Determinant from Paul Garrett's Definition of the Characteristic Polynomial.

Question

$\DeclareMathOperator{\id}{id} \DeclareMathOperator{\End}{End}$ On pg. 390 of Paul Garrett's notes on Algebra, a definition for the characteristic polynomial is given, which I discuss here.

Let $V$ be an $n$-dimensional vector space over a field $F$ and let $T$ be a linear operator on $V$. Consider the $F$-module $M=F[x]\otimes_F V$ and give it an $F[x]$-module structure by extending the scalars via the natural inclusion $F\to F[x]$. Thus $M$ is a free $F[x]$-module of rank $n$. Now we can think of $1\otimes T-x\otimes \id_V$ as a member of $\End_{F[x]}(M)$.

Definition. The characteristic polynomial of $T$ is then defined as the unique polynomial $p(x)\in F[x]$ such that the $n$-th exterior power map $$\Lambda^n_{F[x]}(1\otimes T-x\otimes \id_V):\Lambda^n_{F[x]} M\to \Lambda^n_{F[x]}M$$ is same as multiplication by $p(x)$ with the identity map on $\Lambda^n_{F[x]}M$.

On the other hand, the following is well known:

Definition. Let $\lambda\in F$ is a scalar in $F$, we have a linear operator $T-\lambda I$ on $V$, whose determinant is defined as the unique scalar $\theta\in F$ such that the map $$\Lambda^n_F (T-\lambda I):\Lambda^n_F V\to \Lambda^n_F V$$ is $\theta$ times the identity map on $V$.

Question. How do we see that $\det(T-\lambda I)=p(\lambda)$ under the above definitions.

I have a feeling that this should be rather straightforward from the definitions but I can't see it. I am just beginning to learn exterior algebra over modules.

Answer 1

First of all, notice that exterior powers commute with base change (Eisenbud, Commutative Algebra with a View..., Proposition A2.2, p. 576), hence

$$\Lambda^n_{F[x]} (F[x] \otimes_F V)=F[x] \otimes_F \Lambda^n_{F}V$$

You can easily check, that the following diagram (of $F$-modules) commutes ($m(\lambda)$ is the map $x \mapsto \lambda$ from the other answer):

$\require{AMScd}$ \begin{CD} F[x] \otimes_F V @>{m(\lambda) \otimes \operatorname{id}_V}>> V\\ @V{1 \otimes T - x \otimes \operatorname{id}_V}VV @V{T-\lambda \operatorname{id}_V}VV \\ F[x] \otimes_F V @>{m(\lambda) \otimes \operatorname{id}_V}>> V \end{CD}

Taking the exterior algebra is functorial, hence we get a commutative diagram (I will omit the exterior power symbols for the name of the maps, unless for the $\det$)

$\require{AMScd}$ \begin{CD} F[x] \otimes_F \Lambda^n_{F}V @>{m(\lambda) \otimes \operatorname{id}_V}>> \Lambda^n_{F}V\\ @V{1 \otimes T - x \otimes \operatorname{id}_V}VV @V{\det(T-\lambda \operatorname{id}_V)}VV \\ F[x] \otimes_F \Lambda^n_{F}V @>{m(\lambda) \otimes \operatorname{id}_V}>> \Lambda^n_{F}V \end{CD}

Using the identification $\Lambda^n_{F}V = F$ and the definition for $p(x)$ above (set $P(\lambda)=\det(T-\lambda \operatorname{id}_V)$), this is precisely the following diagram:

$\require{AMScd}$ \begin{CD} F[x] @>{m(\lambda)}>> F\\ @V{\cdot p(x)}VV @V{\cdot P(\lambda)}VV \\ F[x] @>{m(\lambda)}>> F \end{CD}

In particular, if we track the image of $1 \in F[x]$, we get the desired

$$p(\lambda) = P(\lambda).$$

Answer 2

$\DeclareMathOperator{\id}{id} \DeclareMathOperator{\End}{End}$Fix a scalar $\lambda\in F$. There is an evaluation map $m_\lambda:F[x]\rightarrow F$ such that $x\mapsto \lambda$. It induces a commutative diagram

$\require{AMScd}$ \begin{CD} \Lambda^n_{F[x]} M @>{\Lambda^n_{F[x]}(1\otimes T-x\otimes \id_V)}>> \Lambda^n_{F[x]} M\\ @V{m_\lambda}VV @V{m_\lambda}VV \\ \Lambda^n_F V @>{\Lambda^n_F (T-\lambda I)}>> \Lambda^n_F V \end{CD}

In particular $m_\lambda(p(x))=p(\lambda)$, so $\theta=\operatorname{det}(T-\lambda I)=p(\lambda)$.