This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Determinant
In the previous post, we verified that for any free \(A\)-module \(M\), if \(M\) has a basis \((e_i)_{i\in I}\), then the elements \(e_J\) expressed in the form
\[e_J=e_{j_1}\wedge e_{j_2}\wedge\cdots\wedge e_{j_k},\qquad j_1<\cdots < j_k, \quad J=\{j_1,\ldots, j_k\}\]constitute a basis of \(\bigwedge(M)\). In particular, collecting all \(J\) satisfying \(\lvert J\rvert=n\) gives a basis of \(\bigwedge^n(M)\).
Now suppose \(M\) has a finite basis \(e_1,\ldots, e_n\). Then \(\bigwedge^n(M)\) has a basis consisting of the single element \(e_1\wedge\cdots\wedge e_n\). On the other hand, for any \(u\in\End_\rMod{A}(M)\), the functoriality of \(\bigwedge\) induces \(\bigwedge^n(u):\bigwedge^n(M)\rightarrow\bigwedge^n(M)\), and by the above discussion this linear map must be of the form \(x\mapsto \alpha x\).
Definition 1 Suppose a free \(A\)-module \(M\) has a basis \((e_i)_{i\in I}\). Then for any \(u:M \rightarrow M\), the scalar \(\alpha\in A\) obtained from the above discussion is called the determinant of \(u\) and is denoted \(\det u\).
Then the following proposition is obvious from the definition.
Proposition 2 The following hold.
- For any \(u,v\in\End_\rMod{A}(M)\), we have \(\det(u\circ v)=(\det u)(\det v)\).
- \(\det(\id_M)=1\).
- For any \(u\in\Aut_\rMod{A}(M)\), \(\det u\) is invertible in \(A\) and satisfies \(\det(u)^{-1}=\det(u^{-1})\).
Corollary 3 For a finite free \(A\)-module \(M\) and \(u\in\End_\rMod{A}(M)\), the following are equivalent.
- \(u\) is bijective.
- \(\det u\) is invertible in \(A\).
Proof
It suffices to show that condition 2 implies condition 1. To do this, define \(x_i=u(e_i)\) to obtain
\[x_1\wedge \cdots\wedge x_n=\det(u) e_1\wedge\cdots\wedge e_n,\]then multiply both sides by \(\det(u)^{-1}\) and consider the change of basis obtained from this equation.
Fixing a free \(A\)-module \(M\) and its basis \(e_1,\ldots, e_n\), for any elements \(x_1,\ldots, x_n\) of \(M\) there exists a scalar \(\alpha\) such that
\[x_1\wedge \cdots\wedge x_n=\alpha e_1\wedge\cdots\wedge e_n,\]and we write this as \(\det(x_1,\ldots, x_n)\). To actually compute this value, one expresses each \(x_i\) as a linear combination of \(e_1,\ldots, e_n\) and then rearranges everything using \(e_i\wedge e_i=0\) and \(e_i\wedge e_j=-e_j\wedge e_i\). In the case \(A=\mathbb{K}\), this was already examined in [Linear Algebra] §Existence and Uniqueness of the Determinant, Equation (2). To explain in a bit more detail, for any \(X\in\Mat_n(A)\), writing \(X=(x_1,\ldots, x_n)\) using column vectors, there is a unique \(u\in\End_\rMod{A}(M)\) satisfying \(u(e_i)=x_i\), and \(\det(u)\) is well-defined; comparing this with the expression appearing in the proof of Corollary 3, we see that \(\det (x_1,\ldots, x_n)=\det(u)\). From this we can formulate a matrix version of Proposition 2, and the process of computing it is exactly [Linear Algebra] §Existence and Uniqueness of the Determinant, Equation (2). In particular, we obtain \(\det(u^\ast)=\det(u)\).
Minors of a Matrix
Among methods for computing determinants, there is Laplace expansion, which was discussed in [Linear Algebra] §Existence and Uniqueness of the Determinant, ⁋Theorem 13. We will not repeat that computation since it was already covered, but we can generalize the \(\det A^{(i,j)}\) that appeared there.
To this end, suppose we are given an arbitrary \(X=(\xi_{ij})\in\Mat_{I\times J}\). Fixing a total ordering on \(I\) and \(J\), whenever finite subsets \(H\subseteq I\) and \(K\subseteq J\) are given, the submatrix \(X_{H,K}=(\xi_{i,j})_{i\in H,j\in K}\) also inherits a total order on its indices. In particular, consider the case where \(\lvert H\rvert=\lvert K\rvert\). Then the following lemma is obvious.
Lemma 4 Suppose a basis \((e_i)_{i\in I}\) of a free \(A\)-module \(M\) is given, and fix a total ordering on \(I\). Also, for any natural number \(p\), consider the basis of \(\bigwedge^p(M)\)
\[(e_J=e_{j_1}\wedge\cdots\wedge e_{j_p})_{\lvert J\rvert=p}.\]Given any \(p\) elements \(x_1,\ldots, x_p\) of \(M\), write them in the form
\[x_j=\sum_{i\in I} \xi_{ij}e_i\]and define the matrix \(X=(\xi_{ij})_{(i,j)\in I\times p}\in\Mat_{I\times p}(A)\). Then the formula
\[x_1\wedge x_2\wedge\cdots\wedge x_p=\sum_{\lvert J\rvert=p}\det X_{I,J}e_J\]holds.
Proposition 5 Suppose two free \(A\)-modules \(M,N\) and their finite bases \((e_i)_{1\leq i\leq m}\), \((f_j)_{1\leq j\leq n}\) are respectively given. Now for a natural number \(p\) less than \(\min(m,n)\), the matrix representation of \(\bigwedge^p(u):\bigwedge^p(M) \rightarrow\bigwedge^p(N)\) with respect to the bases \((e_I)_{\lvert I\rvert=p}\) and \((f_J)_{\lvert J\rvert=p}\) is given by \((\det(X_{J,I}))\).
Proof
In the given situation, arrange the elements of \(I\) in increasing order as \(i_1<\cdots< i_p\). Then by the definition of \(\bigwedge^p(u)\),
\[{\bigwedge}^p(u)=u(e_{i_1})\wedge\cdots\wedge u(e_{i_p}),\]so it suffices to apply the preceding lemma.
Corollary 6 Suppose a free \(A\)-module \(M\) has a finite basis \((e_i)_{1\leq i\leq n}\). Then for any \(u\in\End_\rMod{A}(M)\) and \(\alpha,\beta\in A\) we obtain the formula
\[\det(\alpha\cdot\id_M+\beta u)=\sum_{k\geq 0}\tr\left({\bigwedge}^k(u)\right)\alpha^{n-k}\beta^k.\]Proof
The left-hand side arises from turning the wedge product
\[(\alpha e_1+\beta u(e_1))\wedge\cdots\wedge(\alpha e_n+\beta u(e_n))\]into a multiple of \(e_1\wedge\cdots\wedge e_n\). Expanding the above expression completely, it equals the sum of the terms \(\alpha^{n-p}\beta^p x_P\) over integers \(p\) with \(0\leq p\leq n\) and subsets \(P\subseteq\{1,\ldots, n\}\) with \(\lvert P\rvert=p\), where \(x_P=x_1\wedge\cdots\wedge x_n\) is the element of \(\bigwedge^n(M)\) defined by the elements
\[x_i=\begin{cases}u(e_i)&\text{if $i\in P$}\\e_i&\text{otherwise.}\end{cases}\]To simplify \(x_P\) of the above form, put \(\{1,\ldots, n\}\setminus P=Q\) and arrange the elements of \(P\) and \(Q\) in increasing order as
\[P=\{i_1< i_2<\cdots< i_p\},\qquad Q=\{j_1< j_2<\cdots < j_{n-p}\}.\]Then by reordering the elements of \(P\) and \(Q\) we can write
\[x_P=\gamma_{P,Q}e_{j_1}\wedge\cdots\wedge e_{j_{n-p}}\wedge u(e_{i_1})\wedge\cdots u(e_{i_{n-p}}).\]Here \(\gamma_{P,Q}\) is the sign arising from reordering these factors, specifically given by the formula
\[\gamma_{P,Q}=(-1)^{\lvert A\rvert},\qquad A=\{(p,q)\in P\times Q\mid p>q\}.\]Then by the definition of \(X\) and Lemma 4,
\[u(e_{i_1})\wedge\cdots\wedge u(e_{i_p})=\sum_{\lvert I\rvert=p}\det(X_{I,Q})e_Q,\]and substituting this gives
\[x_P=\gamma_{P,Q}\sum_{\lvert I\rvert=p}\det X_{I,P} e_Q\wedge e_I\]is obtained. However, since \(\lvert I\rvert=p\) and \(\lvert Q\rvert=n-p\), unless \(I=P\) they always share a common \(e_i\), and therefore the above expression can be written as
\[x_P=\det (X_{P,P} )e_1\wedge e_2\wedge\cdots\wedge e_n.\]By Proposition 5, for a fixed \(p\), the sum of \(\det(X_{p,p})\) over all \(p\) satisfying \(\lvert P\rvert=p\) equals \(\tr\left(\bigwedge^k(u)\right)\), and thus the proof is complete.
In particular, setting \(\alpha=\beta=1\) gives \(\tr(\bigwedge(u))=\det(\id_M+u)\).
Characteristic Polynomial
We now define the characteristic polynomial.
Considering the polynomial algebra \(A[\x]\) and the canonical inclusion \(\iota: A \hookrightarrow A[\x]\), the extension of scalars defines an \(A[\x]\)-module structure on \(\iota_!M=A[\x]\otimes_A M\). ([Algebraic Structures] §Change of Base Ring, ⁋Definition 3) Moreover, whenever \(u\in\End_\rMod{A}(M)\) is given, \(\iota_!u\in\End_\rMod{A[\x]}(\iota_!M)\) is also defined.
For any \(u\in\End_\rMod{A}(M)\), let us introduce the notation
\[u^k=\underbrace{u\circ\cdots\circ u}_\text{$k$ times}.\]Then for any \(p\in A[\x]\) we can regard \(p(u)\) as an element of \(\End_\rMod{A}(M)\), and in this case for any \(p,q\in A[\x]\) we have
\[(pq)(u)=p(u)\circ q(u)=q(u)\circ p(u).\]Therefore, defining an \(A[\x]\)-action on \(M\) by the formula
\[p\bullet x=p(u)(x)\]gives \(M\) an \(A[\x]\)-module structure. That is, via the scalar multiplication just defined,
\[\rho: A[\x]\otimes_A M \rightarrow M;\qquad p\otimes_A x\mapsto p\bullet x\tag{1}\]is defined and is \(A\)-linear.
Moreover, \(\rho\) is an \(A[\x]\)-linear map, because for any \(p\in A[\x]\) and \(q\otimes x\in \iota_!M\),
\[\rho(p(q\otimes_Ax))=\rho((pq)\otimes_Ax)=(pq)\bullet x=(pq)(u)(x)=p(u)(q(u)x)=p\bullet\rho(q\otimes_Ax).\]To avoid confusion, let us write \(M_u\) for \(M\) regarded as an \(A[\x]\)-module. Now viewing \(u\) as a map from \(M_u\) to \(M_u\), the formula
\[u(p\bullet x)=u(p(u)(x))=(\x p)\bullet x=(p\x)\bullet x=p\bullet(u(x))\]holds, so \(u\) becomes an \(A[\x]\)-module endomorphism. Then from the above formula and (1) we know that
\[\rho\circ(\iota_!u)=u\circ\rho\]holds.
Proposition 7 In the above situation, define the \(A[\x]\)-endomorphism \(\psi=\x-\iota_!u\) by the formula
\[\psi(p\otimes_Ax)=(\x p)\otimes_Ax -p\otimes_A u(x).\]Then the sequence of \(A[\x]\)-modules
\[\iota_!M\overset{\psi}{\longrightarrow}\iota_!M\overset{\rho}{\longrightarrow}M_u\longrightarrow 0\]is exact.
Proof
It suffices to show that \(\ker\rho\subseteq \im\psi\). Let \(z\in\ker\rho\) be arbitrary. Then by decomposing \(z\) into a sum of elements of the form \(p\otimes_A x\), and further decomposing the \(p\)’s into linear combinations of \(1,\x,\x^2,\ldots\) and regrouping according to the \(\x^k\)’s, we can write
\[z=\sum_k \x^k\otimes_A x_k,\qquad x_k\in M.\]Then the condition \(z\in\ker\rho\) yields
\[\rho(z)=\sum_k u^k(\x_k)=0.\]Now since \(\sum 1\otimes u^k(x_k)=0\), we obtain
\[z=\sum_k (\x^k\otimes_A x_k-1\otimes_A u^k(x_k))=\sum_k (\x^k-\iota_!u^k))(1\otimes x_k),\]and since in \(\iota_!M=A[\x]\otimes_A M\) the element \(\x\) acts on the \(A[\x]\) factor and \(\iota_!u\) acts on \(M\), these multiplications commute. That is, the above expression can be written as
\[\sum_k (\x-\iota_!u)\circ\left(\sum_{j=0}^{k-1} \x^j (\iota_!u)^{k-j-1}\right),\]which completes the proof.
On the other hand, considering the determinant of \(\psi\), from Corollary 6 we obtain
\[\det (\x-\iota_!u)=\sum_{k=0}^n (-1)^k\tr\left({\bigwedge}^j(\iota_!u)\right)\x^{n-k}.\]Moreover, the matrix representation \([u]_\mathcal{B}^\mathcal{B}\) of \(u\) equals the matrix representation \([\iota_!u]_{\mathcal{B}'}^{\mathcal{B}''}\) of \(\iota_!u\) with respect to the \(A[\x]\)-basis \(\mathcal{B}'=(1\otimes e_i)_{1\leq i\leq n}\) of \(M[\x]\), so the above formula can be written as
\[\det (\x-\iota_!u)=\sum_{k=0}^n (-1)^k\tr\left({\bigwedge}^j(u)\right)\x^{n-k}.\]Definition 8 The polynomial \(\det(\x-\iota_!u)\) defined above is called the characteristic polynomial of \(u\) and is denoted \(\chi_u(\x)\).
Then from the preceding formula, we see that in the characteristic polynomial the coefficient of \(\x^n\) is \(1\), the coefficient of \(\x^{n-1}\) is \(-\tr(u)\), and the constant term is \((-1)^n\det(u)\).
Proposition 9 (Cayley–Hamilton) \(\chi_u(u)=0\).
Proof
We need to show that \(\chi_u(u)(x)=0\) for every \(x\in M\). Now using equation (1), \(\chi_u(u)(x)\) equals \(\rho(\chi_u(\x)\otimes_Ax)\). Observe that
\[\chi_u(\x)\otimes_Ax=\chi_u(\x)(1\otimes_Ax)=\det(\x-\iota_!u)(1\otimes_Ax).\]But thinking of Laplace expansion, for any matrix \(X\) and its cofactor matrix \(Y\) we have \(XY^t=(\det X)I\); hence there exists a suitable \(v\in\End_\rMod{A[\x]}(\iota_!M)\) such that
\[\det(\x-\iota_!u)(1\otimes_Ax)=(\x-\iota_!u)(v(1\otimes_A x)),\]and therefore we obtain the desired result by Proposition 7.
댓글남기기