# Linear Algebra: Minimum Polynomials for Operators on $\mathbb{R}[x,y]$.

I've been working through some problems set by my University over the past few years, and have encountered this problem.

Problem Let $n > 1$ 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 $x^n, x^{n-1}y, \ldots y^n$. Let $P$ and $Q$ be the linear transformations on $V_n$ defined for $f$ in $V_n$ by

$$Pf=x\frac{\delta f}{\delta y} \quad \quad \quad Qf=y\frac{\delta f}{\delta x}$$

What are the minimum polynomials of $P$, $Q$ and ($PQ-QP$) and which of these are diagonalisable?

Progress

We denote the minimum polynomial of any endomorphism $T$ on $V_n$ to be $m_T(X)$.

Let us first consider $P$; we look to find $m_P$ such that $m_P(P)=0$. Certainly the the polynomial $m_P(X)=X^{n+1}$ satisfies the condition as this reduces all the $y$ terms to zero. As such, $m_P(X)|X^{n+1}$. Similarly $m_Q(X)|X^{n+1}$.

Can anyone offer assistance with progressing further? I realise I haven't made much ground. Regards.

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for $f=y^n, P^nf=n!x^n$ so that the minimum polynomial must be $X^{n+1}$. similarly for $Q$ on $f=x^n$. –  yoyo Dec 23 '11 at 17:51

From what you've done, you've shown that $m_P(X)$ and $m_Q(X)$ are both of the form $X^k$ for some $k$. However, if $m_P(P)=0$, then $m_P(P)(y^{n})=0$. Since each application of $P$ shifts things in the list to the left (up to a constant), the minimal polynomial must be $X^{n+1}$. Similarly for $m_Q$. Since $X^{n+1}$ has repeated roots, $P$ and $Q$ are NOT diagonalizable.
For $PQ-QP$, note that each of the elements $x^ky^{n-k}$ is an eigenvector (which can easily seen by thinking of $P$ and $Q$ as moving you left and right). Computing the eigenvalues will give you the minimal polynomial.
Many thanks! I get the eigenvalues to be $(2k-n)$ for $k=1,\ldots, n-1$. Does this give us the minimum polynomial as $m_{PQ-QP}(X)=\prod_{i=1}^{n-1} (X-(2i-n))$? –  Mathmo Dec 23 '11 at 18:13
@TJO Yes, except that $k$ should go from $0$ to $n$, and so if the edge cases give different values (I have not checked), you need include them. In general, if you have a basis of eigenvectors $v_1,\ldots v_n$ with unique eigenvalues $\lambda_1,\ldots, \lambda_m$ (note $m$ will be less than $n$ if there are repeated eigenvalues), the minimal polynomial will be $\prod_{i=1}^m (X-\lambda_i$)\$. –  Aaron Dec 24 '11 at 3:49