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

abstract-algebradeterminantlinear algebramatrices

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.

Best Answer

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!

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