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The setup is as follows,

Let $p(x) = x^n+a_{n-1}x^{n-1}+...+a_0 \in F[x]$ be a non-constant polynomial of degree $n$. Then $V$ is the quotient $\frac{F[x]}{(p(x))}$, which is an $n$ dimensional vector space over $F$. We let $m_x$ denote the linear map of $V$ given by multiplication by $x$, and we want to show that that $\det(t \cdot id_V-m_x) =p(t)$.

Here $id_V$ is just the identity map of $V$. I don't have any idea where to start here. I've seen the determinant as an element of $\Lambda^k(V)$ where this denotes the set of alternating and $k$-multilinear forms from $V^k$ into $F$, as well as the explicit definition as the sum over permutations of $S_n$, but I don't see how I'm supposed to use this information here. Anything hint or insight is appreciated.

Irving Rabin
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1 Answers1

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One approach is as follows.

First consider the minimal polynomial $\mu(x)$. Because $p(m_x) = 0$, it follows that $\mu\mid p$. On the other hand, for any polynomial $q$ with $\deg(q) < n$, we have $p(m_x) = m_{p(x)} \neq 0$, which means that $\deg(\mu) \geq n$, and by definition $\mu$ is monic. Thus, $\mu = p$

As a consequence of the Cayley-Hamilton theorem, we know that $\mu$ must divide the characteristic polynomial $\chi(t) = \det(t\,\mathrm{id}_V-m_x)$. Moreover, $\deg(\chi) = n$ and $\chi$ is monic. It follows that $\mu = \chi$.

So, we have $\chi = p$, which is what we wanted.

An approach without C-H: We find that the matrix of $m_x$ relative to the basis $\mathcal B = (1,x,\dots,x^{n-1})$ is given by $$ M := [m_x]_{\mathcal B} = \pmatrix{0 & &&-a_0\\ 1 & \ddots& &-a_1\\ &\ddots & 0&\vdots\\ &&1&-a_{n-1}}. $$ Using a Laplace expansion along the final column, we can show that $\det(t I - M) = p(t)$.

Ben Grossmann
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