# Finding minimal polynomial

• May 6th 2010, 09:07 AM
Mimi89
Finding minimal polynomial
Hi everyone,

Let $n \geq 1$ and let $V_{n}$ be the subspace of $\mathbb{R} [x,y]$ of dimension n+1 consisting of homogeneous polynomials of degree n, that is, the subspace spanned by $x^n, \ x^{n-1}y, \ ..., \ y^n$. Let P and Q be linear transformations on $V_{n}$ defined for f in $V_{n}$ by

$Pf=x \frac{ \partial f}{ \partial y}$ and $Qf=y \frac{\partial f}{\partial x}$

Find the minimal polynomial of (PQ-QP).

Now, if my calculations are correct, for an arbitrary basis vector of the form $x^{n-k}y^k$, $(PQ-QP)(x^{n-k}y^k)=(n-2k)x^{n-k}y^k$.

Now, the minimal polynomial (denote m) is the unique monic polynomial of least degree s.t. m(PQ-QP)=0 (0 transformation). So in particular, I want $m((PQ-QP)(x^{n-k}y^k))=m((n-2k)x^{n-k}y^k)=0$, and moreover any linear combination of $(n-2k)x^{n-k}y^k$ to be equal to zero. How am I supposed to achieve this?

Thanks a lot for all the help.
• May 6th 2010, 10:14 PM
NonCommAlg
Quote:

Originally Posted by Mimi89
Hi everyone,

Let $n \geq 1$ and let $V_{n}$ be the subspace of $\mathbb{R} [x,y]$ of dimension n+1 consisting of homogeneous polynomials of degree n, that is, the subspace spanned by $x^n, \ x^{n-1}y, \ ..., \ y^n$. Let P and Q be linear transformations on $V_{n}$ defined for f in $V_{n}$ by

$Pf=x \frac{ \partial f}{ \partial y}$ and $Qf=y \frac{\partial f}{\partial x}$

Find the minimal polynomial of (PQ-QP).

Now, if my calculations are correct, for an arbitrary basis vector of the form $x^{n-k}y^k$, $(PQ-QP)(x^{n-k}y^k)=(n-2k)x^{n-k}y^k$.

Now, the minimal polynomial (denote m) is the unique monic polynomial of least degree s.t. m(PQ-QP)=0 (0 transformation). So in particular, I want $m((PQ-QP)(x^{n-k}y^k))=m((n-2k)x^{n-k}y^k)=0$, and moreover any linear combination of $(n-2k)x^{n-k}y^k$ to be equal to zero. How am I supposed to achieve this?

Thanks a lot for all the help.

what you've got so far is correct. so $\{n-2k, \ 0 \leq k \leq n \}$ is the set of eigenvalues of $PQ-QP.$ since these eigenvalues are pairwise distinct, the characteristic and minimal polynomials of

$PQ-QP$ are equal. thus the minimal polynomial of $PQ-QP$ is $\prod_{k=0}^n (x-n+2k).$