Is there a matrix invariant or property that is multiplicative, i.e.,
$$f(AB) = f(A) f(B)$$
other than the determinant? In addition, some matrix norms are submultiplicative, but is there a supermultiplicative property?
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13$\begingroup$ If for example you demand $f$ to be continuous, then it is enough to see what $f$ does to invertible matrices. Then, since $GL_n(K)/[GL_n(K),GL_n(K)] \cong K^{\times}$ by using the determinant, such a function $f$ will necessarily be the composition of the determinant with some group homomorphism from $K^{\times}$ to $K^{\times}$. If $f$ is Zariski continuous, you only get integer powers of the determinant. $\endgroup$– Ehud MeirCommented Jan 18, 2018 at 14:04
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$\begingroup$ Regarding your last point: $1/\|A\|$ is supermultiplicative. $\endgroup$– Federico PoloniCommented Jan 19, 2018 at 13:30
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2$\begingroup$ The determinant is a special case of the wider family of multiplicative functions, e.g., look at "multiplicative compounds" en.wikipedia.org/wiki/Compound_matrix (which is based on exterior powers of matrices) $\endgroup$– SuvritCommented Jan 19, 2018 at 14:36
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1$\begingroup$ The identically-$0$ and identically-$1$ functions …. $\endgroup$– LSpiceCommented Jan 19, 2018 at 15:57
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7$\begingroup$ Weird that a question so imprecise and open-ended and basic gets so many positive votes. MO has changed a lot. Matrix means square matrix here, but this is not said. Also, matrices with what kind of coefficients: complex numbers, a field, a commutative ring, any ring? invariant with values in what? What about the identity from $M_n(R)$ to $M_n(R)$? Isn't it true anymore that $AB = AB$ on MO? And moreover this invariant is super-multiplicative for the trivial preorder structure on $M_n(R)$! And it satisfies a nice universal property: every other invariant factors through it. $\endgroup$– JoëlCommented Jan 19, 2018 at 19:01
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2 Answers
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It depends on what is the target space. Linear representations of ${\bf Gl}_n(k)$ do satisfy $\rho(AB)=\rho(A)\rho(B)$ by definition, and they often extend in a natural way to ${\bf M}_n(k)$.
On another hand, invariant scalar functions $f$ are necessarily of the form $f(A)=(\det A)^s$, $(\det A)_+^s$, $(\det A)_-^s$ or $|\det A|^s$.
answered Jan 18, 2018 at 14:06
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8$\begingroup$ Or a combination of (det A)<sup>s</sup> and |det A|<sup>s</sup>, e.g. sign(det A). And over the complex numbers, conjugation is multiplicative. $\endgroup$ Commented Jan 18, 2018 at 17:59
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2$\begingroup$ The last part of the answer is wrong as stated - in order to make sense of it, we have to assume $k=\mathbb R$ or $\mathbb C$, which doesn't seem to be assumed in the question, and even then you need to assume continuity, otherwise you can compose $\det A$ with some multiplicative function $\mathbb R\to\mathbb R$ coming from a nonlinear solution to Cauchy equation. $\endgroup$– WojowuCommented Jan 19, 2018 at 14:32
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1$\begingroup$ It also assumes that we are mapping into $k^\times$. $\endgroup$– LSpiceCommented Jan 19, 2018 at 15:58
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With the exception of $GL_2(\mathbb{F}_2)$, the commutator subgroup of $GL_n(k)$ is $SL_n(k)$ (for $k$ a field). So any multiplicative map from $GL_n(k)$ to an abelian group factors through determinant. This is with no hypotheses on continuity.
EDIT Johannes asks about noninvertible matrices. Let $M$ be an abelian monoid and $\mathrm{Mat}_{n \times n}(k) \to M$ a multiplicative map. Then $GL_n$ must map to the group of units of $M$, so $SL_n(k)$ must map to $1$ by the above (except for $n=2$, $k = \mathbb{F}_2$). Now, if $X$ and $Y \in \mathrm{Mat}_{n \times n}(k)$ are noninvertible matrices of the same rank then there are matrices $U$ and $V$ in $SL_n(k)$ with $UXV=Y$. So our function must be constant on functions of the same rank. Let $e_r$ be its value on matrices of rank $r$. Now, rank $n-1$ idempotents exist. So $e_{n-1}^2=e_{n-1}$. But any noninvertible matrix is a product of rank $n-1$ matrices, so $e_{n-k}$ is a power of $e_{n-1}$ and we deduce $e_{n-1} = e_{n-2} = \cdots = e_1 = e_0$. In short, a multiplicative map to an abelian monoid must take the same value on all noninvertible matrices.
answered Jan 19, 2018 at 14:54
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1$\begingroup$ Yes, but the question asked only about "matrix invariants or properties". The continuity assumption was used to extend from GL to all matrices in case OP is interested in non-invertible matrices as well. $\endgroup$ Commented Jan 19, 2018 at 15:54