Let A be a matrix with independent rows. A. Show that AAT is invertible. B. Show that if b is any vector in Col(A), then the equation Ax = b has a solution in Row(A) given by xR = AT (AAT )−1 b. (The matrix AT (AAT )−1 is called the pseudoinverse of A, and is usually denoted by A+ . It is a right inverse of A, and coincides with the two-sided inverse A−1 if A is square.) c. The mapping of Col(A) to Row(A) given by xR = AT (AAT )−1 b is an isomorphism

Attached below

Step-by-step explanation:

Given that A is a matrix with independent rows

A) Prove is invertible

rank of A = number of rows of Matrix A

This shows that | A | ≠ 0  ( i.e. A has a full rank )

also   ≠ 0

and = | A |   ≠ 0

Hence we can conclude that is invertible

B) Prove that b is any vector in Col( A )

Attached below is the detailed solution

C) Mapping of Col ( A ) to Row ( A )

b is a non-zero vector hence AXr = 0 ( i.e. there is no solution )

also the kernel of the mapping will be Null

show that mapping preserves the operation

attached below is the detailed solution

step-by-step explanation: you take each of the numbers and divide by the amount then you take the range