# Linear Algebra – Non-Trivial Surjective Multiplicative Homomorphism from 4×4 to 2×2 Matrices

It's well known the determinant provides a surjective homomorphism between $$n \times n$$ real matrices and $$\mathbb{R}$$ with the operation of multiplication.

I was curious of inter-matrix multiplicative homorphisms.

The trivial idea of viewing a $$4\times 4$$ matrix as a $$2\times 2$$ matrix with entries in the $$2\times 2$$ matrices suggests a determinant formula (of the usual type) but unfortunately that doesn't actually seem to work at all as a homomorphism after investigating it with sympy.

Doing some reading I stumbled upon: Quasideterminants for non-commutative rings.

But i'm a bit dissatisfied. I would like to think such a homomorphism should exist but I just don't know how to find it.

Claim: There exists no surjective multiplicative morphism from $$M_r(k) \to M_s(k)$$ for $$r > s > 1$$.

Proof: Assume the contrary, let $$\varphi: M_r(k) \to M_s(k)$$ be such a morphism. For all $$A \in M_r(k)$$, we have $$\varphi(A) = \varphi(A)\varphi(I_r)$$. The surjectivity of $$\varphi$$ allows us to pick $$A$$ such that $$\varphi(A)$$ is invertible. Thus, $$\varphi(I_r) = I_s$$. Furthermore, for all invertible $$A \in M_r(k)$$, we have $$\varphi(A)\varphi(A^{-1}) = \varphi(I_r) = I_s$$. Thus, $$\varphi$$ maps invertible transformations to invertible transformations.

Key Observation: For any $$A, B \in M_r(k)$$ with the same rank, there exist invertible $$U, V \in M_r(k)$$ satisfying $$A = UBV$$. Then, $$\varphi(A) = \varphi(U)\varphi(B)\varphi(V)$$. Since $$\varphi(U)$$ and $$\varphi(V)$$ are invertible, $$\varphi(A)$$ and $$\varphi(B)$$ have the same rank. In other words, the rank of $$\varphi(A)$$ is determined by the rank of $$A$$.

Let $$N \in M_r(k)$$ be nilpotent with rank $$(r-1)$$. The elements $$I_r, N, \ldots, N^r$$ have ranks $$r, r-1, \ldots, 0$$ respectively. Thus, the ranks of $$I_s, \varphi(N), \ldots, \varphi(N)^r$$ must include all possible integers from $$0$$ to $$s$$ by surjectivity. Furthermore, $$\varphi(N)^r = 0_s$$, so $$\varphi(N)$$ must be nilpotent. This forces $$\varphi(N)$$ to be nilpotent of rank $$(s-1)$$. We have consequently shown that for all $$A \in M_r(k)$$, $$\mathrm{rank} \, \varphi(A) = \mathrm{max}(\mathrm{rank}\, A - (r-s), 0)$$.

Now, let $$e_1, \ldots, e_r$$ be any basis for $$k^r$$. There are $$t = {r \choose {r-s+1}} > s$$ possible projections of rank $$r - s + 1$$ with respect to this basis. Note that we are using both $$r > s$$ and $$s > 1$$ for this to hold true. Denote these projections by $$P_1, \ldots, P_t$$. By our formula above, the images $$Q_i = \varphi(P_i)$$ have rank $$1$$ and they are projections by multiplicativity.

For each $$1 \leqslant i \leqslant t$$, pick an arbitrary eigenvector $$v_i$$ of eigenvalue $$1$$ for the projection $$Q_i$$ (so $$v_i$$ is determined up to scaling). Since $$t > s$$, the vectors $$\{v_i\}$$ must have some linear dependence. Assume without loss that $$v_1$$ is in the span of the rest. For each $$i \neq 1$$, since $$P_1P_i = 0_r$$, we have $$Q_1Q_i = 0_s$$. This implies that $$Q_1Q_iv_i = Q_1v_i = 0$$. However, $$v_1$$ is in the span of $$\{v_i : i > 1\}$$, so $$Q_1v_1 = 0$$. This yields the required contradiction!