Is the inverse of a linear mapping linear?
Exercise 1. I) Let denote unit vectors. They possess properties
(1) for all for all
II) For any we have the representation
III) In (2) the coefficients can be found as
Proof. I) (1) is proved by direct calculation. II) To prove (2) we write
III) If we have (2), then it's easy to see that by (1)
Definition 1. Any system of vectors that satisfies (1) is called an orthonormal system. An orthonormal system is called complete if for any we have the decomposition (3). Exercise 1 shows that our system of unit vectors is complete orthonormal. A complete orthonormal system is called an orthonormal basis.
Analyzing a linear mapping
Exercise 2. Let be a matrix of size Suppose you don't know the elements of but you know the products for all How can you reveal the elements of from How do you express using the elements you define?
Solution. Let us partition into rows and suppose the elements are known. Let us try unit vectors as
Using (2) and (4) one can check that Hence, from (3) we have the answer to the second question:
The above calculation means that when are unknown, we can define them by and then the action of on will be described by the last expression in (5).
We know that a mapping generated by a matrix is linear. The converse is also true: a linear mapping is given by a matrix:
Exercise 3. Suppose a mapping is linear: for any numbers and vectors Then there exists a matrix of size such that for all
Proof. Based on (4) in our case put . Applying (3) to we get
(the summation index is replaced by on purpose). Plugging (2) in (6)
(both and scalar product are linear)
The last equation is the definition of .
Exercise 4. An inverse of a linear mapping is linear (and given by a matrix by Exercise 3).
Proof. Let be a linear mapping and suppose its inverse in the general sense exists. Then for all Let us put for arbitrary numbers and vectors Then we have or, using linearity of
Putting we get
Thus, is linear.