11
$\begingroup$

It well known that, under standard matrix multiplication $\det(AB) = \det(A)\det(B)$, or in other words, that $\det : \mathbb{R}^{n \times n} \rightarrow \langle\mathbb{R}, * \rangle$ is a monoid homomorphism.

In a similar vein, given any two matrices $A,B \in \mathbb{R}^{n \times n}$ , is there an operation $A\star B$ such that $\det(A \star B) = \det(A) + \det(B)$?

$\endgroup$
1
  • 3
    $\begingroup$ Not exactly what was asked for, but $tr(A+B)=tr(A)+tr(B)$, i.e. the trace. $\endgroup$
    – vadim123
    May 20, 2016 at 4:37

4 Answers 4

23
$\begingroup$

Let $A\star B$ be the diagonal matrix whose diagonal entries are all equal to 1 except the topmost leftmost one, which is equal to $\det A+\det B$.

$\endgroup$
1
  • 8
    $\begingroup$ Clever! This shows that the OP needs to put much more thought into what properties he/she wants $\star$ to have. Presumably, the goal is to find a list of natural-looking constraints such that there is a unique operation $\star$ satisfying those constraints. $\endgroup$ May 20, 2016 at 6:07
10
$\begingroup$

If $A,B$ are invertible, decompose them as $\det(A)^{1/n} \bar A$ where $\bar A$ has determinant $1$ and define $A \star B = (\det A + \det B)^{1/n}\bar A \bar B$. For $A$ singular, $B$ invertible define $A \star B = B \star A = B.$ When both are singular define $A \star B = 0$.

It ain't pretty. It probably isn't even associative.

Actually, this is overcomplicated - just define $A \star B = {\rm diag}( x, 1, 1, \ldots, 1)$ where $x=\det A + \det B$. This is associative, but it's trivializing the question: you can write this operation as $+ \circ d$ where $d(A) = {\rm diag}(\det A, 1 \ldots).$ Thus what this is essentially doing is taking the determinant first and identifying this one-dimensional space of matrices with the reals, so that the homomorphism we desire is really just the identity map on $\mathbb R$.

Is there a meaningful operation with this property? I doubt it.

$\endgroup$
10
$\begingroup$

This is actually impossible for $n=0$, since there the determinant is always equal to $1$, and $1+1\neq1$. (Recall that the determinant of an identity matrix is always $1$; this is true even for the $0\times0$ matrix.) You might call this an edge case, but failure in such a simple case might be a warning that you are on a wrong track.

$\endgroup$
2
  • $\begingroup$ You just need to consider $0\times 0$ matrices over $\mathbb{Z}/1\mathbb{Z}$, then $1+1=1$ and all is good ;-) $\endgroup$ May 20, 2016 at 13:38
  • $\begingroup$ @JeppeStigNielsen Of course it is (and in fact for any matrix size), since addition and multiplication are the same in that ring. $\endgroup$ May 20, 2016 at 13:55
0
$\begingroup$

Someone posted this, but later deleted their post, but I thought it was clever, and probably the closest to what I was looking for, so I thought I'd reproduce it here for reference:

Let $A \star B = \pmatrix{A & -I \\ B & I}$, then by the block determinant formula $det(A \star B) = det(A)det(I) - det(-I)det(B) = det(A) + det (B)$.

$\endgroup$
3
  • 1
    $\begingroup$ You need to assume $n$ is odd $\endgroup$
    – M Turgeon
    May 23, 2016 at 22:29
  • $\begingroup$ Ahh, I suppose that's why he deleted it then. $\endgroup$ May 23, 2016 at 22:32
  • 2
    $\begingroup$ You can remedy this by replacing $-I$ with the permutation matrix which swaps the first two rows. This works for any $n>1$. $\endgroup$ May 23, 2016 at 22:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .