Linearly independent & inverse of a square matrix

* Linearly independent라는 개념은 AI 알고리즘을 build up 하는데 있어서 상당히 중요!!

 

 

Example

A and B are two NxN matrices which are invertible.

Show that: (AB)^(−1) =B^(−1)A^(−1) 

 

* 증명문제 Tip

1. 정의 이용하기

2. 가정이 잘못되었다고 설립한 뒤 그 결과가 모순됨을 보여주면 됨

 

여기서는 '1. 정의 이용하기'를 통해 해결해보자..!

 

(AB) (B^(−1)A^(−1)) Is this equal to I(Identity matrix)?

= A (B B^(−1)) A^(−1)

= A I A^(−1) = A A^(−1) = I

 

Thus: (AB)^(−1) =B^(−1)A^(−1) 

 

Linearly independent vectors

- 한 벡터를 다른 벡터들의 linear combination으로 표현할 수 있으면 Linearly dependent!!(Not Linearly Independent!)

 

- Definition :

First! Form the homogeneous system

c1w1 + c2w2 + ... + cpwp = 0

 

If c1 = c2 = ... = cp = 0 is the only solution of the system then the vectors are linearly independent.!

 

Why does this work?

 

For example, if we can find c1 which is not zero then we can write : 

w1 = -c2/c1 w2 -c3/c1 w3 - ... - cp/c1 wp

-------> w1이 나머지 벡터들의 linear combination으로 표현되었으므로 Not Linearly independent!

 

SO!!!!!

 

If c1 = c2 = ... = cp = 0 is the only solution of the system then the vectors are linearly independent.!

 

Inverse of a square matrix

- B(NxN) is said to be an inverse of A(NxN) if the relation AB=BA=I(NxN Identity) is true

 

How many inverses can a square matrix have?

- At most one!

 

If a square matrix A has an inverse there is only one inverse it can have

Let B be an inverse of A -----> AB = BA =I

Let C be an inverse of A -----> AC = CA =I

 

AB = BA

ABC = BAC (행렬 연산은 Commutative하지 않으므로 같은 쪽? 에 곱해줘야 한다--> 여기서는 오른쪽)

(AB)C = B(AC)

IC = BI

C = B

 

Conclusion : inverse가 있다면 오직 하나만 있고 없는 경우도 있다!

 

If A has an inverse we say that A is invertible

 

Solve Inverse problem

- Form the tableau A | I and perforn legitimate row operations on it

- A | I ---> I | V(if possible)

- If A can be reduced to I via row operations then V is the inverse of A

- If A cannot be reduced to I then A is not invertible

 

* Determinant = 0 ------> No inverse matrix

 

Consider the system AX = B

- If A^(-1) exists then we can write:

A^(-1)AX = A^(-1)B

IX = A^(-1)B

X = A^(-1)B ----------> This is the unique solution of the system

 

Note that finding the inverse of A requires more work than reducing  AX=B to UX=C

 

Thus, in solving AX=B, it is usually not recommendable to find the inverse of A first

However, if the inverse of A is already known, it can be used to find the unique solution of the system

 

-------> 지금도 계속 A^(-1)를 어떻게 하면 빠르게 구할지 연구중이다!!