determinants and polynomials in matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:10:10Z http://mathoverflow.net/feeds/question/98997 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98997/determinants-and-polynomials-in-matrices determinants and polynomials in matrices Kjetil B Halvorsen 2012-06-07T00:24:39Z 2012-12-12T16:27:44Z <p>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):</p> <p>" 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 $k$.</p> <p>Where otherwise, can I find this result and discussions of it?</p> http://mathoverflow.net/questions/98997/determinants-and-polynomials-in-matrices/98998#98998 Answer by Gjergji Zaimi for determinants and polynomials in matrices Gjergji Zaimi 2012-06-07T01:10:24Z 2012-06-07T20:24:53Z <p>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.</p> <p>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 <a href="http://www.jstor.org/stable/2312885" rel="nofollow">"Scalar valued mappings of squared matrices"</a>.</p> http://mathoverflow.net/questions/98997/determinants-and-polynomials-in-matrices/98999#98999 Answer by Francois Ziegler for determinants and polynomials in matrices Francois Ziegler 2012-06-07T01:35:45Z 2012-06-09T23:29:02Z <p>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 $\mathrm{det}$.</p> <p>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 <em><a href="http://ams.org/mathscinet-getitem?mr=2265844" rel="nofollow">Lie Groups</a></em> lists <em>all</em> 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.</p> http://mathoverflow.net/questions/98997/determinants-and-polynomials-in-matrices/99056#99056 Answer by Ralph for determinants and polynomials in matrices Ralph 2012-06-07T19:31:56Z 2012-06-07T19:31:56Z <p>Using some group theory, the result can be easily generalized as follows: </p> <blockquote> <p>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 $n\ge 0$. </p> </blockquote> <p>Examples for $R$ are all (infinite) local rings (in particular fields) and principal ideal domains. </p> <p>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. </p> http://mathoverflow.net/questions/98997/determinants-and-polynomials-in-matrices/116160#116160 Answer by Peter Michor for determinants and polynomials in matrices Peter Michor 2012-12-12T08:26:34Z 2012-12-12T16:27:44Z <p>Francois Ziegler's answer is not massive overkill. The proof is simple. </p> <p>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. </p>