determinants and polynomials in matrices - MathOverflow

determinants and polynomials in matrices

2012-06-07T00:24:39Z

Muirhead (1982, "Aspects of Multivariate Statistical Theory") references on page 59 a result (from MacDuffee, 1943, chap 3, "Vectors and Matrices") a book I cannot find):

" The only polynomials in the elements of a matrix satisfying $p(I)=1$ and $p(AB)=p(A)p(B)$ for all matrices, are the integer powers of det B: $p(B) = (\det B)^k$ for some integer $k$.

Where otherwise, can I find this result and discussions of it?

Answer by Gjergji Zaimi for determinants and polynomials in matrices

2012-06-07T01:10:24Z

S. Cater proved that every $\mathbb F$ valued map $f$ on square matrices which satisfies $f(ABC)=f(CBA)$ can be written as $f(X)=\pi(\det(X))$ for a unique map $\pi:\mathbb F\to \mathbb F$. Also $f$ is multiplicative iff $\pi$ is multiplicative.

When you assume $\mathbb F=\mathbb R$ and $f$ is continuous, for example, then a continuous multiplicative $\pi$ is of the form $x^{r}$ for some $r$. And the result you quote is a corollary,(because polynomials are continuous). This general result is proved in "Scalar valued mappings of squared matrices".

Answer by Francois Ziegler for determinants and polynomials in matrices

2012-06-07T01:35:45Z

This is going to sound like massive overkill, but it is "very well known" that the only 1-dimensional polynomial representations of $GL(V)$ (which is what you're looking at) are the nonnegative powers of $\mathrm{det}$.

Reference (I assume from the mention of statistics that you are OK working with base field $\mathbf{R}$ or $\mathbf{C}$): e.g. Procesi on p.278 of Lie Groups lists all irreducible rational representations as all $$ S_\lambda(V)\otimes\mathrm{det}^k,\qquad k\in\mathbf{Z}, $$ where $\lambda$ runs over a certain set of partitions or Young tableaux; and on p.270 he gives a dimension formula for $S_\lambda(V)$ which is $>1$ unless $S_\lambda(V)$ is trivial.

Answer by Ralph for determinants and polynomials in matrices

2012-06-07T19:31:56Z

Using some group theory, the result can be easily generalized as follows:

If $R$ is an infinite commutative ring such that $SL_n(R)$ is perfect (i.e. $SL_n(R)$ is its own commutator) then the only polynomial functions $p: R[x_{11},...,x_{nn}] \to R$ satisfying the required identities are $p(X) = \det(X)^n$ for some $n\ge 0$.

Examples for $R$ are all (infinite) local rings (in particular fields) and principal ideal domains.

Proof: $p$ induces a group homomorphism $p: GL_n(R) \to R^\times$ those kernel contains the commutator subgroup. Since $SL_n(R)$ is perfect, $SL_n(R) \le \ker(p)$. Define a group hom. $$f: R^\times \to R^\times,\; x \mapsto p\big(\operatorname{diag}(x,1,...,1)\big).$$ If $A \in GL_n(R)$, set $B := \operatorname{diag}(\det(A),1,...,1)$. Then $AB^{-1} \in SL_n(R)$ implies $$p(A)=p(B)=f(\det(A)).\hspace{70pt}(\ast)$$ Since $p$ is a polynomial function, $f(x)$ is a polynomial function in $x$ satisfying $$f(xy)=f(x)f(y).\hspace{110pt}(\ast\ast)$$ As $R$ is infinite, it's easy to see that the only polynomial functions with $(\ast\ast)$ are $f(x)=x^n$. Now the result follows from $(\ast)$. q.e.d.

Answer by Peter Michor for determinants and polynomials in matrices

2012-12-12T08:26:34Z

Francois Ziegler's answer is not massive overkill. The proof is simple.

Suppose you have a continuous multiplicative mapping $P: \operatorname{Mat}_n(\mathbb R) \to (\mathbb R, \cdot)$ as you started with, then it restricts to a continuous group homomorphism $P:GL(n)\to (\mathbb R\setminus\{0\}, \cdot)$, which is analytic (using $\exp$). Its derivative at $\mathbb I_n$ is a Lie algebra homomorphism $P':\mathfrak g\mathfrak l(n)\to \mathbb R$ which must vanish on each commutator. The space of all commutators is the codimension 1 Lie subalgebra $\mathfrak s\mathfrak l(n)$. Since $P'$ is also linear, it is of the form $P'(X) = k.\operatorname{Trace}(X)$ for some $k$. This integrates to $P(A) = \det(A)^k$. Here $k$ must be integral if the ground field is $\mathbb C$. In the real case any $k$ works if $\det(A)$ is always $\ge 0$, and integral generally.