## Is the inverse of a linear mapping linear?

### Orthonormal basis

**Exercise 1**. I) Let denote unit vectors. They possess properties

(1) for all for all

II) For any we have the representation

(2)

III) In (2) the coefficients can be found as

(3)

**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 try unit vectors as

(4)

One can check that Hence, from (3) we have the answer to the second question:

(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

(6)

(the summation index is replaced by on purpose). Plugging (1) in (6)

(both and scalar product are linear)

(applying (5))

**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.

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