Suppose that \(T:U \rightarrow V\) is a linear transformation. Say two vectors from \(U\), \(x\) and \(y\), are related exactly when \(T(x)=T(y)\).
We are asked to prove that related is an equivalence relation on \(U\). Let us denote the relation as \(x \sim y\). The underlying reason this works is because related depends on equality being an equivalence relationship in a vector space.
For any \(x \in U\), we must prove \(x \sim x\). It follows because \(T(x) = T(x)\).
If \(x \sim y\) for \(x,y \in U\), then we must prove \(y \sim x\). For our assumption, \(T(x) = T(y)\). By symmetry of equality in \(V\), this implies \(T(y) = T(x)\) which implies \(y \sim x\).
If \(x \sim y\) and \(y \sim z\) for \(x,y,z \in U\), then we obtain \(T(x) = T(y)\) and \(T(y)=T(z)\). By transitivity of equality in \(V\), we obtain \(T(x) = T(z)\). This implies \(x \sim z\) which proves the result.
We will prove that the pre-images of a linear transformation \(T:U \rightarrow V\) is a partition on \(U\).
The first step is to prove every \(x \in U\) is contained in some pre-image. Since \(T(x)\) is defined for every \(x\), this implies that \(x\) belong to its own pre-image \(T^{-1}(T(x)) = { u \in U | T(u) = T(x)}\).
The second step is that two different pre-images do not have any elements in common. Suppose two pre-images are both non-empty. Then \(A = T^{-1}(a)\) and \(B = T^{-1}(b)\) for \(a,b \in V\). Moreover, we assume $ A B$. This implies there exists \(x\) where \(x \in A \triangle B \neq \emptyset\). Without loss of generality, we can assume \(x \in A - B\). This means \(T(x) = a\) but \(T(x) \neq b\). Therefore, \(a \neq b\). Now we will show the intersection of two non-empty images is empty. We argue by contradiction. Suppose there is \(c \in A \cap B \neq \emptyset\). Then \(T(c) = a\) since \(c\) in the preimage of \(a\). Likewise, \(T(c) = b\). By transitivity, \(a = b\). But this contradicts our earlier result that \(a \neq b\). This completes the proof.