[UnSA] An Integers Matrix Lemma that sometimes being forgotten

21:53:00   UnSA

I am going to prove the following useful lemma and wrapping the proof of inverse-adjugate formula inside my proof. I hope I put enough details on it.

Lemma
Let $A\in \mathbb{Z}^{n\times n}$ with $\text{det } A \neq 0$. Then there exists a unique matriks $B\in \mathbb{Q}^{n\times n}$ such that $AB=BA=(\text{det } A) I$, and moreover $B$ has integer entries.

Proof:
Suppose that $A=(a_{ij})$, where $a_{ij}\in \mathbb{Z}$. Let $C_{ij}$ denotes the $(i,j)$ cofactor of $A$, that is $C_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column of $A$. Since $A$ has integers entries, $M_{ij}$ is the determinant of a matrix with integers entries, thus $M_{ij}$ is also an integer for all $i,j \in \{1,2,\cdots,n\}$.  So we have $C_{ij} \in \mathbb{Z}$ for all $i,j\in\{1,2,\cdots,n\}$.

Define the matrix $C=(C_{ij})$, that is the $ij$-th entry of $C$ is $C_{ij}$, and let $B=C^{T}$. Since all $C_{ij}$ are integers,  we conclude that $B$ has integer entries.

Next we try to compute $AB$. Denote the $ij$-th entry of $AB$ by $z_{ij}$, therefore by the rule of multiplication of matrix we have
$$z_{ij}=\sum_{k=1}^n a_{ik} C_{jk}$$
If $i=j$ then by cofactor expansion for determinant of $A$ we have
$$z_{ii}=\sum_{k=1}^n a_{ik} C_{ik}= \text{det } A$$
If $i\neq j$, consider the matrix
$$A_{ij}=\left(\begin{array}{ccccc} a_{11} &a_{12} &\cdots &\cdots &a_{1n}\\ \vdots &\vdots &\vdots &\vdots &\vdots \\ a_{i1} &a_{i2} &\cdots &\cdots &a_{in} \\ \vdots &\vdots &\vdots &\vdots &\vdots \\  a_{i1} &a_{i2} &\cdots &\cdots &a_{in} \\ \vdots &\vdots &\vdots &\vdots &\vdots \\ a_{n1} &a_{n2} &\cdots &\cdots &a_{nn}\end{array}\right)$$
here is the explanation of the above matrix, the two identical rows are the $i$-th row and the $j$-th row, and since we have these two identical rows, the determinant of this new matrix is zero. Also we have $C_{jk}(A_{ij})=C_{jk}$, that is the cofactor of $A_{ij}$ amongst $j$-th row is the same as the cofactor of $A$ amongst $j$-th row. Using these facts and cofactor expansion, for $i\neq j$ we have
$$0=\text{det }(A_{ij})=\sum_{k=1}^n a_{ik} C_{jk}(A_{ij})=\sum_{k=1}^n a_{ik} C_{jk}=z_{ij}$$
We conclude that
$$z_{ij} = \left\{ \begin{array}\text{det } A & i= j \\ 0 & i\neq j \end{array}\right.$$

and we have managed to prove that $AB=(\text{det } A) I$, where $B$ has integer entries. Now, since $\text{det } (A) \neq 0$ the inverse matrix $A^{-1}$ exists, thus
$$\begin{align*}AB=(\text{det } A) I &\Rightarrow ABA= (\text{det } A)IA=A (\text{det } A) \\ &\Rightarrow A^{-1}A(BA)=A^{-1}A(\text{det } A) \end{align*}$$
so that we have $BA=(\text{det } A) I$ as well. It remains to prove that $B$ is unique.
Suppose there exists $B_1\neq B$ such that $AB_1 = B_1A= (\text{det } A) I$. Therefore $$AB_1 = (\text{det } A) I = AB$$
multiplying both sides by $A^{-1}$ we have $B_1=B$, a contradiction, thus $B$ is unique and we are done.

One Corollary that follows from this lemma is:
The inverse of the integer matrix with determinant equals $+1$ or $-1$ is again the integer matrix.

[sometimes we just forgot we have this fact]

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