Block Structure of Semimagic Matrices and Split Quaternions

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1. The structural problem

Let K be a field of characteristic other than 2 or 3, and let SM₃(K) be the algebra of semimagic 3×3 matrices. The preceding article gives every element uniquely as S(E,x,y,z,w), so dim SM₃(K)=5. To understand multiplication in this algebra, separate the direction carrying the common row and column sum from the action on the zero-sum subspace.

Structure theorem

\operatorname{SM}_3(K)\cong K\oplus M_2(K) \cong K\oplus\mathbb H_{\mathrm{split}}(K).

The four-dimensional summand consists exactly of semimagic matrices with common sum zero. This summand is the split quaternion algebra.

The one-dimensional summand records the common sum, while all nontrivial noncommutative structure lies in the four-dimensional summand. The following sections derive this decomposition and then construct a quaternion basis in the four-dimensional ideal.

2. The invariant line and zero-sum plane

Let e=(1,1,1)ᵀ and consider the plane W in K³ consisting of vectors whose coordinates sum to zero:

L=K\mathbf e,\qquad W=\{(a,b,c)^{\mathsf T}\in K^3:a+b+c=0\}.

If a semimagic matrix A has common row and column sum T, then

A\mathbf e=T\mathbf e,\qquad \mathbf e^{\mathsf T}A=T\mathbf e^{\mathsf T}.

The first equality makes L an invariant line. From the second, every v∈W satisfies eᵀAv=T eᵀv=0, so W is invariant as well. In the basis

\mathbf e=(1,1,1)^{\mathsf T},\qquad u=(1,-1,0)^{\mathsf T},\qquad v=(1,0,-1)^{\mathsf T}

the matrix A has block-diagonal form

[A]_{(\mathbf e,u,v)} = \begin{pmatrix} T&0\\ 0&B \end{pmatrix}, \qquad B\in M_2(K).

Conversely, any pair (T,B) defines such an operator and, after returning to the standard basis, gives a semimagic matrix. The change-of-basis determinant is 3, so this proof only requires 3 to be invertible. Under multiplication the blocks multiply independently:

(T_1,B_1)(T_2,B_2)=(T_1T_2,B_1B_2).

3. The coordinate form of the isomorphism

For A=S(E,x,y,z,w), the scalar block is the common sum T=3E, while the restriction A|W in the basis (u,v) has matrix

B(x,y,z,w)= \begin{pmatrix} 3z+3w+x+y&2x+2y\\ -3w+x-2y&3z-3w-x-y \end{pmatrix}.

Hence the structure isomorphism is given by

\Phi\bigl(S(E,x,y,z,w)\bigr) =\bigl(3E,B(x,y,z,w)\bigr).

To prove bijectivity, write the inverse map explicitly. If the row-major entries of B are a,b,c,d, all five coordinates are recovered uniquely:

\begin{aligned} E&=\frac{T}{3},& z&=\frac{a+d}{6},& w&=\frac{a-d-b}{6},\\ x&=\frac{a-d+b+2c}{6},& y&=\frac{-a+d+2b-2c}{6}. \end{aligned}

Substitution into B recovers a,b,c,d. Thus Φ is bijective and, because restriction of a product is the product of the restrictions, it is an algebra isomorphism.

4. Two central idempotents

Let J be the all-ones matrix. Define

H=\frac13J=M\left(\frac13,0,0\right), \qquad Q=I_3-H=C\left(0,\frac13,0\right).

Direct multiplication gives

H^2=H,\qquad Q^2=Q,\qquad HQ=QH=0,\qquad H+Q=I_3.

The matrix H is the identity of the one-dimensional scalar ideal KH. The matrix Q has zero row and column sums and is the internal identity of the four-dimensional ideal

\mathcal I_0 =\{S(0,x,y,z,w):x,y,z,w\in K\}.

The two identities must be distinguished: the identity of the full algebra is I₃=H+Q, whereas the element denoted by 1 in the quaternion basis of the ideal is Q.

5. An explicit basis 1,i,j,k

Choose the following four matrices in the ideal 𝓘₀:

\mathbf 1_0=Q

\frac13\begin{pmatrix} 2&-1&-1\\-1&2&-1\\-1&-1&2 \end{pmatrix}

i=C(0,0,\frac13)

\frac13\begin{pmatrix} -1&-1&2\\-1&2&-1\\2&-1&-1 \end{pmatrix}

j=M(0,\frac23,\frac13)

\frac13\begin{pmatrix} 2&-1&-1\\-3&0&3\\1&1&-2 \end{pmatrix}

k=M(0,\frac13,\frac23)

\frac13\begin{pmatrix} 1&1&-2\\-3&0&3\\2&-1&-1 \end{pmatrix}

Their multiplication table is

·	$\mathbf1_0$	$i$	$j$	$k$
$\mathbf1_0$	$\mathbf1_0$	$i$	$j$	$k$
$i$	$i$	$\mathbf1_0$	$k$	$j$
$j$	$j$	$-k$	$\mathbf1_0$	$-i$
$k$	$k$	$-j$	$i$	$-\mathbf1_0$

In particular,

i^2=j^2=\mathbf1_0,\qquad ij=-ji=k,\qquad k^2=-\mathbf1_0.

This is the quaternion algebra (1,1) over K, which is split and isomorphic to M₂(K). Here 1₀ and i are charming, while j and k are magic squares. Thus the previously proved decomposition

\mathcal I_0 =\operatorname{span}\{\mathbf1_0,i\} \oplus \operatorname{span}\{j,k\}

is precisely the even–odd grading of the split quaternions. The four laws CC→C, CM→M, MC→M, and MM→C are exactly the multiplication table of its even and odd parts.

6. The split quaternion algebra

The multiplication table defines a split quaternion algebra rather than a division algebra. Over ℝ, Hamilton's quaternions satisfy i²=j²=k²=−1 and form a division algebra, whereas here two generators square to +1. Moreover,

(\mathbf1_0+i)(\mathbf1_0-i) =\mathbf1_0-i^2=0,

although both factors are nonzero. Thus 𝓘₀ contains zero divisors and cannot be isomorphic to ℍ. The term “split” means precisely that the quaternion algebra is isomorphic to the full matrix algebra M₂(K), rather than to a division algebra.

7. Coordinates and the split norm

Every element of the ideal has a unique expression

S(0,x,y,z,w) =\alpha\mathbf1_0+\beta i+\gamma j+\delta k,

where the two coordinate systems are related by

\alpha=3z,\qquad \beta=3w,\qquad \gamma=2x-y,\qquad \delta=2y-x.

Quaternion conjugation changes the signs of i,j,k. The corresponding reduced norm is

N_{\mathrm{split}} (\alpha+\beta i+\gamma j+\delta k) =\alpha^2-\beta^2-\gamma^2+\delta^2.

Substituting our coordinates gives

N_{\mathrm{split}} =3\bigl(3z^2-3w^2-x^2+y^2\bigr) =\det B(x,y,z,w).

The determinant of the original matrix therefore factors into the determinants of its two blocks:

\det S(E,x,y,z,w) =(3E)\det B =9E\bigl(3z^2-3w^2-x^2+y^2\bigr).

Consequently, the determinant of a semimagic matrix has a precise structural meaning: it is the product of the one-dimensional block norm and the reduced norm of a split quaternion.

8. Center, ideals, and theorem boundaries

Since the center of M₂(K) consists of scalar matrices, the center of the full semimagic algebra is two-dimensional:

Z(\operatorname{SM}_3(K)) =KH\oplus KQ.

Because M₂(K) is simple, there are exactly four two-sided ideals over a field:

0,\qquad KH,\qquad \mathcal I_0,\qquad \operatorname{SM}_3(K).

The abstract block decomposition only requires 3 to be invertible. The M,C,S coordinates used on this site and the usual anticommuting quaternion presentation are treated here under char K≠2,3.

Over ℚ and ℝ all displayed formulas give a literal isomorphism. Over ℤ the matrices H and Q contain the denominator 3 and do not belong to the integral semimagic lattice. The integral algebra is therefore a lattice inside ℚ⊕M₂(ℚ), but it does not split in the same way as a direct product of ℤ-algebras. This distinction belongs to a separate article.

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