TITLE: determinants and polynomials in matrices
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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.