NOTE & BLOG

Definition ($k$-th Order Minor): In an $n$-th order determinant $D$, a $k$-th order minor $M$ is a $k \times k$ determinant formed by selecting any $k$ rows and $k$ columns from $D$ ($k \le n$), and taking the $k^2$ elements at the intersections of these rows and columns in their original order.

When $k < n$, after removing these $k$ rows and $k$ columns from $D$, the remaining $(n-k)$-th order determinant $M'$ formed by the leftover elements is called the cofactor of the $k$-th order minor $M$.

Example:

$$D=\begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55} \end{vmatrix}, \\ M=\begin{vmatrix} a_{12} & a_{13} & a_{15} \\ a_{22} & a_{23} & a_{25} \\ a_{42} & a_{43} & a_{45} \end{vmatrix}, \\ M'=\begin{vmatrix} a_{31} & a_{34} \\ a_{51} & a_{54} \end{vmatrix}$$

Note: In the above example, the algebraic cofactor of $M$ is

$$(-1)^{(1+2+4)+(2+3+5)}M'=-M'$$

Definition (Algebraic Cofactor): Let $M$ be a $k$-th order minor of $D$ with row and column indices $i_{1},i_{2},\cdots,i_{k}; \ j_{1},j_{2},\cdots,j_{k}$, respectively. The cofactor $M'$ of $M$ is multiplied by the sign $(-1)^{(i_{1}+i_{2}+\cdots+i_{k})+(j_{1}+j_{2}+\cdots+j_{k})}$ to form the algebraic cofactor of $M$.

Lemma: Every term in the product of any minor $M$ of determinant $D$ and its algebraic cofactor $A$ corresponds to a term in the expansion of determinant $D$, with the same sign.

Proof: First, consider the case where $M$ is located in the upper-left corner of determinant $D$:

$$\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1k} & |& a_{1,k+1} & \cdots & a_{1n}\\ \vdots & \vdots & M & \cdots & | & \vdots & &\vdots \\ a_{k1} & a_{k2} &\cdots & a_{kk}& | & a_{k,k+1} & \cdots & a_{kn}\\ -&- &- &- &-|-&- &- &- \\ a_{k+1,1}& a_{k+1,2} &\cdots & a_{k+1,k} & |& a_{k+1,k+1} & \cdots & a_{k+1,n}\\ \vdots & \vdots & & \vdots& |& \vdots & M' & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nk} & |& a_{n,k+1} & \cdots & a_{nn}\end{vmatrix}$$

In this case, the algebraic cofactor of $M$ is:

$$A=(-1)^{(1+2+\cdots +k)+(1+2+\cdots+k)}M'=M'$$

Each term in $M$ can be written as $a_{1\alpha_{1}}a_{2\alpha_{2}}\cdots a_{k\alpha_{k}}$, where $\alpha_{1}\alpha_{2}\cdots\alpha_{k}$ is a permutation of $1,2,\cdots,k$. The sign of this term is $(-1)^{\tau(\alpha_1 \alpha_2\cdots\alpha_k)}$.

Each term in $M'$ can be written as $a_{k+1,\beta_{k+1}}a_{k+2,\beta_{k+2}}\cdots a_{n\beta_{n}}$, where $\beta_{k+1}\beta_{k+2}\cdots\beta_{n}$ is a permutation of $k+1,k+2,\cdots,n$, and its sign is $(-1)^{\tau((\beta_{k+1}-k)(\beta_{k+2}-k)\cdots(\beta_{n}-k))}$.

The product of these two terms is $a_{1\alpha_{1}}a_{2\alpha_{2}}\cdots a_{k\alpha_{k}}a_{k+1,\beta_{k+1}}\cdots a_{n\beta_{n}}$, and the combined sign is:

$$(-1)^{\tau(\alpha_{1}\alpha_{2}\cdots\alpha_{k})+\tau((\beta_{k+1}-k)(\beta_{k+2}-k)\cdots(\beta_{n}-k))}= (-1)^{\tau(\alpha_{1}\alpha_{2}\cdots \alpha_{k}\beta_{k+1}\cdots \beta_{n}) }$$

Thus, this product is a term in the expansion of determinant $D$ with the same sign.

Now, we will prove the general case.

Let minor $M$ be located in rows $i_1, i_2, \cdots, i_k$ and columns $j_1, j_2, \cdots, j_k$ of $D$, where $i_1 < i_2 < \cdots < i_k$, and $j_1 < j_2 < \cdots < j_k$.

We rearrange the rows and columns of determinant $D$ such that $M$ is moved to the top-left corner.

First, swap the $i_1$-th row with rows $i_1-1, i_1-2, \cdots, 2, 1$, which takes $i_1-1$ swaps to move the $i_1$-th row to the first row. Then, swap the $i_2$-th row $i_2-2$ times to move it to the second row, and so on. In total, this requires

$$(i_1-1)+(i_2-2)+\cdots+(i_k-k)=(i_1+i_2+\cdots+i_k)-(1+2+\cdots+k)$$

row swaps to move rows $i_1, i_2, \cdots, i_k$ to rows $1, 2, \cdots, k$.

Using similar column swaps, we move the columns of $M$ to columns $1, 2, \cdots, k$, requiring

$$(j_1-1)+(j_2-2)+\cdots+(j_k-k)=(j_1+j_2+\cdots+j_k)-(1+2+\cdots+k)$$

column swaps.

Let $D_1$ denote the new determinant after these transformations. We have:

$$D_1 = (-1)^{(i_1+i_2+\cdots+i_k)-(1+2+\cdots+k)+(j_1+j_2+\cdots+j_k)-(1+2+\cdots+k)}D \\ = (-1)^{i_1+i_2+\cdots+i_k+j_1+j_2+\cdots+j_k}D$$

Thus, the expansion of $D_1$ contains the same terms as that of $D$, but with each term having a sign difference of $(-1)^{i_1+i_2+\cdots+i_k+j_1+j_2+\cdots+j_k}$.

Now, $M$ is located in the top-left corner of $D_1$, and the cofactor and algebraic cofactor of $M$ in $D_1$ are both $M'$. Therefore, each term in the product $MM'$ corresponds to a term in $D_1$ with the same sign.

The algebraic cofactor $A$ of $M$ in $D$ gives $MA = (-1)^{i_1+i_2+\cdots+i_k+j_1+j_2+\cdots+j_k}MM'$, which is a term in $D_1$ with a sign difference of $(-1)^{i_1+i_2+\cdots+i_k+j_1+j_2+\cdots+j_k}$. Hence, $MA$ is a term in $D$ with the same sign.

Theorem (Laplace's Theorem): For any $k$ ($1 \le k \le n-1$) rows chosen from determinant $D$, the sum of the products of all $k$-th order minors formed from these rows and their respective algebraic cofactors equals the determinant $D$.

Proof: Let the minors formed from the chosen $k$ rows of $D$ be $M_1, M_2, \cdots, M_t$, with algebraic cofactors $A_1, A_2, \cdots, A_t$. The theorem claims that:

$$D = M_1A_1 + M_2A_2 + \cdots + M_tA_t$$

where

$$t = \text{C}_{n}^{k} = \dfrac{n!}{k!(n-k)!}$$

By the lemma, each term in $M_iA_i$ corresponds to a term in $D$ with the same sign, and there are no common terms between $M_iA_i$ and $M_jA_j$ ($i \ne j$).

The left-hand side of the equation contains $n!$ terms, and the right-hand side contains

$\sum\limits_{i=1}^{t}k!(n-k)! = k!(n-k)!\dfrac{n!}{k!(n-k)!} = n!$

terms. Hence, the theorem is proven.

Example: In the determinant $D=\begin{vmatrix} 1 & 2 & 1 &4 \\ 0 & -1 & 2 & 1\\ 1 & 0 & 1 & 3\\ 0 & 1& 3 & 1\end{vmatrix}$, selecting the first two rows yields six minors:

$$M_{1}= \begin{vmatrix} 1& 2\\ 0 &-1 \end{vmatrix}=-1, \quad M_{2}= \begin{vmatrix} 1 &1 \\ 0& 2\end{vmatrix}=2, \quad M_{3}= \begin{vmatrix} 1 & 4\\ 0 &1 \end{vmatrix}=1,$$ $$M_{4}= \begin{vmatrix} 2& 1\\ -1&2 \end{vmatrix}=5, \quad M_{5}= \begin{vmatrix} 2& 4\\ -1 &1 \end{vmatrix}=6, \quad M_{6}= \begin{vmatrix} 1& 4\\ 2&1 \end{vmatrix}=-7.$$

Their algebraic cofactors are:

$$A_1=-8, \quad A_2=3, \quad A_3=-1, \quad A_4=1, \quad A_5=-3, \quad A_6=1 \Rightarrow D=-7.$$

Theorem: The product of two $n$-order determinants

$$D_{1}=\begin{vmatrix} a_{11} & a_{12} & \cdots &a_{1n} \\ a_{21} &a_{22} &\cdots & a_{2n}\\ \vdots & \vdots & &\vdots \\ a_{n1} & a_{n2} &\cdots & a_{nn}\end{vmatrix}, \\ D_{2}=\begin{vmatrix} b_{11} & b_{12} & \cdots &b_{1n} \\ b_{21} &b_{22} &\cdots & b_{2n}\\ \vdots & \vdots & &\vdots \\ b_{n1} & b_{n2} &\cdots & b_{nn}\end{vmatrix}$$

is equal to a determinant

$$C = \begin{vmatrix} c_{11} &c_{12} & \cdots &c_{1n} \\ c_{21} &c_{22} &\cdots & c_{2n}\\ \vdots & \vdots & &\vdots \\ c_{n1} & c_{n2} &\cdots & c_{nn}\end{vmatrix}$$

where each element $c_{ij}$ is the sum of the products of elements from the $i$-th row of $D_1$ and the corresponding elements from the $j$-th row of $D_2$,

i.e.

$$c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}.$$

Proof: Construct a $2n$-order determinant

$$D = \begin{vmatrix} a_{11} & a_{12} & \cdots &a_{1n} &0&0&0&0\\ a_{21} &a_{22} &\cdots & a_{2n}&0&0&0&0\\ \vdots & \vdots & &\vdots &0&0&0&0\\ a_{n1} & a_{n2} &\cdots & a_{nn} &0&0&0&0\\ -1&0&\cdots &0&b_{11}&b_{12}&\cdots&b_{1n}\\ 0&-1&\dotso&0&b_{21}&b_{22}&\cdots&b_{2n}\\ \vdots & \vdots & &\vdots&\vdots & \vdots &&\vdots \\ 0&0&\cdots&-1 &b_{n1}&b_{n2}&\cdots &b_{nn} \end{vmatrix}$$

According to Laplace's theorem, expanding $D$ by the first $n$ rows, all of the minors formed by the first $n$ rows and other elements will be zero except for the minor at the top left, so

$$D = \begin{vmatrix} a_{11} & a_{12} & \cdots &a_{1n} \\ a_{21} &a_{22} &\cdots & a_{2n}\\ \vdots & \vdots & &\vdots \\ a_{n1} & a_{n2} &\cdots & a_{nn}\end{vmatrix}\cdot \begin{vmatrix} b_{11} & b_{12} & \cdots &b_{1n} \\ b_{21} &b_{22} &\cdots & b_{2n}\\ \vdots & \vdots & &\vdots \\ b_{n1} & b_{n2} &\cdots & b_{nn}\end{vmatrix}=D_1D_2.$$

Next, we show that $D = C$. Perform elementary row operations on $D$: add the $a_{11}$-multiple of the $(n+1)$-th row, the $a_{12}$-multiple of the $(n+2)$-th row, $\dots$, and the $a_{1n}$-multiple of the $(2n)$-th row to the first row to get

$$D = \begin{vmatrix} 0&0&\cdots&0&c_{11}&c_{12}&\cdots&c_{1n}\\ a_{21} &a_{22} &\cdots & a_{2n}&0&0&0&0\\ \vdots & \vdots & &\vdots &0&0&0&0\\ a_{n1} & a_{n2} &\cdots & a_{nn} &0&0&0&0\\ -1&0&\cdots &0&b_{11}&b_{12}&\cdots&b_{1n}\\ 0&-1&\dotso&0&b_{21}&b_{22}&\cdots&b_{2n}\\ \vdots & \vdots & &\vdots&\vdots & \vdots &&\vdots \\ 0&0&\cdots&-1 &b_{n1}&b_{n2}&\cdots &b_{nn} \end{vmatrix}.$$

Similarly, successively add the $a_{k1}$-multiple of the $(n+1)$-th row, the $a_{k2}$-multiple of the $(n+2)$-th row, $\dots$, and the $a_{kn}$-multiple of the $(2n)$-th row to the $k$-th row to obtain

$$D = \begin{vmatrix} 0&0&\cdots&0& a_{11} & a_{12} & \cdots &a_{1n} \\ 0&0&\cdots&0&a_{21} &a_{22} &\cdots & a_{2n}\\ \vdots &\vdots & &\vdots &\vdots & \vdots & &\vdots \\ 0&0&\cdots&0&a_{n1} & a_{n2} &\cdots & a_{nn} \\ -1&0&\cdots &0&b_{11}&b_{12}&\cdots&b_{1n}\\ 0&-1&\dotso&0&b_{21}&b_{22}&\cdots&b_{2n}\\ \vdots & \vdots & &\vdots&\vdots & \vdots &&\vdots \\ 0&0&\cdots&-1 &b_{n1}&b_{n2}&\cdots &b_{nn} \end{vmatrix}.$$

In this determinant, the first $n$ rows contain only one non-zero $n$-order minor. Therefore, by Laplace's theorem,

$$D = \begin{vmatrix} c_{11} &c_{12} & \cdots &c_{1n} \\ c_{21} &c_{22} &\cdots & c_{2n}\\ \vdots & \vdots & &\vdots \\ c_{n1} & c_{n2} &\cdots & c_{nn}\end{vmatrix} ~\\ ~\\ \ast (-1)^{(1+2+\cdots+n)+(n+1+n+2+\cdots +2n)} ~\\ ~\\ \ast \begin{vmatrix} -1 & 0 & \cdots & 0\\ 0 & -1 & \cdots & 0\\ \vdots & \vdots & &\vdots \\ 0& 0 & \cdots &-1 \end{vmatrix} = C.$$