[Linear Algebra 05] Properties and Applications of Determinants

   Continuing the notes of MIT, this article sorts out the 10 properties of the determinant, of which the first 3 properties serve as the basis for the subsequent properties. Applications are grouped into Cramer's Law, Inverse Matrix and Volume. Note that only square matrices have determinants.

nature

1. $\mathrm{the}I=1$
2. Swap two rows of the determinant, change the sign of the determinant
   Note: exchange once to change number once, exchange $i$ times, multiply by$(-1{)}^i$ , so the even-numbered transformation and the odd-numbered transformation are in their own independent transformation categories.
3. Linearity of "rows":a) Homogeneity: the coefficient of a row of the determinant is proposed as the coefficient of the whole determinant; b) Additivity: when a row of the determinant can be split into two parts, it is equal to the two determinants after splitting of and.Use $2\times 2$行列式表示即为：
   $$\left|\begin{array}{cc}2a\_+3e& 2b+3f\_\\ c& d\end{array}\right|=\left|\begin{array}{cc}2a\_& 2b\\ c& d\end{array}\right|+\left|\begin{array}{cc}3e& 3f\_\\ c& d\end{array}\right|=2\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|+3\left|\begin{array}{cc}e& f\\ c& d\end{array}\right|$$
   Note: The linearity of the determinant is for the "row", and the linearity of the matrix is ​​for the entire matrix. Therefore $|2A|=2^n|A|$

4. When two rows are equal, the determinant is 0. It can be deduced from property 2.
   $$A=\left|\begin{array}{cc}a& b\\ a& b\end{array}\right|\underrightarrow{swap\; two}-A=\left|\begin{array}{cc}a& b\\ a& b\end{array}\right|\Rightarrow A=0$$
   Note: Further by nature 3, when there are two rows in a multiple relationship, the determinant is also 0.

5. When one row subtracts k times another row, the determinant does not change.
   Note: The proof similar to property 4 is based on the reasonable extrapolation of property 3. The summary of the determinant is 0 is:When there is a linear combination of other rows in a certain row, the determinant is 0, which is equivalent to requiring the corresponding matrix to be full rank。

6. If one row is 0, the determinant is 0.
   Note: 0 row plus any row is equal to two rows of the original determinant. It can be deduced based on properties 4 and 5.

7. After the elimination method becomes a triangular matrix, such as $A=LU$ transform,Triangular array UUThe product of elements on the main column of$U\; (that\; is,\; on\; the\; main\; diagonal)\; is\; the\; value\; of\; the\; determinant.$
   Note: The elimination method does not change the value of the determinant.

8. When the matrix is ​​singular (that is, the rank is not satisfied), the corresponding determinant is 0 .
   Note: It is equivalent to the summary in Property 5.

9. The determinant of the product of matrices is equal to the product of the determinants , that is$\mathrm{the}AB=\mathrm{the}A\mathrm{the}B$
   Note: Since all matrices can be transformed into diagonal matrices, if the product of diagonal matrices is considered, obviously the result is valid. Further we can know: $\mathrm{the}{A}^{-1}=1/\mathrm{the}A$

10. The determinant is equal to the transposed determinant. Give a sufficient and concise proof:
    $$|A^T|=|A|\Rightarrow |U^T{L}^T|=|LU|\Rightarrow |U^T||L^T|=|L||U|(property9)\surd Note$$
    : This shows thatThe properties of the rows can be reasonably extrapolated to the columns, for example, when a column is a linear combination of other columns, the determinant will also be 0, and if two columns are exchanged, the sign of the determinant will also change.

  The determinant is equal to the product of the pivots , which implies the importance of these pivots (i.e. eigenvalues).

Expansion of the determinant

  Remember the following formula, namelyThe determinant can be expanded by taking a certain row：
$$\mathrm{the}A=a_i{C}_i+a_{i2}{C}_{i2}+\cdots +a_{in}{C}_{in}$$
  Among them, C is called the algebraic cofactor , which is the cofactor after taking the transformed sign . That is,
$$C_{ij}=(-1{)}^{i+j}\mathrm{the}{M}_{ij}$$
  The remainder formula is to delineate $a_{ij}$After the elements in the row and column, the remaining elements are arranged into a matrix.
$$\left|\begin{array}{c}\begin{array}{cc}{+}_{a_{ij}}& {-}_{...}\\ {|}_{...}& M_{ij}\end{array}\end{array}\right|$$
  Take a $3\times 3.$ Take the calculation of the determinant as an example. Use the det command in Matlab to find the value of the matrix determinant:
$$\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|=1(-1{)}^2\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|+2(-1{)}^3\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|+3(-1{)}^4\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|=-3+12-9=0$$

>> A = [1,2,3;4,5,6;7,8,9];
>> det(A) % 行列式

ans =

  -9.5162e-16

>> rank(A) % 秩

ans =

     2

>> rref(A) % 最简行形式

ans =

     1     0    -1
     0     1     2
     0     0     0

  The key to prove the idea is to take a certain line and expand it one by one, it is easy to find that when taking $a_{ij}$elements, the elements of the rest of the rows cannot be selected from the rows and columns that have been crossed out, thus becoming a permutation and combination problem , that is, for $n\times In\; terms\; of\; the\; determinant\; of\; n$ , there are a total of$n!$ sub-items, arranged in a determinant, becomes$n\times n-1\times \cdots The\; level-by-level\; decomposition\; of\; 1$ is the level-by-level expansion of the determinant.

application

Inverses (inverse matrix)

  Explicit representation of the inverse matrix:The inverse matrix is ​​the transpose of the algebraic cofactor matrix divided by the determinant of the original matrix。 即
$$A^{-1}=C^T/\mathrm{the}The\; derivation\; of\; the\; inverse\; matrix\; of\; A$$
  starts with the expansion of the determinant, that is to say,$A{C}^T=\left(\mathrm{it}A\right)I$ ratio:
$$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\left[\begin{array}{ccc}{C}_{11}& C_{21}& C_{31}\\ C_{12}& C_{22}& C_{32}\\ C_{13}& C_{23}& C_{33}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{the}A& 0& 0\\ 0& \mathrm{the}A& 0\\ 0& 0& \mathrm{the}A\end{array}\right]$$
  Focusing on the elements of the diagonal line, it is not difficult to find that it is the expression of the determinant expansion, and how to explain the remaining zeros? This is a great question. Let's look at an example:
$$？a_{21}{C}_{11}+a_{22}{C}_{12}+a_{23}{C}_{13}=0$$
  First quote the explanation in the textbook:

This is the cofactor rule for a new matrix, when the second row of $A$ is copied into its first row. The new matrix $A^{\ast }$ has two equal rows, so det $A^{\ast }$ = 0. Notice that $A^{\ast }$ has the same cofactors $C_{11}$, $C_{12}$, $C_{13}$ as $A$-because all rows agree after the first row. Thus the remarkable multiplication is correct.

  In other words, by construction, the algebraic remainders corresponding to the first row of the following two matrices are the same :
$$A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]withA{}^{\ast }=\left[\begin{array}{ccc}{a}_{21}& a_{22}& a_{23}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\Rightarrow \mathrm{the}{A}^{\ast }=a_{21}{C}_{11}+a_{22}{C}_{12}+a_{23}{C}_{13}=0$$

Cramer's Rule

  For the equation $Ax=As\; far\; as\; b$ is concerned, Cramer's law is a "fancy but not useful" solution formula (not as good as the elimination method). Its main significance lies inexplicitly giving the expression of the solution of the equation, although it will be very cumbersome to do. When$The\; determinant\; of\; A$ is not 0 (or when reversible),$Ax=b$字性的解的
$$x_1=\frac{\mathrm{the}{B}_1}{\mathrm{the}A}{x}_2=\frac{\mathrm{the}{B}_2}{\mathrm{the}A}\cdots x_n=\frac{\mathrm{the}{B}_n}{\mathrm{the}A}$$
  Among them $B_i$listed as $A$ secondiibbfor$i$The result after replacement in column$b\; .$

Idea 1

  There are two ways of thinking to explain this result. The first is to directly substitute into the expression of the inverse matrix, that is,
$$x=A^{-1}b=\frac{C^{T}b}{\mathrm{the}A}$$
  take $x_1$For example, the corresponding $C^{T}b$is
$$b_1{C}_{11}+b_2{C}_{21}+b_3{C}_{31}=0$$
  Since the value of the determinant remains unchanged after the transposition, it can be expanded by taking a certain row, and of course it can also be expanded by taking a certain column. And when a certain column is taken to expand, we know that thealgebraic cofactors corresponding to the first column of the following two matrices are the same:
$$A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]withB{}_1=A^{\ast }=\left[\begin{array}{ccc}{b}_1& a_{12}& a_{13}\\ b_2& a_{22}& a_{23}\\ b_3& a_{32}& a_{33}\end{array}\right]\Rightarrow \mathrm{the}{A}^{\ast }=b_1{C}_{11}+b_2{C}_{21}+b_3{C}_{31}$$
  Therefore, for $C^{}$Corresponding to xi x_iin${}^T$$b$$x_i$In terms of the row, it can be used respectively by $B_i$Substitute and get the same result.

Idea 2

  This idea is more ingenious. With the help of the identity matrix, we do the following transformation:
$$\left[\begin{array}{ccc}& & \\ a_1& a_2& a_3\\ & & \end{array}\right]\left[\begin{array}{ccc}{x}_1& 0& 0\\ x_2& 1& 0\\ x_3& 0& 1\end{array}\right]=\left[\begin{array}{ccc}& & \\ b& a_2& a_3\\ & & \end{array}\right]=B_1$$
  Taking the determinant on both sides, with the help of properties 7 and 9 of the determinant, there is:
$$(\mathrm{it}A)x_1=\mathrm{the}{B}_1$$

Volumes

  We have come into contact with such formulas when learning the cross product of vectors, such as calculating vector $\vec{a}=(1,2,3)$ b ⃗ =$\vec{b}=(4,5,6)$的叉乘：
$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ \vec{i}& \vec{j}& \vec{k}\end{array}\right|=(12-15)\vec{i}+(12-6)\vec{i}+(5-8)\vec{k}=(-3,6,-3)$$

>> a=[1,2,3];b=[4,5,6];
>> cross(a,b)

ans =

    -3     6    -3

  Further, we find the above result and $\vec{c}=(0,1,0)$ , that is, the triple product, which representsthe volume of the parallelepiped formed by these three vectors:
$$\vec{a}\times \vec{b}\cdot \vec{c}=\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 0& 1& 0\end{array}\right|=(12-15)+(12-6)1+(5-8)0=6$$

>> c=[0,1,0];
>> dot(c,cross(a,b))

ans =

     6

  Based on the understanding of the triple product, we can derive the formula for the area of ​​a triangle on the plane, that is, it can be regarded as half the volume of a parallelepiped with a height of 1, so, there is:
$$\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\frac{1}{2}[(x_2{y}_3-x_3{y}_2)+(x_3{y}_1-x_1{y}_3)+(x_1{y}_2-x_2{y}_1)]=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$
  In particular, when$(x_3,y_3)=(0,0)$ , the area formula can be further simplified as:
$$\frac{1}{2}\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|=\frac{1}{2}(x_1{y}_2-x_2{y}_1)$$
  It is worth noting that the orthogonal matrix$Q$ , since$Q{Q}^T=I$ , so when taking the determinant on both sides, there is$(\mathrm{it}Q{)}^2=1$ , which means thatthe rotation matrix Q only changes the angle, but does not change the volume. This knowledge helps to enhance the understanding of operator operations in electromagnetism and phase rotation in signals and systems. In the next lecture on eigenvalues, we will more specifically discover the rich meanings brought by matrix operations.