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The naive quaternionic formula $ad-bc$ is not row-alternating: a matrix with equal rows can have value $2k\ne0$

Statement refuted

The commutative-ring formula $ad-bc$ does not define a row-alternating determinant for quaternionic $2\times2$ matrices. A matrix with two equal rows can have naive value $2k\ne0$ .

Facts & Assumptions

Given: The quaternionic matrix $A=\begin{pmatrix}i&j\\i&j\end{pmatrix}$ and the naive expression $D(A)=ad-bc$ .

[L1]

The quaternions have basis elements $i,j,k$ and real scalar multiplication (The quaternions $\mathbb{H}$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[L2]

Quaternion multiplication satisfies $ij=k$ , $ji=-k$ , and $k\ne-k$ ( $\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1} = \bar q / N(q)$ for $q \ne 0$ , while $ij = k$ and $ji = -k$ ).

[L3]

Over a commutative ring the genuine $2\times2$ determinant formula is $ad-bc$ (The Leibniz formula gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$ ).

Counterexample

technique · direct

1.1

The two rows of $A$ are both $(i,j)$ , so a row-alternating function would vanish on $A$ .

L1L2L3

2.1

The naive expression instead gives $D(A)=ij-ji=k-(-k)=2k$ . If $2k=0$ , then $k=-k$ , contrary to [L2]; hence $D(A)\ne0$ .

step 1.1L2algebra∎

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Sources

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Ch. 4 (standard reference, not scraped)

The naive quaternionic formula ad−bcad-bcad−bc is not row-alternating: a matrix with equal rows can have value 2k≠02k\ne02k=0