Prove the antisymmetric property of the function $\det$.

Question

Suppose $A$ is a matrix. The $(i, j)$-entry of $A$ is denoted as $[A]_{i,j}$.

Suppose that $a_1$, $a_2$, $\dots$, $a_n$ are $m \times 1$ matrices. Then $[a_1, a_2, \dots, a_n]$ is the $m \times n$ matrix whose $(i, j)$-entry is equal to $[a_j]_{i,1}$. The piece of notation allows one to display a matrix using its columns.

Determinants are defined here.

How does one prove the antisymmetric property?

Suppose that $q$ is a positive integer less than or equal to $n$. Suppose that $p$ is a positive integer less than $q$. Suppose that $a_1$, $a_2$, $\dots$, $a_n$ are $n \times 1$ matrices. Then $$ \det {[a_1, \dots, \overset{\text{column}\,p}{a_q}, \dots, \overset{\text{column}\,q}{a_p}, \dots, a_n]} = -\det {[\dots, a_p, \dots, a_q, \dots]}. $$

Let the proposition $P(n)$ be as follows:

For any $n \times 1$ matrices $a_1$, $a_2$, $\dots$, $a_n$, any positive integer $q$ less than or equal to $n$, any positive integer $p$ less than $q$, $$ \det {[a_1, \dots, \overset{\text{column}\,p}{a_q}, \dots, \overset{\text{column}\,q}{a_p}, \dots, a_n]} = -\det {[\dots, a_p, \dots, a_q, \dots]}. $$

It is easy to check that $P(2)$ is true: $$ \det {\begin{bmatrix} a & c \\ b & d \\ \end{bmatrix}} = ad - bc = -(cb - da) = -\det {\begin{bmatrix} c & a \\ d & b \\ \end{bmatrix}}. $$

Suppose that $P(n-1)$ is true (in which $n \geq 3$). I need to show that $P(n)$ is true.

I know that it suffices to prove that swapping two adjacent columns changes the sign. Suppose that one has proved it. Suppose that $p < q$. It can be shown that swapping columns $p$ and $q$ amounts to swapping two adjacent columns $2(q - p) - 1$ times, which ultimately changes the sign: $$ \begin{aligned} 0\colon & \dots, p, p+1, p+2, \dots, q-1, q, \dots \\ 1\colon & \dots, p+1, p, p+2, \dots, q-1, q, \dots \\ {} & \vdots \\ q-p\colon & \dots, p+1, p+2, \dots, q-1, q, p, \dots \\ q-p+1\colon & \dots, p+1, p+2, \dots, q, q-1, p, \dots \\ {} & \vdots \\ q-p+(q-p+1)\colon & \dots, q, p+1, p+2, \dots, q-1, p, \dots \\ \end{aligned} $$

However, I do not know how to prove the special case.

Answer 1

This proof is direct, which is based on the fact that the determinant of a square matrix can be computed by expansion about any column.

I will prove that $P(n)$ (in the description of the question) is true for $n = 2$, $3$, $\dots$ by mathematical induction.

It is easy to check that $P(2)$ is true.

Suppose that $P(n-1)$ is true (in which $n \geq 3$). I will prove that $P(n)$ is true under the hypothesis.

Choose a positive integer $q$ less than or equal to $n$. Choose a positive integer $q$ less than $p$. Define $$ \begin{aligned} & A = [\dots, a_p, \dots, a_q, \dots], \\ & B = [\dots, a_q, \dots, a_p, \dots], \end{aligned} $$ in which $B$ is obtained from $A$ by interchanging the location of columns $p$ and $q$.

Choose a positive integer $u$ less than or equal to $n$ so that $u \neq p$ and that $u \neq q$. Note that $B(i|u)$ can be seen as the matrix obtained from $A(i|u)$ by interchanging the location of two columns (for $i = 1$, $2$, $\dots$, $n$). By the hypothesis, $\det {(B(i|u))} = -\det {(A(i|u))}$. Note that $[B]_{i,u} = [A]_{i,u}$. Hence $$ \begin{aligned} \det {(B)} = {} & \sum_{i = 1}^{n} {(-1)^{i+u} [B]_{i,u} \det {(B(i|u))}} \\ = {} & \sum_{i = 1}^{n} {(-1)^{i+u} [A]_{i,u} (-\det {(A(i|u))})} \\ = {} & {-\sum_{i = 1}^{n} {(-1)^{i+u} [A]_{i,u} \det {(A(i|u))}}} \\ = {} & {-\det {(A)}}. \end{aligned} $$

Hence by mathematical induction, $P(n)$ is true for $n = 2$, $3$, $\dots$.

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Note that the proof works with any two columns, which are not necessarily adjacent.

Answer 2

The proof uses the fact that the determinant of a square matrix can be computed by expansion about the first two columns: $$ \begin{aligned} \det {(A)} = \sum_{1 \leq i < k \leq n} {\det { \begin{bmatrix} [A]_{i,1} & [A]_{i,2} \\ [A]_{k,1} & [A]_{k,2} \\ \end{bmatrix} } (-1)^{i+k+1+2} \det {(A(i,k|1,2))}}. \end{aligned} $$ Here is a proof, which just uses the definition.

I will prove that $P(n)$ (in the description of the question) is true for $n = 2$, $3$, $\dots$ by mathematical induction.

It is easy to check that $P(2)$ is true.

Suppose that $P(n-1)$ is true (in which $n \geq 3$). I will prove that $P(n)$ is true under the hypothesis.

Let $A = [a_1, a_2, \dots, a_n]$ be an $n \times n$ matrix.

As is mentioned in the description of the question, it suffices to prove that swapping two adjacent columns changes the sign.

Let $p$ be a positive integer less than $n$. Put $q = p + 1$. Let $B$ be the matrix obtained from $A$ by interchanging the location of columns $p$ and $q$. Then $[A]_{i,p} = [B]_{i,p+1}$, $[A]_{i+1,p} = [B]_{i,p}$, and for $\ell \neq p, p+1$, $[A]_{i,\ell} = [B]_{i,\ell}$.

Suppose first that $p > 1$. Then $B(i|1)$ can be seen as the square matrix obtained from $A(i|1)$ by interchanging the location of columns $p-1$, $p$. By the hypothesis, $\det {(B(i|1))} = -\det {(A(i|1))}$. Hence $$ \begin{aligned} \det {(B)} = {} & \sum_{i = 1}^{n} {(-1)^{i+1} [B]_{i,1} \det {(B(i|1))}} \\ = {} & \sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} (-\det {(A(i|1))})} \\ = {} & {-\sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}}} \\ = {} & {-\det {(A)}}. \end{aligned} $$

Suppose then that $p = 1$. The definition does not work as it just did; it will work differently. Note that $A(i,k|1,2) = B(i,k|1,2)$ for distinct integers $i$, $k$. Hence $$ \begin{aligned} \det {(B)} = {} & \sum_{1 \leq i < k \leq n} {\det { \begin{bmatrix} [B]_{i,1} & [B]_{i,2} \\ [B]_{k,1} & [B]_{k,2} \\ \end{bmatrix} } (-1)^{i+k+1+2} \det {(B(i,k|1,2))}} \\ = {} & \sum_{1 \leq i < k \leq n} {\det { \begin{bmatrix} [A]_{i,2} & [A]_{i,1} \\ [A]_{k,2} & [A]_{k,1} \\ \end{bmatrix} } (-1)^{i+k+1+2} \det {(A(i,k|1,2))}} \\ = {} & \sum_{1 \leq i < k \leq n} {\left(-\det { \begin{bmatrix} [A]_{i,1} & [A]_{i,2} \\ [A]_{k,1} & [A]_{k,2} \\ \end{bmatrix} }\right) (-1)^{i+k+1+2} \det {(A(i,k|1,2))}} \\ = {} & {-\det {(A)}}. \end{aligned} $$

Hence by mathematical induction, $P(n)$ is true for $n = 2$, $3$, $\dots$.

Answer 3

The proof is based on the fact that the determinant of a square matrix can be computed by expansion about the first row.

I will prove that $P(n)$ (in the description of the question) is true for $n = 2$, $3$, $\dots$ by mathematical induction.

It is easy to check that $P(2)$ is true.

Suppose that $P(n-1)$ is true (in which $n \geq 3$). I will prove that $P(n)$ is true under the hypothesis.

Let $A = [a_1, a_2, \dots, a_n]$ be an $n \times n$ matrix.

As is mentioned in the description of the question, it suffices to prove that swapping two adjacent columns changes the sign.

Let $p$ be a positive integer less than $n$. Put $q = p + 1$. Let $B$ be the matrix obtained from $A$ by interchanging the location of columns $p$ and $q$. Then $[A]_{i,p} = [B]_{i,p+1}$, $[A]_{i+1,p} = [B]_{i,p}$, and for $\ell \neq p, p+1$, $[A]_{i,\ell} = [B]_{i,\ell}$.

Note that $A(1|p) = B(1|p+1)$ and that $A(1|p+1) = B(1|p)$. Note that for $\ell \neq p, p+1$, $B(1|\ell)$ can be seen as the square matrix obtained from $A(1|\ell)$ by interchanging the location of two (adjacent) columns. By the hypothesis, $\det {(B(1|\ell))} = -\det {(A(1|\ell))}$. Hence $$ \begin{aligned} & \det {(B)} \\ = {} & \sum_{k = 1}^{n} {(-1)^{1 + k} [B]_{1,k} \det {(B(1|k))}} \\ = {} & \hphantom{{} + {}} (-1)^{1 + p} [B]_{1,p} \det {(B(1|p))} + (-1)^{1 + p+1} [B]_{1,p+1} \det {(B(1|p+1))} \\ & + \sum_{\substack{1 \leq k \leq n \\k \neq p, p+1}} {(-1)^{1 + k} [B]_{1,k} \det {(B(1|k))}} \\ = {} & \hphantom{{} + {}} (-1)^{1 + p} [A]_{1,p+1} \det {(A(1|p+1))} + (-1)^{1 + p+1} [A]_{1,p} \det {(A(1|p))} \\ & + \sum_{\substack{1 \leq k \leq n \\k \neq p, p+1}} {(-1)^{1 + k} [A]_{1,k} (-\det {(A(1|k))})} \\ = {} & \hphantom{{} + {}} {-(-1)^{1 + p+1}} [A]_{1,p+1} \det {(A(1|p+1))} - (-1)^{1 + p} [A]_{1,p} \det {(A(1|p))} \\ & - \sum_{\substack{1 \leq k \leq n \\k \neq p, p+1}} {(-1)^{1 + k} [A]_{1,k} \det {(A(1|k))}} \\ = {} & {-\sum_{k = 1}^{n} {(-1)^{1 + k} [A]_{1,k} \det {(A(1|k))}}} \\ = {} & {-\det {(A)}}. \end{aligned} $$

Hence by mathematical induction, $P(n)$ is true for $n = 2$, $3$, $\dots$.