bmatrix}\,+c\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \end{bmatrix}\,+d\begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0\\ 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{bmatrix}=aM^{e_1}+bM^{e_2}+cM^{e_3}+dM^{e_4}$ where $e_i$ are integer exponents and $a,b,c,d$ positive integers.

Answer 1

I think that the analysis below is interesting, so I will leave it there. However, the fact that no such matrix $M$ can exist (even if we allow complex-number entries) can be quickly argued with an appeal to group theory.

Keeping the notation of the work below, I will denote $$ I = \pmatrix{1 &0\\0&1}, \quad J = \pmatrix{0&1\\-1&0},\\ M_0 = I \otimes I, \quad M_1 = I \otimes J, \quad M_2 = J \otimes I, \quad M_3 = J \otimes J, $$

where $\otimes$ denotes a Kronecker product. First of all, note that the matrices $M_0, M_1,M_2,M_3$ (with coefficients $a,b,c,d$ respectively) generate the following multiplicative group: $$ G = \{J^p \otimes J^q : p,q \in \Bbb Z\} = \{\pm M_j: j = 0,1,2,3\} \cong \Bbb Z_{2}\times \Bbb Z_4. $$ Suppose (for the purpose of contradiction) that $M$ were such that $M^{e_j} = M_j$ for $j = 0,1,2,3$. It would follow that $M$ is invertible and that $G$ must be a subgroup of the multiplicative group generated by $M$, $\langle M\rangle = \{M^p : p\in \Bbb Z\}$. This would mean that a cyclic group contains $G$, a non-cyclic group, as a subgroup. However, every subgroup of a cyclic group is cyclic, so this is impossible.

Note: Of course, the nature of $G$ can be found without building an explicit isomorphism since we can count and check the order of elements. That said, I found that building an isomorphism between $G$ and $\Bbb Z_2 \times \Bbb Z_4$ was a surprisingly counterintuitive exercise. Here is one isomorphism that can be constructed:

$$\phi:G \to \Bbb Z_2 \times \Bbb Z_4,\quad \phi(J^p \otimes J^q) = (p,q-p) \\ \text{ for } p = 1,2 \text{ and } q = 1,2,3,4 $$

The inverse is a bit more intuitive:

$$\psi: \Bbb Z_2 \times \Bbb Z_4 \to G, \quad \psi(p,q) = J^p \otimes J^{p+q} = (J \otimes J)^p (I\otimes J)^q.$$

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The analysis of these matrices is made significantly easier with the help of the Kronecker product. In particular, if we define $$ I = \pmatrix{1 &0\\0&1}, \quad J = \pmatrix{0&1\\-1&0}, $$ then your linear combination can be written in the form $$ a(I \otimes I) + b(I \otimes J) + c(J \otimes I) + d(J \otimes J). $$ for convenience, I will denote the four matrices in the sum above as $$ M_0 = I \otimes I, \quad M_1 = I \otimes J, \quad M_2 = J \otimes I, \quad M_3 = J \otimes J. $$ Using the properties of the Kronecker product, it's easy to see that these matrices will commute, i.e. that $M_iM_k = M_kM_j$ for all $1 \leq j,k \leq 4$. Another convenient fact is that we can simulatenously diagonalize these matrices using a diagonalization of $J$. Let $U$ and $D$ denote the unitary and diagonal (respectively) matrices given below: $$ U = \frac 1{\sqrt{2}}\pmatrix{-i&1\\1&-i}, \quad D = \pmatrix{i &0\\0& -i}. $$ We find that $U^*JU = D$, where $U^*$ denotes the conjugate-tranpose of $U$ (which in this case is simply the conjugate of $U$). Accordingly, we find that $(U \otimes U)^*M_j (U \otimes U)$ is diagonal for all $j$. More specifically, if we define $P_j = (U \otimes U)^*M_j (U \otimes U)$, we find that $$ P_0 = I \otimes I = M_0, \\ P_1 = I \otimes D = \operatorname{diag}(i,i,-i,-i),\\ P_2 = D \otimes I = \operatorname{diag}(i,-i,i,-i),\\ P_3 = D \otimes D = \operatorname{diag}(-1,1,1,-1), $$ where $\operatorname{diag}(a,b,c,d)$ denotes the diagonal matrix with diagonal entries $a,b,c,d$.

Now, if a matrix $M$ exists such that $M^{e_j} = M_j$ for each $j = 1,\dots,4$ and some integer exponents $e_j$, it would follow that the matrix $P = (U \otimes U)^*M(U \otimes U)$ is such that $P^{e_j} = P_j$ for each $j = 1,\dots,4$. If such a matrix $P$ exists, we can deduce the following:

• From the fact that $PP_j = P_jP$ for $j = 1,2$, we can deduce that $P$ must be a diagonal matrix.
• From the fact that $P^{e_0} = I$, we can deduce that each eigenvalue $\lambda$ of $P$ (i.e. each diagonal entry of $P$) satisfies $\lambda^{e_0} = 1$ where $e_0$.
• From the fact that some power of $P$ has $i = \exp(\pi i/2)$ as an eigenvalue, we can deduce that $e_0$ is divisible by $4$. That is, $e_0 = 4n$ for some integer $n$.
• Because each diagonal entry of $P$ satisfies $\lambda^{4n} = 1$, we can deduce that each such entry has the form $\exp(2 \pi m_k i/e_0)$ for some integers $m_1,m_2,m_3,m_4$.

From there, we deduce that the exponents $e_1,e_2,e_3$ satisfy the follows equations: $$ e_1 m_1 \equiv n \quad e_1m_2 \equiv n \quad e_1m_3 \equiv 3n \quad e_1m_4 \equiv 3n\\ e_2 m_1 \equiv n \quad e_2m_2 \equiv 3n \quad e_2m_3 \equiv n \quad e_2m_4 \equiv 3n\\ e_3 m_1 \equiv 2n \quad e_3m_2 \equiv 0 \quad e_3m_3 \equiv 0 \quad e_3m_4 \equiv 2n, $$ where $\equiv$ in the above denotes equality modulo $4n$.