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

Question

Determinants are defined here.

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.

How does one prove the multilinear property?

Suppose that $j$ is a positive integer less than or equal to $n$. Suppose that $a_1$, $\dots$, $a_{j-1}$, $a_{j+1}$, $\dots$, $a_n$ are $n \times 1$ matrices. Suppose that $x$, $y$ are $n \times 1$ matrices. Suppose that $s$, $t$ are numbers. Then $$ \begin{aligned} & \det {[a_1, \dots, a_{j-1}, \overset{\text{column}\,j}{sx + ty}, a_{j+1}, \dots, a_n]} \\ = {} & s \det {[a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n]} + t \det {[a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n]}. \end{aligned} $$

Answer 1

The proof is indirect, which is based on no more than the definition of the determinant of a square matrix.

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

For any positive integer $j$ less than or equal to $n$, for any $n \times 1$ matrices $a_1$, $\dots$, $a_{j-1}$, $a_{j+1}$, $\dots$, $a_n$, for any $n \times 1$ matrices $x$, $y$ and for any numbers $s$, $t$, $$ \begin{aligned} & \det {[a_1, \dots, a_{j-1}, \overset{\text{column}\,j}{sx + ty}, a_{j+1}, \dots, a_n]} \\ = {} & s \det {[a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n]} + t \det {[a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n]}. \end{aligned} $$

It will be proved that $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$ by mathematical induction.

$P(1)$ is true, which is left as an exercise.

Suppose that $P(n-1)$ is true. It will be proved that $P(n)$ is true under the hypothesis.

Define $$ \begin{aligned} & A = [a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n], \\ & B = [a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n], \\ & C = [a_1, \dots, a_{j-1}, sx + ty, a_{j+1}, \dots, a_n]. \end{aligned} $$ If $k \neq j$, then $[A]_{i,k} = [B]_{i,k} = [C]_{i,k}$. Hence $A(i|j) = B(i|j) = C(i|j)$. Note that $[C]_{i,j} = s[A]_{i,j} + t[B]_{i,j}$.

Suppose first that $j = 1$. Then $$ \begin{aligned} & \det {(C)} \\ = {} & \sum_{i = 1}^{n} { (-1)^{i+1} [C]_{i,1} \det {(C(i|1))} } \\ = {} & \sum_{i = 1}^{n} { (-1)^{i+1} (s[A]_{i,1} + t[B]_{i,1}) \det {(C(i|1))} } \\ = {} & s\sum_{i = 1}^{n} { (-1)^{i+1} [A]_{i,1} \det {(C(i|1))} } + t\sum_{i = 1}^{n} { (-1)^{i+1} [B]_{i,1} \det {(C(i|1))} } \\ = {} & s\sum_{i = 1}^{n} { (-1)^{i+1} [A]_{i,1} \det {(A(i|1))} } + t\sum_{i = 1}^{n} { (-1)^{i+1} [B]_{i,1} \det {(B(i|1))} } \\ = {} & s \det{(A)} + t \det{(B)}. \end{aligned} $$

Suppose then that $j > 1$. Note that for $\ell \neq j - 1$, column $\ell$ of $A(i|1)$, that of $B(i|1)$ and that of $C(i|1)$ are equal. Suppose that column $j-1$ of $A(i|1)$, that of $B(i|1)$ and that of $C(i|1)$ are $p$, $q$ and $r$, respectively. Then $r = sp + tq$. By the hypothesis, one has $$ \det {(C(i|1))} = s\det {(A(i|1))} + t\det {(B(i|1))}. $$ Hence $$ \begin{align*} & \det {(C)} \\ = {} & \sum_{i = 1}^{n} {(-1)^{i+1} [C]_{i,1} \det {(C(i|1))}} \\ = {} & \sum_{i = 1}^{n} {(-1)^{i+1} [C]_{i,1} (s\det {(A(i|1))} + t\det {(B(i|1))})} \\ = {} & s \sum_{i = 1}^{n} {(-1)^{i+1} [C]_{i,1} \det {(A(i|1))}} + t \sum_{i = 1}^{n} {(-1)^{i+1} [C]_{i,1} \det {(B(i|1))}} \\ = {} & s \sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}} + t \sum_{i = 1}^{n} {(-1)^{i+1} [B]_{i,1} \det {(B(i|1))}} \\ = {} & s \det {(A)} + t \det {(B)}. \end{align*} $$

Hence by mathematical induction, $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$.

Answer 2

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.

Define $$ \begin{aligned} & A = [a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n], \\ & B = [a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n], \\ & C = [a_1, \dots, a_{j-1}, sx + ty, a_{j+1}, \dots, a_n]. \end{aligned} $$ If $k \neq j$, then $[A]_{i,k} = [B]_{i,k} = [C]_{i,k}$. Hence $A(i|j) = B(i|j) = C(i|j)$. Since $[C]_{i,j} = s[A]_{i,j} + t[B]_{i,j}$, one has $$ \begin{aligned} & \det {(C)} \\ = {} & \sum_{i = 1}^{n} { (-1)^{i+j} [C]_{i,j} \det {(C(i|j))} } \\ = {} & \sum_{i = 1}^{n} { (-1)^{i+j} (s[A]_{i,j} + t[B]_{i,j}) \det {(C(i|j))} } \\ = {} & s\sum_{i = 1}^{n} { (-1)^{i+j} [A]_{i,j} \det {(C(i|j))} } + t\sum_{i = 1}^{n} { (-1)^{i+j} [B]_{i,j} \det {(C(i|j))} } \\ = {} & s\sum_{i = 1}^{n} { (-1)^{i+j} [A]_{i,j} \det {(A(i|j))} } + t\sum_{i = 1}^{n} { (-1)^{i+j} [B]_{i,j} \det {(B(i|j))} } \\ = {} & s \det{(A)} + t \det{(B)}. \end{aligned} $$

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. It is, however, surprisingly easier, compared with the one that is based on the definition (expansion about the first column).

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

For any positive integer $j$ less than or equal to $n$, for any $n \times 1$ matrices $a_1$, $\dots$, $a_{j-1}$, $a_{j+1}$, $\dots$, $a_n$, for any $n \times 1$ matrices $x$, $y$ and for any numbers $s$, $t$, $$ \begin{aligned} & \det {[a_1, \dots, a_{j-1}, \overset{\text{column}\,j}{sx + ty}, a_{j+1}, \dots, a_n]} \\ = {} & s \det {[a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n]} + t \det {[a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n]}. \end{aligned} $$

It will be proved that $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$ by mathematical induction.

$P(1)$ is true, which is left as an exercise.

Suppose that $P(n-1)$ is true. It will be proved that $P(n)$ is true under the hypothesis.

Define $$ \begin{aligned} & A = [a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n], \\ & B = [a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n], \\ & C = [a_1, \dots, a_{j-1}, sx + ty, a_{j+1}, \dots, a_n]. \end{aligned} $$ If $k \neq j$, then $[A]_{i,k} = [B]_{i,k} = [C]_{i,k}$. Hence $A(i|j) = B(i|j) = C(i|j)$. Note that $[C]_{i,j} = s[A]_{i,j} + t[B]_{i,j}$.

Let $k$ be a positive integer less than or equal to $n$, and unequal to $j$. Let $\ell$ be $j$ if $k < j$, or $j-1$ if $k > j$. Suppose that column $\ell$ of $A(1|k)$, that of $B(1|k)$ and that of $C(1|k)$ are $p$, $q$, $r$, respectively. Then $r = sp + tq$. By the hypothesis, one has $$ \det {(C(1|k))} = s \det {(A(1|k))} + t \det {(B(1|k))}. $$ Note that $$ [C]_{1,j} = s[A]_{1,j} + t[B]_{1,j}. $$ Hence $$ \begin{aligned} & \det {(C)} \\ = {} & \sum_{k = 1}^{n} {(-1)^{1 + k} [C]_{1,k} \det {(C(1|k))}} \\ = {} & (-1)^{1 + j} [C]_{1,j} \det {(C(1|j))} + \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [C]_{1,k} \det {(C(1|k))}} \\ = {} & \hphantom{{} + {}} (-1)^{1 + j} (s[A]_{1,j} + t[B]_{1,j}) \det {(C(1|j))} \\ & + \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [C]_{1,k} (s \det {(A(1|k))} + t \det {(B(1|k))})} \\ = {} & \hphantom{{} + {}} s (-1)^{1 + j} [A]_{1,j} \det {(C(1|j))} + t (-1)^{1 + j} [B]_{1,j} \det {(C(1|j))} \\ & + s \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [C]_{1,k} \det {(A(1|k))}} \\ & + t \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [C]_{1,k} \det {(B(1|k))}} \\ = {} & \hphantom{{} + {}} s (-1)^{1 + j} [A]_{1,j} \det {(A(1|j))} + t (-1)^{1 + j} [B]_{1,j} \det {(B(1|j))} \\ & + s \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [A]_{1,k} \det {(A(1|k))}} \\ & + t \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [B]_{1,k} \det {(B(1|k))}} \\ = {} & \hphantom{{} + {}} s (-1)^{1 + j} [A]_{1,j} \det {(A(1|j))} + s \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [A]_{1,k} \det {(A(1|k))}} \\ & + t (-1)^{1 + j} [B]_{1,j} \det {(B(1|j))} + t \sum_{\substack{1 \leq k \leq n \\k \neq j}} {(-1)^{1 + k} [B]_{1,k} \det {(B(1|k))}} \\ = {} & s \sum_{k = 1}^{n} {(-1)^{1 + k} [A]_{1,k} \det {(A(1|k))}} + t \sum_{k = 1}^{n} {(-1)^{1 + k} [B]_{1,k} \det {(B(1|k))}} \\ = {} & s \det {(A)} + t \det {(B)}. \end{aligned} $$

Hence by mathematical induction, $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$.