Kunze Linear Algebra · Problem 5.2.13

Determinants

Mathematics — Methods · multiplicative-determinant-functions

Problem

Let RR be the field of real numbers, and let DD be a function on 2×22 \times 2 matrices over RR, with values in RR, such that D(AB)=D(A)D(B)D(AB) = D(A)D(B) for all AA, BB. Suppose also that D([0110])≠D([1001]).D\left(\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\right) \neq D\left(\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}\right). Prove the following. ((a)) D(0)=0D(0) = 0; ((b)) D(A)=0D(A) = 0 if A2=0A^2 = 0; ((c)) D(B)=−D(A)D(B) = -D(A) if BB is obtained by interchanging the rows (or columns) of AA; ((d)) D(A)=0D(A) = 0 if one row (or one column) of AA is 00; ((e)) D(A)=0D(A) = 0 whenever AA is singular.
Intuitive explanation

The key insight is that DD is a multiplicative function on matrices—it respects the product structure—so it must behave like a determinant. The condition that DD distinguishes the identity from a row-swap matrix tells us DD is not trivial, which forces it to capture something essential about matrix structure. Use the multiplicative property strategically: if you can write a matrix as a product involving nilpotent or singular factors, or express it in terms of matrices whose DD-values you know, you can deduce DD of the original. The path to each part involves finding clever factorizations or relationships that leverage D(AB)=D(A)D(B)D(AB) = D(A)D(B).

Intuitive explanation

The key insight is that DD is a multiplicative function on matrices—it respects the product structure—so it must behave like a determinant. The condition that DD distinguishes the identity from a row-swap matrix tells us DD is not trivial, which forces it to capture something essential about matrix structure. Use the multiplicative property strategically: if you can write a matrix as a product involving nilpotent or singular factors, or express it in terms of matrices whose DD-values you know, you can deduce DD of the original. The path to each part involves finding clever factorizations or relationships that leverage D(AB)=D(A)D(B)D(AB) = D(A)D(B).

Related theory

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