12. 행렬식의 성질Properties of Determinants

Proof. (1) We claim $\det(A) = A_{11}A_{22}\cdots A_{nn}$, the product of the diagonal entries. We prove this by induction on $n$.

If $n=1$, then $A=(A_{11})$ and $\det(A)=A_{11}$, so the claim holds.

Suppose the claim holds for triangular matrices of size $n-1$. Suppose first that $A$ is upper triangular, so $A_{i1}=0$ for all $i>1$. Expanding along the first column, $$\det(A) = A_{11}\tilde A_{11},$$ where $\tilde A_{11} = (-1)^{1+1}\det(A(1|1)) = \det(A(1|1))$ and $A(1|1)$ is the $(n-1)\times(n-1)$ matrix obtained by deleting row 1 and column 1 of $A$. Since $A$ is upper triangular, so is $A(1|1)$, with diagonal entries $A_{22},\ldots,A_{nn}$. By the induction hypothesis, $\det(A(1|1)) = A_{22}\cdots A_{nn}$, hence $\det(A) = A_{11}A_{22}\cdots A_{nn}$.

If $A$ is lower triangular, then $A_{1j}=0$ for all $j>1$, and expanding along the first row gives the same conclusion by an identical argument.

(2) By a basic property of determinants, $A$ is invertible if and only if $\det(A)\neq 0$. By (1), $\det(A) = A_{11}A_{22}\cdots A_{nn}$, and since $F$ is a field, this product is nonzero if and only if $A_{ii}\neq 0$ for every $i=1,\ldots,n$. Therefore $A$ is invertible if and only if all of its diagonal entries are nonzero.

Proof. (1) Suppose $A^k = O$ for some $k\geq 1$. By multiplicativity of the determinant, $$\det(A)^k = \det(A^k) = \det(O) = 0.$$ Since $F$ is a field, it has no zero divisors, so $\det(A)^k=0$ implies $\det(A)=0$.

(2) Suppose $A^t=-A$, $F=\mathbb C$, and $n$ is odd. Since the determinant is unchanged by transposition, $\det(A^t)=\det(A)$. Also, $-A$ is obtained from $A$ by multiplying each row by the scalar $-1$, so $$\det(-A) = (-1)^n\det(A).$$ Since $n$ is odd, $(-1)^n=-1$, and hence $$\det(A) = \det(A^t) = \det(-A) = -\det(A).$$ Thus $2\det(A)=0$, and since $\mathbb C$ has characteristic $0$, $\det(A)=0$.

(3) Suppose $F=\mathbb R$ and $AA^t=I$. Then $$\det(A)^2 = \det(A)\det(A^t) = \det(AA^t) = \det(I) = 1,$$ so $\det(A) = \pm 1$.

(4) Suppose $F=\mathbb C$ and $AA^*=I$. We first show $\det(A^*) = \overline{\det(A)}$. Since $A^* = \overline{A^t}$ (conjugate each entry, then transpose — equivalently, transpose then conjugate), and the determinant is a polynomial (a signed sum of products) in the entries, conjugating every entry conjugates the determinant: $$\det(\overline A) = \overline{\det(A)},$$ where $\overline A$ denotes the matrix with each entry of $A$ replaced by its complex conjugate. Since transposition does not change the determinant, $$\det(A^*) = \det\!\left(\overline{A}^{\,t}\right) = \det(\overline A) = \overline{\det(A)}.$$ Therefore $$|\det(A)|^2 = \det(A)\overline{\det(A)} = \det(A)\det(A^*) = \det(AA^*) = \det(I) = 1,$$ and since $|\det(A)|\geq 0$, we conclude $|\det(A)|=1$.

Proof. Since $A$ and $B$ are similar, there exists an invertible matrix $Q\in M_{n\times n}(F)$ such that $B=Q^{-1}AQ$. By multiplicativity of the determinant, $$\det(B) = \det(Q^{-1})\det(A)\det(Q).$$ Since $QQ^{-1}=I$, we have $\det(Q)\det(Q^{-1})=\det(I)=1$, i.e., $\det(Q^{-1})=\det(Q)^{-1}$. As multiplication in $F$ is commutative, $$\det(B) = \det(Q)^{-1}\det(A)\det(Q) = \det(A).$$

Proof. By multiplicativity of the determinant, $$\det(A)\det(B) = \det(AB) = \det(-BA) = (-1)^n\det(BA) = (-1)^n\det(B)\det(A).$$ Since $n$ is odd, $(-1)^n=-1$, and since multiplication in $\mathbb R$ is commutative, $$\det(A)\det(B) = -\det(A)\det(B).$$ Hence $2\det(A)\det(B)=0$, and since $\mathbb R$ has characteristic $0$, $\det(A)\det(B)=0$. Therefore $\det(A)=0$ or $\det(B)=0$, and by a basic property of determinants, $A$ or $B$ is not invertible.

Solution. Let the coefficient matrix be $A=\begin{pmatrix}2&1&-3\\1&-2&1\\3&4&-2\end{pmatrix}$ and the constant vector $b=\begin{pmatrix}5\\10\\0\end{pmatrix}$. For each $i=1,2,3$, let $A_i$ be the matrix obtained from $A$ by replacing column $i$ with $b$; by Cramer's rule, $x_i = \det(A_i)/\det(A)$.

First, $$\det(A) = 2\det\begin{pmatrix}-2&1\\4&-2\end{pmatrix} - 1\det\begin{pmatrix}1&1\\3&-2\end{pmatrix} - 3\det\begin{pmatrix}1&-2\\3&4\end{pmatrix} = 2(0) -1(-5) -3(10) = -25.$$

$$\det(A_1) = \det\begin{pmatrix}5&1&-3\\10&-2&1\\0&4&-2\end{pmatrix} = 5(0) -1(-20) -3(40) = -100,$$ $$\det(A_2) = \det\begin{pmatrix}2&5&-3\\1&10&1\\3&0&-2\end{pmatrix} = 2(-20) -5(-5) -3(-30) = 75,$$ $$\det(A_3) = \det\begin{pmatrix}2&1&5\\1&-2&10\\3&4&0\end{pmatrix} = 2(-40) -1(-30) +5(10) = 0.$$

Hence $$x_1 = \frac{\det(A_1)}{\det(A)} = \frac{-100}{-25} = 4, \quad x_2 = \frac{\det(A_2)}{\det(A)} = \frac{75}{-25} = -3, \quad x_3 = \frac{\det(A_3)}{\det(A)} = \frac{0}{-25} = 0.$$

Check: $2(4)+(-3)-3(0)=5$, $4-2(-3)+0=10$, and $3(4)+4(-3)-2(0)=0$, so all three equations are satisfied. Thus the solution is $x_1=4,\ x_2=-3,\ x_3=0$.

Proof. Let $A_{ij}$ denote the $(i,j)$ entry of $A$ and $\tilde A_{ij} = (-1)^{i+j}\det(A(i|j))$ its cofactor, where $A(i|j)$ is the $(n-1)\times(n-1)$ matrix obtained by deleting row $i$ and column $j$ of $A$. By the definition given, the $(j,i)$ entry of $C$ is $C_{ji} = \tilde A_{ij}$.

The $(p,q)$ entry of $AC$ is $$(AC)_{pq} = \sum_{k=1}^n A_{pk}C_{kq} = \sum_{k=1}^n A_{pk}\tilde A_{qk}$$ (the first equality is the definition of matrix multiplication, and the second uses $C_{kq}=\tilde A_{qk}$).

If $p=q$, this is exactly the cofactor expansion of $\det(A)$ along row $p$: $$\sum_{k=1}^n A_{pk}\tilde A_{pk} = \det(A).$$

Now suppose $p\neq q$. Let $A'$ be the matrix obtained from $A$ by replacing row $q$ with the entries of row $p$, leaving all other rows unchanged. The cofactors $\tilde A_{qk}$ are determined by the submatrix $A(q|k)$ obtained by deleting row $q$, hence do not depend on the contents of row $q$; so the cofactors along row $q$ of $A'$ are also $\tilde A_{qk}$. Since row $q$ of $A'$ consists of the entries $A_{pk}$ ($k=1,\ldots,n$), expanding $\det(A')$ along row $q$ gives $$\det(A') = \sum_{k=1}^n A_{pk}\tilde A_{qk} = (AC)_{pq}.$$ But $A'$ has row $p$ and row $q$ both equal to row $p$ of $A$, i.e., two identical rows, so by the alternating property of the determinant, $\det(A')=0$. Hence $(AC)_{pq}=0$ when $p\neq q$.

Combining both cases, $(AC)_{pq} = \det(A)$ if $p=q$ and $0$ if $p\neq q$, so $AC = (\det A)I$.

Solution. Saying that $\delta$ is 3-linear means that, with the other two rows held fixed, $\delta$ is linear in each row.

(1) $\delta(A)=1$ is not 3-linear. For example, let $A'$ be the matrix obtained from $A$ by replacing its first row with the zero vector. Linearity in the first row would require $\delta(A') = 0\cdot\delta(A) = 0$. But $\delta$ is the constant function, so $\delta(A')=1\neq 0$. This contradiction shows $\delta$ is not 3-linear.

(2) $\delta(A)=A_{11}A_{21}A_{32}$ is 3-linear. Fixing rows 2 and 3, $\delta(A)=(A_{21}A_{32})A_{11}$ is linear in $A_{11}$, hence linear in row 1. Similarly, fixing rows 1 and 3, $\delta(A)=(A_{11}A_{32})A_{21}$ is linear in row 2, and fixing rows 1 and 2, $\delta(A)=(A_{11}A_{21})A_{32}$ is linear in row 3. Since $\delta$ is linear in each row, it is 3-linear.

(3) $\delta(A) = A_{11}A_{22}A_{33} - A_{11}A_{21}A_{32}$ is 3-linear. Each of the two terms $A_{11}A_{22}A_{33}$ and $A_{11}A_{21}A_{32}$ is, by the same reasoning as in (2), a product of exactly one entry from each of rows 1, 2, 3, hence 3-linear individually. A difference of 3-linear functions is again 3-linear (see EX 3.2), so $\delta$ is 3-linear.

Solution. For any $A=\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix}\in M_{2\times 2}(F)$, regard the rows of $A$ as vectors in $F^2$ and write them in terms of the standard basis $e_1=(1,0), e_2=(0,1)$: $$(A_{11},A_{12}) = A_{11}e_1 + A_{12}e_2, \qquad (A_{21},A_{22}) = A_{21}e_1 + A_{22}e_2.$$ If $\delta$ is 2-linear, i.e., linear in each row, then $$\delta(A) = \delta(A_{11}e_1+A_{12}e_2,\ A_{21}e_1+A_{22}e_2)$$ $$= A_{11}A_{21}\delta(e_1,e_1) + A_{11}A_{22}\delta(e_1,e_2) + A_{12}A_{21}\delta(e_2,e_1) + A_{12}A_{22}\delta(e_2,e_2).$$ Writing $\delta_{ij} := \delta(e_i,e_j)\in F$ for $i,j\in\{1,2\}$, we obtain $$\delta(A) = A_{11}A_{21}\delta_{11} + A_{11}A_{22}\delta_{12} + A_{12}A_{21}\delta_{21} + A_{12}A_{22}\delta_{22}.$$ That is, every 2-linear function is determined by the four scalars $\delta_{11},\delta_{12},\delta_{21},\delta_{22}\in F$ as above.

Conversely, for any fixed scalars $\delta_{11},\delta_{12},\delta_{21},\delta_{22}\in F$, the function $\delta$ defined by the formula above is 2-linear: each term is a product of exactly one entry from row 1 and one entry from row 2, so fixing row 2 makes $\delta(A)$ linear in $(A_{11},A_{12})$, and fixing row 1 makes it linear in $(A_{21},A_{22})$.

Hence the 2-linear functions $\delta:M_{2\times 2}(F)\to F$ are exactly the functions of the form $$\delta(A) = A_{11}A_{21}\delta_{11} + A_{11}A_{22}\delta_{12} + A_{12}A_{21}\delta_{21} + A_{12}A_{22}\delta_{22}$$ for arbitrary $\delta_{11},\delta_{12},\delta_{21},\delta_{22}\in F$.

Proof. Let $\mathcal F(M_{n\times n}(F), F)$ denote the set of all functions from $M_{n\times n}(F)$ to $F$; this is a vector space over $F$ under pointwise addition and scalar multiplication. Let $\Lambda$ denote the set of all $n$-linear functions, so $\Lambda \subseteq \mathcal F(M_{n\times n}(F),F)$. It suffices to show that $\Lambda$ is a subspace of $\mathcal F(M_{n\times n}(F),F)$.

(i) The zero function $0$ (with $0(A)=0$ for all $A$) is trivially linear in each row, so $0\in\Lambda$; in particular $\Lambda\neq\varnothing$.

(ii) Let $\delta_1,\delta_2\in\Lambda$ and $c\in F$. Write $\alpha_i$ for the $i$-th row of $A$, and suppose $\alpha_i = kv+w$ for some $k\in F$ and $v,w\in F^n$, with all other rows fixed (only $\alpha_i$ is shown below). Then $$(c\delta_1+\delta_2)(\ldots,kv+w,\ldots) = c\,\delta_1(\ldots,kv+w,\ldots) + \delta_2(\ldots,kv+w,\ldots)$$ $$= c\big[k\delta_1(\ldots,v,\ldots)+\delta_1(\ldots,w,\ldots)\big] + \big[k\delta_2(\ldots,v,\ldots)+\delta_2(\ldots,w,\ldots)\big]$$ $$= k\big[c\delta_1(\ldots,v,\ldots)+\delta_2(\ldots,v,\ldots)\big] + \big[c\delta_1(\ldots,w,\ldots)+\delta_2(\ldots,w,\ldots)\big]$$ $$= k(c\delta_1+\delta_2)(\ldots,v,\ldots) + (c\delta_1+\delta_2)(\ldots,w,\ldots).$$ Since this holds for every $i=1,\ldots,n$, the function $c\delta_1+\delta_2$ is linear in each row, i.e., $n$-linear, so $c\delta_1+\delta_2\in\Lambda$.

By (i) and (ii), $\Lambda$ is a subspace of $\mathcal F(M_{n\times n}(F),F)$, and hence $\Lambda$ is itself a vector space over $F$.