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How would one prove the following theorem?

$p(A)$ is a nonzero polynomial of the entries of $A$ and satisfies $p(AB)=p(A)p(B)$, for all square matrices $A$ and $B$ of complex numbers. Prove $p(A)=(\det\,A)^k$ for some nonnegative integer $k$.


My attempt:

A matrix $A$ can be Shur-decomposed into $A=QTQ^\dagger$ where $T$ is a triangular matrix and $Q$ a unitary matrix. Now $p(Q)p(Q^\dagger)=p(QQ^\dagger)=p(I)=1$ where the last equation comes from the second paragraph below. Now $p(A)=p(Q)p(T)p(Q^\dagger)=p(T)$. If we can show $p(T)=(\Pi_i\ T_{i,i})^k$ for some integer $k$, then we are done.

Now if $T$ is diagonal with its diagonal entries all equal to $1$ except one being a complex variable $x$, we can show $p(T)=x^k$ for some nonnegative integer $k$. Let $p(T)=\sum_{i=0}^k a_ix^i$ for some nonnegative integer $k$ and $a_k\ne0$. $\sum_{i=0}^k a_ix^{2i}=p(T^2)=p(T)^2=\big(\sum_{i=0}^k a_ix^i\big)^2$. Expand the last expression and collecting coefficients of $\{x^i\}_{i=0}^k$. By the linear independence of $\{x^i\}_{i=0}^k$, the coefficient of the left hand side and that of the right hand side of the same order terms have to match. Considering the coefficents of the terms of order no less than $k$, we draw the following conclusion. The coefficients of the odd order terms have to be zero. $a_k=1$. Recursively we conclude $a_i=0,\,\forall i<k$.

As $p(T_1T_2)=p(T_1)p(T_2)$, $p(T)=\Pi_i\ T_{i,i}$ for any diagonal matrix $T$. But I am unable to proceed further to the triangular matrix.

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    $\begingroup$ What about $p(A)=\det(A^k)$ ? $\endgroup$ – Surb Sep 23 '18 at 21:32
  • $\begingroup$ I do not get your point. $\det(A^k)=(\det A)^k$. So your example does not contradict the conclusion. Am I missing something? $\endgroup$ – Hans Sep 23 '18 at 21:40
  • $\begingroup$ Well in your title it is stated that you want to solve a functional equation. So I looked at the properties you wrote and suggested a solution. But apparently, I misunderstood your question which, now that I read it again, is really unclear. There seems to be no theorem and if this is the case, it is unclear what are the assumptions and what should be proved. $\endgroup$ – Surb Sep 23 '18 at 21:52
  • $\begingroup$ @Surb: "We have ... " usually means what follows is the conclusion and is to be proved. I have edited the statement and explicitly declared the conclusion with an imperative clause. How is it now? $\endgroup$ – Hans Sep 23 '18 at 22:02
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    $\begingroup$ Thinking about that. Whenever the field is algebraically closed Marco's answer survives when using Zariski topology for the last type of elementary matrices. And using Zariski is surely overkill. I'm sure the conclusion is the same for most fields, if not all! I asked about the field of coefficients because superficially Marco is using the topology of real/complex numbers, but may be not :-) $\endgroup$ – Jyrki Lahtonen Sep 24 '18 at 4:10
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Let $D_x=(x;1;\ldots;1)$ be a diagonal matrix and define $f(x)=p(D_x)$. Then $f(x)$ is a polynomial in $x$ and $f(xy)=f(x)f(y)$. It follows that $f(x)=x^k$ for some integer $k$. Similarly, let $E_x=(1;x;1;\ldots;1)$ and let $E$ be the elementary matrix that is obtained from the identity matrix by interchanging rows 1 and 2. Since $E^2=I$, one has $p(E)p(E)=p(I)=1$. Since $ED_xE=E_x$ we must have $p(E_x)=p(D_x)=x^k$. It follows that for every diagonal matrix $p(D)=(\det D)^k$.

Let $G$ be an elementary matrix obtained from interchanging rows $i$ and $j$. Then $G^2=I$ and so $p(G)=\pm 1$.

Let $F$ be an elementary matrix obtained from adding $r$ times row $i$ to row $j \neq i$. We show that $p(F)=1$. Let $D(x)$ be a diagonal matrix with entry $x$ on the row $i$ and 1 in other diagonal entries. One has $D(x)FD(1/x) \rightarrow I$ as $x\to\infty$. By the continuity of $p$, we must have $p(F)=1=\det(F)^k$.

Every matrix is a product of elementary matrices. So $p(A)=\pm (\det A)^k$. Since $p(A)$ is a polynomial in the entries of $A$, a choice of sign must be fixed, but $p(A)=-(\det A)^k$ does not satisfy the property. It follows that $p(A)=(\det A)^k$.

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  • $\begingroup$ Nothing wrong with this (+1 now that the OP specified the field). Just thinking whether we can use the fact that $F(r)F(s)=F(r+s)$ instead of a continuity argument. $\endgroup$ – Jyrki Lahtonen Sep 24 '18 at 4:14
  • $\begingroup$ +1 and accepted. Should ''of an identity matrix" be appended to the end of "adding $r$ times row $i$ to row $j\ne i$"? $\endgroup$ – Hans Sep 24 '18 at 4:31
  • $\begingroup$ Marco, is there a way to prove p(A)=p(AT) without the continuity property? $\endgroup$ – Hans Sep 25 '18 at 17:10
  • $\begingroup$ @Hans, I can't think of a way. I don't think it is true without that property. $\endgroup$ – Marco Sep 25 '18 at 18:15

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