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Let $F$ be the field ${\mathbb Q}(x,y)$ and let

$n>0$ be an integer. Consider the following two matrices

in $M_2(F)$ :

$$D=\left(\begin{array}{cc} x & 0 \\ 0 & y\end{array}\right),

P=\left(\begin{array}{cc} y^n & -x^n \\ y^n & -x^n\end{array}\right)$$

One has $PD^nP=0$, and I would like to show that no product of smaller

length of $P$’s and $D$’s is zero.

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**My thoughts :** Suppose that we have an identity of the form

$M_1M_2\ldots M_r=0$, with $r>0$, $M_i\in\lbrace D ;P \rbrace$ and $r$ minimal.

Since $D$ is invertible, we must have $M_1=M_r=P$. Since

$P^2=(y^n-x^n)P$, the sequence $(M_1,M_2,\ldots,M_r)$

cannot contain two successive $P$’s. So the product must

be of the form $PD^{i_1}PD^{i_2}P\ldots PD^{i_s}P$. Then I am stuck.

I encountered this while trying to answer another

recent MSE question of mine.

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Write

$$P_m:=\left(\begin{array}{cc} y^m & -x^m \\ y^m & -x^m\end{array}\right)\ \ (m\ge 0),$$

so $P=P_n$.

Then $P_m D = xy P_{m-1}\ (m>0)$ and $P_k P_m=(y^k-x^k) P_m$, so, if $i_1$, $i_2$, $\dots$, $i_s$ are in $\{0,\ldots,n-1\}$,

\begin{eqnarray*}P D^{i_1} P D^{i_2} \cdots D^{i_s} P &=& x^\Sigma y^\Sigma P_{n-i_1} P_{n-i_2} \cdots P_{n-i_s} P_n\\

&= &x^\Sigma y^\Sigma(y^{n-i_1}-x^{n-i_1})\cdots (y^{n-i_s}-x^{n-i_s}) P_n\\

&\ne& 0,

\end{eqnarray*}

where $\Sigma:=i_1+\cdots+i_s$. Therefore, to be zero, any word $P D^{i_1} P D^{i_2} \cdots D^{i_s} P$ must contain a subword $D^n$. Together with your argument above, this shows what you want.

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