3.2.3. Affine Subspace

The solution to the preimage problem does not form a linear subspace because, e.g., *<sup>W</sup>*(*v* + *v*) = 2*b* which is generally not equal to *b*. It is a linear subspace of *A* if and only if *b* = 0, i.e., if and only if it is equal to null *W* (right nullspace, or kernel, of a matrix *W* ∈ R*l*×*<sup>L</sup>* contains all vectors *a* ∈ <sup>R</sup>*L*, s.t. *Wa* = 0, it is a linear subspace of R*<sup>L</sup>*). In the other words, if *b* = 0, then the preimage problem, *Wv* = *b*, becomes a homogeneous system of linear equations, making its solution be a linear subspace of *A*.

Coming back to the general case where the vector *b* ∈ *B* is arbitrary, we obtain that the solution to the preimage problem forms an affine subspace of *A*.

**Lemma 9.** *Let b* ∈ *B and v* ∈ *A be a particular solution to the preimage problem Wv* = *b, then the set of all vectors that solve this problem, forms an affine subspace* {*v* + *v*0 | *v*0 ∈ null *W*} *of A.*

**Proof.** For the vector *<sup>v</sup>*, it holds that *Wv* = *b* and for any vector *v*0 ∈ null *W*, it holds that *Wv*0 = 0. Hence, for any element *a* = *v* + *v*0 of the set A = {*v* + *v*0 | *v*0 ∈ null *<sup>W</sup>*}, it holds that *Wa* = *Wv* + *Wv*0 = *b* + 0 = *b*, so every element of A is a solution to the preimage problem.

Conversely, consider a vector *a* ∈ *A*, s.t. *Wa* = *b* and *a* ∈ A. Then *<sup>W</sup>*(*a* − *v*) = *Wa* − *Wv* = *b* − *b* = 0, so the vector *a* − *v* ∈ null *W*. As the vector *a* can be expressed as *v* + *a* − *<sup>v</sup>*, we see that *a* ∈ A. This contradiction proves that the set A contains exactly all solutions to the preimage problem. Moreover, it proves that the set A is unique, i.e., if we express the solution to the preimage problem as B = {*v* + *v*0 | *v*0 ∈ null *W*} for any vector *v* ∈ *A*, s.t. *Wv* = *b*, then A = B.

As null *W* ⊆⊆ *A*, A is an affine subspace of *A* with the displacement vector *<sup>v</sup>*.

Using the SVD of *W* as above, we can express the solution to the preimage problem by right singular vectors corresponding to zero singular values of *W* as

$$\left\{ v' + \sum\_{i=k+1}^{L} c\_i z\_i \, \middle| \, c\_i \in \mathbb{R} \right\}.$$

The dimension of that affine subspace is then dim(null *W*) which is zero (and hence exists only one vector solving the preimage problem) if and only if *k* = *l* = *L*, i.e., the closeness matrix *W* is square and regular. If *W* is rectangular, we have infinitely many vectors *a* ∈ *A* solving the preimage problem.

**Corollary 2.** *Any weighted mean of elements of the set of all vectors solving the preimage problem with respect to weights with nonzero sum is again an element of this set.*

**Proof.** Let *n* ∈ N and ∀*i* = 1, ... , *n*: *Wv<sup>i</sup>* = *b*, i.e., each *vi* ∈ *A* be a solution to the preimage problem. Let *c*1,..., *cn* ∈ R, s.t. *n* ∑ *i*=1*ci* = 0. Then it holds that

$$\mathcal{W}\left(\frac{\sum\_{i=1}^{n}c\_{i}\upsilon^{i}}{\sum\_{i=1}^{n}c\_{i}}\right) = \frac{\sum\_{i=1}^{n}c\_{i}\mathcal{W}\upsilon^{i}}{\sum\_{i=1}^{n}c\_{i}} = \frac{\left(\sum\_{i=1}^{n}c\_{i}\right)b}{\sum\_{i=1}^{n}c\_{i}} = b\_{\mathcal{A}}$$

which proves the claim.

In this subsection, we described the solution to the preimage problem from three different perspectives: using weighted means of functional values (determining an equivalence class of mutually similar functions), using right inverses of the closeness matrix (that can be obtained also by SVD) and using the notion of affine subspace (its displacement vectors can be also obtained by SVD of *W*, based on Theorem 1).

In the next section, we show the connection between the inverse F-transform and a solution to the preimage problem as the rows of *W* are formed by the basic functions of the fuzzy partition.
