Determinant 2

Proof of properties.

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Notation

$$det(A)=|A|=D({\mathbf{v}}_1,\cdots ,{\mathbf{v}}_n)$$

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Definition

• Determinant of identity matrix is one
  
  $$\begin{array}{}\text{(1)}& det(I_n)=1\end{array}$$
• Antisymmetry
  
  $$\begin{array}{}\text{(2)}& D(\cdots ,{\mathbf{v}}_j,\cdots ,{\mathbf{v}}_k,\cdots )=-D(\cdots ,{\mathbf{v}}_k,\cdots ,{\mathbf{v}}_j,\cdots )\end{array}$$
• Linearity in each argument:
  
  $$\begin{array}{}\text{(3)}& D(\cdots ,\alpha {\mathbf{v}}_k,\cdots )=\alpha D(\cdots ,{\mathbf{v}}_k,\cdots )\end{array}$$
  and
  
  $$\begin{array}{}\text{(4)}& D(\cdots ,{\mathbf{u}}_k+{\mathbf{v}}_k,\cdots )=D(\cdots ,{\mathbf{u}}_k,\cdots )+D(\cdots ,{\mathbf{v}}_k,\cdots )\end{array}$$

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Properties

1. Two equal columns, then determinant equals to zero.
   • pf: By antisymmetry, Eq.~(2),
     
     $$D(\cdots ,\mathbf{v},\cdots ,\mathbf{v},\cdots )=-D(\cdots ,\mathbf{v},\cdots ,\mathbf{v},\cdots ).$$
     Therefore
     
     $$\begin{array}{}\text{(5)}& D(\cdots ,\mathbf{v},\cdots ,\mathbf{v},\cdots )=0.\end{array}$$
2. A column of zero, then determinant equals to zero.

   • pf: By linearity, Eq.~(3),
     

     $$\alpha D(\cdots ,\mathbf{0},\cdots )=D(\cdots ,\alpha \mathbf{0},\cdots )=D(\cdots ,\mathbf{0},\cdots ).$$

     Therefore (choose

     $\alpha =2$),
     $$\begin{array}{}\text{(6)}& D(\cdots ,\mathbf{0},\cdots )=0.\end{array}$$

3. Preservation under “column replacement”
   
   $$\begin{array}{}\text{(7)}& D(\cdots ,{\mathbf{v}}_j+\alpha {\mathbf{v}}_k,\cdots ,{\mathbf{v}}_k,\cdots )=D(\cdots ,{\mathbf{v}}_j,\cdots ,{\mathbf{v}}_k,\cdots ).\end{array}$$
   • pf: By linearity, Eq.~(4),
     
     $$D(\cdots ,{\mathbf{v}}_j+\alpha {\mathbf{v}}_k,\cdots ,{\mathbf{v}}_k,\cdots )=D(\cdots ,{\mathbf{v}}_j,\cdots ,{\mathbf{v}}_k,\cdots )+D(\cdots ,\alpha {\mathbf{v}}_k,\cdots ,{\mathbf{v}}_k,\cdots ),$$
     where (use (3) and (5))
     
     $$D(\cdots ,\alpha {\mathbf{v}}_k,\cdots ,{\mathbf{v}}_k,\cdots )=\alpha D(\cdots ,{\mathbf{v}}_k,\cdots ,{\mathbf{v}}_k,\cdots )=0.$$
4. If
   $A$ is singular, then
   $\text{det}(A)=0$.
   • pf: If
     $A$ is singular, one of
     ${\mathbf{v}}_j$ is linear combination of other columns, then we can use "column replacement" to zero-out this column, i.e.,
     
     $$D(\cdots ,{\mathbf{v}}_j,\cdots )=D(\cdots ,\mathbf{0},\cdots )=0.$$
5. Determinant of upper triangular matrix is the product of its diagonal elements, i.e.,
   
   $$\begin{array}{}\text{(8)}& \left|\begin{array}{cccc}{d}_1& \ast & \cdots & \ast \\ 0& d_2& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & d_n\end{array}\right|=d_1{d}_2\cdots d_n.\end{array}$$
   • pf:
     • If
       $d_1=0$, we have a column of zero and
       $|A|=0$ by (6). So LHS equals to RHS, true.
     • If
       $d_1\ne 0$, we can apply "column replacement" to have
       
       $$\left|\begin{array}{cccc}{d}_1& \ast & \cdots & \ast \\ 0& d_2& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & d_n\end{array}\right|=\left|\begin{array}{cccc}{d}_1& 0& \cdots & 0\\ 0& d_2& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & d_n\end{array}\right|$$
     • Use induction, either
       $d_k=0$ for some
       $k$ and
       $|A|=0$, or we have
       
       $$\left|\begin{array}{cccc}{d}_1& \ast & \cdots & \ast \\ 0& d_2& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & d_n\end{array}\right|=\left|\begin{array}{cccc}{d}_1& 0& \cdots & 0\\ 0& d_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & d_n\end{array}\right|=d_1{d}_2\cdots d_n\left|\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 1\end{array}\right|=d_1{d}_2\cdots d_n.$$
       In either case, we have LHS equals to RHS. So the statement is true.
   • Remark: Same conclusion applied to lower triangular matrices.
6. $\text{det}(AB)=\text{det}(A)\text{det}(B)$.
   • pf:
     • If
       $A$ is singular,
       $AB$ is singular, and
       $\text{det}(AB)=\text{det}(A)=0$. So the equality is true.
     • If
       $A$ is non-singular,
       $|A|\ne 0$, we define a function
       $\tilde{D}:M_{n\times n}\to \mathbb{R}$ as
       
       $$\begin{array}{}\text{(9)}& \tilde{D}(B)=\frac{|AB|}{|A|}.\end{array}$$
       • Properties of
         $\tilde{D}$:
         a.
         
         $$\tilde{D}(I_n)=\frac{|A{I}_n|}{|A|}=1$$
         b. Let
         $B=[\cdots ,{\mathbf{v}}_j,\cdots ,{\mathbf{v}}_k,\cdots ]$, then
         $[\cdots ,{\mathbf{v}}_k,\cdots ,{\mathbf{v}}_j,\cdots ]=BP$, where
         $P$ is the permutation matrix that exchanges columns
         $j$ and
         $k$. Also we have
         
         $$|A(BP)|=|(AB)P|=-|AB|.$$
         So
         
         $$\tilde{D}(\cdots ,{\mathbf{v}}_k,\cdots ,{\mathbf{v}}_j,\cdots )=\tilde{D}(BP)=-\tilde{D}(B)=-\tilde{D}(\cdots ,{\mathbf{v}}_j,\cdots ,{\mathbf{v}}_k,\cdots ).$$
         c. Let
         $B=[\cdots ,{\mathbf{v}}_k,\cdots ]$, then
         $[\cdots ,\alpha {\mathbf{v}}_k,\cdots ]=BJ$, where
         $J$ is the identity matrix except
         $J_{k,k}=\alpha$. Then
         
         $$\tilde{D}(\cdots ,\alpha {\mathbf{v}}_k,\cdots )=\tilde{D}(BJ)=\frac{|ABJ|}{|A|}=\frac{\alpha |AB|}{|A|}=\alpha \tilde{D}(B)=\alpha D(\cdots ,{\mathbf{v}}_k,\cdots ).$$
         Finally, recall that we have
         $A[\cdots ,{\mathbf{u}}_k+{\mathbf{v}}_k,\cdots ]=[\cdots ,A{\mathbf{u}}_k+A{\mathbf{v}}_k,\cdots ]$, so
         
         $$\begin{gathered} \text{det}(A[\cdots ,{\mathbf{u}}_k+{\mathbf{v}}_k,\cdots ])=\text{det}([\cdots ,A{\mathbf{u}}_k+A{\mathbf{v}}_k,\cdots ]) \\ =\text{det}([\cdots ,A{\mathbf{u}}_k,\cdots ])+\text{det}([\cdots ,A{\mathbf{v}}_k,\cdots ]). \end{gathered}$$
         It is then easy to show that
         
         $$\tilde{D}(\cdots ,{\mathbf{u}}_k+{\mathbf{v}}_k,\cdots )=\tilde{D}(\cdots ,{\mathbf{u}}_k,\cdots )+\tilde{D}(\cdots ,{\mathbf{v}}_k,\cdots ).$$
       • Summary: It turns out that
         $\tilde{D}$ satisfies exactly the same
         $3$ definitions as
         $D$, so
         $\tilde{D}=D$ and
         $\tilde{D}(B)=|B|$. Therefore,
         $|AB|=|A||B|$.
7. $\text{det}(A^T)=\text{det}(A)$.
   • pf:
     • Observation: If
       $P$ is a permutation matrix, then
       $|P|=|P^T|$ and
       $|P|=\pm 1$, i.e., either both are
       $1$ or both are
       $-1$.
     • Let
       $PA=LU$, where
       $P$ is the permutation matrix,
       $L$ is lower triangular that has all the diagonals being
       $1$ and
       $U$ is upper triangular. Then
       $|L|=|L^T|=1$,
       $|U|=|U^T|$, and
       
       $$|A|=|P^{T}LU|=|P^T||L||U|=|U^T||L^T||P|=|U^T{L}^{T}P|=|A^T|.$$

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References

• Linear Algebra Done Wrong: Ch2