Transpose and inverse of [linear algebra 01] matrix

   I have always wanted to write a summary in this regard , Let's start with the transpose and inverse of the matrix . The content of this article is compiled from the webpage .

Transpose and inverse of matrix

Give the protagonist of this part of the narrative , matrix A And matrices B
$$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]B=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]$$

Matrix creation

stay Python in With the help of Numpy Of array Object can create a two-dimensional array （ Matrix ）.

>>> import numpy as np
>>> A = np.array([[1,2],[3,4]])#python use numpy Medium array Create a matrix print(A)
>>> print(A)
[[1 2]
 [3 4]]
>>> B = np.array([[1,2,3],[4,5,6]])
>>> print(B)
[[1 2 3]
 [4 5 6]]

stay matlab This creation process is easier , Use commas or spaces to separate columns , Use semicolons to separate lines .

>> A = [1,2;3,4]

A =

     1     2
     3     4

>> B = [1 2 3;4 5 6]

B =

     1     2     3
     4     5     6

The transpose of the matrix

The transpose of a matrix is Make symmetry around the main diagonal , In other words, the row subscripts and column subscripts of matrix elements are interchanged .
$$A^T=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]B^T=\left[\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right]$$
Numpy in Provides transpose Function to realize transpose .

>>> At = np.transpose(A)
>>> print(At)
[[1 3]
 [2 4]]
>>> Bt = np.transpose(B)
>>> print(Bt)
[[1 4]
 [2 5]
 [3 6]]

matlab in Transpose can be achieved by directly using a prime （ Conjugate transpose is implemented by default ）.

>> At = A'

At =

     1     3
     2     4

>> Bt = B'

Bt =

     1     4
     2     5
     3     6

The matrix of the inverse

Unlike the transpose of a matrix, which is valid for all matrices , The first object of the inverse of a matrix is a square matrix . For a square array , These concepts are equivalent ：

The full rank of a reversible matrix can be transformed into an identity matrix through a finite elementary row transformation, and the determinant of a nonsingular matrix is not 0

From definition , Inverse matrix means that the result of matrix multiplication with the original matrix will be the identity matrix , namely ：
$$A{A}^{-1}=A^{-1}A=E$$

The augmented matrix solves the inverse matrix

Manually calculate the inverse of a matrix , We generally have three ways , The method of undetermined coefficient 、 Adjoint matrix method and elementary transformation method . Because reversible matrix “ Through finite elementary row transformation, it can be transformed into an identity matrix ” The nature of , So we often use augmented matrix to solve the inverse of general matrix .

With A For example , Easy to calculate A The determinant of is -2, Not for 0, A matrix is reversible , The following shows how to solve with augmented matrix A The steps of the inverse matrix of .
$$A=\left[\begin{array}{ccccc}1& 2& |& 1& 0\\ 3& 4& |& 0& 1\end{array}\right]$$
Easy to see , First subtract the second row from the first row and multiply by 3：
$$A\Rightarrow \left[\begin{array}{ccccc}1& 2& |& 1& 0\\ 0& -2& |& -3& 1\end{array}\right]$$
then , Add the first line back to the second line ：
$$A\Rightarrow \left[\begin{array}{ccccc}1& 0& |& -2& 1\\ 0& -2& |& -3& 1\end{array}\right]$$
Last , Multiply the second line by negative half ：
$$A\Rightarrow \left[\begin{array}{ccccc}1& 0& |& -2& 1\\ 0& 1& |& \frac{3}{2}& -\frac{1}{2}\end{array}\right]$$
So there is A The inverse matrix of is
$$A^{-1}=\left[\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right]$$
Easy to see $A{A}^{-1}=E$, The inverse matrix is indeed correct .

inv() Function to solve the inverse matrix

Numpy Medium linear algebra Library linalg Provides inv() Function to realize the inverse matrix solution of nonsingular matrix .

>>> Ai = np.linalg.inv(A)
>>> print(Ai)
[[-2.   1. ]
 [ 1.5 -0.5]]

matlab Similar functions are also provided in inv().

>> Ai = inv(A)

Ai =

   -2.0000    1.0000
    1.5000   -0.5000

Pseudo inverse

Obviously, nonsingular matrices are in the minority , So for singular matrices and non square matrices , There is no inverse matrix , But we can introduce pseudo inverse matrix as the generalized form of inverse matrix . The mathematical definition is as follows ：

If there is a relation with A The transpose matrix of A’ A matrix of the same type X, And satisfy ：AXA=A,XAX=X. here , Symmetric matrix X For matrix A The pseudo inverse of , Also called generalized inverse matrix .

Easy to verify Inverse matrix is a special case of generalized inverse matrix , namely
$$A{A}^{-1}A=EA=A,A^{-1}A{A}^{-1}=E{A}^{-1}=A^{-1},amongX=A^{-1}$$

Satisfy $A^{L}A=E$, But not satisfied $A{A}^L=E$ Matrix $A^L$ It's called a matrix A The left inverse matrix of . Allied , Satisfy $A{A}^R=E$, But not satisfied $A^{R}A=E$ Matrix $A^R$ It's called a matrix A Right inverse matrix of . We discuss it in three ways （ It can be seen that the matrix with smaller rank is always taken ）：

1. When m≥n when , Full rank , matrix $A_{m\times n}$ There is a left inverse matrix ,$A_{n\times m}^L=(A^{T}A{)}^{-1}{A}^T$
2. When n≥m when , Row full rank , matrix $A_{m\times n}$ There is a right inverse matrix ,$A_{n\times m}^R=A^T(A{A}^T{)}^{-1}$
3. When n =m when ,$A_{m\times n}$ The rank of is $r\le m=n$, Yes A Singular value decomposition $A=UD{V}^T$ ,A The pseudo inverse matrix of is $A^{+}=V{D}^{+}{U}^T$, among $D^{+}$ Is the inverse matrix of singular value matrix , That is, the elements on the protagonist line take the reciprocal .（ It should be pointed out that , Singular value decomposition is a general method , Leave it for discussion when writing singular value decomposition ）

We deal with relatively simple B For example , It is known that B Yes ：
$$B=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]B^T=\left[\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right]$$
therefore $B{B}^T$ by
$$B{B}^T=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]\times \left[\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right]=\left[\begin{array}{cc}14& 32\\ 32& 77\end{array}\right]$$
Calculate the determinant of this formula , It is easy to find that its value is -54, Inverse matrix exists . Next, use row transformation to solve the inverse matrix , First the (2,1) Element zeroing ：
$$B{B}^T\Rightarrow \left[\begin{array}{ccccc}14& 32& |& 1& 0\\ 0& 3.857& |& -2.286& 1\end{array}\right]$$
then (1,2) Element zeroing ：
$$B{B}^T\Rightarrow \left[\begin{array}{ccccc}14& 0& |& 19.967& -8.297\\ 0& 3.857& |& -2.286& 1\end{array}\right]$$
Finally, return the diagonal elements to 1：
$$B{B}^T\Rightarrow \left[\begin{array}{ccccc}1& 0& |& 1.426& -0.593\\ 0& 1& |& -0.593& 0.259\end{array}\right]$$

So there is B The right inverse matrix of is ( There is a certain error )：
$$B^R=B^T(B{B}^T{)}^{-1}=\left[\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right]\times \left[\begin{array}{cc}1.426& -0.593\\ -0.593& 0.259\end{array}\right]=\left[\begin{array}{cc}-0.946& -0.443\\ -0.113& 0.109\\ -0.720& -0.225\end{array}\right]$$
Can verify $B^R$ Is to meet $B{B}^R=E$, But not satisfied $B^{R}B=E$ Matrix , This is what “ Right ” The meaning of .

pinv() Function to solve pseudo inverse matrix

Compared with the complicated discussion of violations , In actual programming, seeking violation is just a line of code . Note that although the inverse matrix is a special case of the inverse matrix , but inv() Ratio pinv() Higher solving efficiency , We solve it separately A and B The generalized inverse matrix of .

Numpy Medium linear algebra Library linalg Provides pinv() Function to solve the violation matrix .

>>> Bpi = np.linalg.pinv(B)
>>> print(Bpi)
[[-0.94444444  0.44444444]
 [-0.11111111  0.11111111]
 [ 0.72222222 -0.22222222]]

Matlab Similar functions are also provided in pinv().

>> Bpi = pinv(B)

Bpi =

   -0.9444    0.4444
   -0.1111    0.1111
    0.7222   -0.2222