Exercise 4.5.16

Fixed an alternating $n$-linear function $\delta$. Let $k$ be the value of $\delta (I)$. We want to chaim that

$$\delta (M)=k\det <!--\; FUNCTION\; APPLICATION\; -->(M).$$

First we know that if $M$ has rank less than $n$, then $\delta (M)=0=\det <!--\; FUNCTION\; APPLICATION\; -->(M)$ by Corollary 2 after Theorem 4.10. So the identity holds. Second if $M$ is full-rank, we can write $M=E_s\cdots E_2{E}_{1}I$ as product of elementary matrices and identity matrix $I$. And it’s lucky now I can copy and paste the text in Exercise 4.5.12.

This time we will claim that

$$\delta (\mathit{EA})=\det <!--\; FUNCTION\; APPLICATION\; -->(E)\delta (A)$$

for all elementary matrix $E$ and all matrix $A$. First, if $E$ is the elementary matrix of type 1 meaning interchangine the $i$-th and the $j$-th rows, we have $\mathit{EM}$ is the matrix obtained from $A$ by interchanging the $i$-th and the $j$-th rows. By Theorem 4.10(a) we know that

$$\delta (\mathit{EM})=-\delta (A)=\det <!--\; FUNCTION\; APPLICATION\; -->(E)\delta (A).$$

Second, if $E$ is the elementary matrix of type 2 meaning multiplying the $i$-th row by a scalar $k$, we have $\mathit{EM}$ is the matrix obtained from $M$ by multiplying the $i$-th row by scalar $k$. Since the function $\delta$ is $n$-linear, we have

$$\delta (\mathit{EM})=\mathit{k\delta }(A)=\det <!--\; FUNCTION\; APPLICATION\; -->(E)\delta (A).$$

Finally, if $E$ is the elementary matrix of type 3 meaning adding $k$ times the $i$-th row to the $j$-th row, we have $\mathit{EM}$ is the matrix obtained from $M$ by adding $k$ times the $i$-th row to the $j$-th row. By Corollary 1 after Theorem 4.10, we have

$$\delta (\mathit{EM})=\delta (A)=\det <!--\; FUNCTION\; APPLICATION\; -->(E)\delta (A).$$

This completes the proof since

$$\delta (M)=\delta (E_s\cdots E_2{E}_{1}I)=\det <!--\; FUNCTION\; APPLICATION\; -->(E_s)\cdots \det <!--\; FUNCTION\; APPLICATION\; -->(E_2)\det <!--\; FUNCTION\; APPLICATION\; -->(E_1)\delta (I)$$

$$=k\det <!--\; FUNCTION\; APPLICATION\; -->(E_s)\cdots \det <!--\; FUNCTION\; APPLICATION\; -->(E_2)\det <!--\; FUNCTION\; APPLICATION\; -->(E_1)=k\det <!--\; FUNCTION\; APPLICATION\; -->(M).$$