## Geometry of linear equations: orthogonal complement and equation solvability

### From orthogonality of vectors to orthogonality of subspaces

**Definition 1**. Let be a linear subspace of Its **orthogonal complement** is defined as the set of *all* vectors orthogonal to *all* elements of

**Exercise 1**. Let be the axis on the plane. Find

**Solution**. The condition for all implies The component is free. Thus, is the axis.

Similarly, the orthogonal complement of the axis is the axis. This fact is generalized as follows:

**Theorem** (*on second orthocomplement*) For any linear subspace one has .

This statement is geometrically simple but the proof is rather complex and will be omitted (see Akhiezer & Glazman, Theory of Linear Operators in Hilbert Space, Dover Publications, 1993, Chapter 1, sections 6 and 7). It's good to keep in mind what makes it possible. For any set its **orthogonal complement** can be defined in the same way: As in Exercise 1, you can show that if contains just the unit vector of the axis, then is the axis and therefore is the axis. Thus, in this example we have a strict inclusion The equality is achieved only for linear subspaces.

**Exercise 2**. In consider the axis Can you tell what is?

**Exercise 3**. and

### Link between image of a matrix and null space of its transpose

**Exercise 4**. The rule for a transpose of a product implies that for any

(1)

**Exercise 5** (*second characterization of matrix image*)

**Proof**. Let us prove that

(2) .

To this end, we first prove the inclusion Let . For an arbitrary we have Hence by (1) Since is arbitrary, we can put and obtain This implies and

Conversely, to prove suppose Then and for an arbitrary we have Since runs over this means that

Passing to orthogonal complements in (2), by the theorem on second orthocomplement we get what we need.

The importance of Exercise 5 is explained by the fact that the null space of a matrix is easier to describe analytically than the image. For the next summary, you might want to review the conclusion on the role of the null space.

**Summary**. 1) In order to see if has solutions, check if is orthogonal to In particular, if is one-to-one, then and has solutions for all (see Exercise 3 above).

2) If is one-to-one, then may have only unique solutions.

3) If both and are one-to-one, then has a unique solution for any

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