Motivation
I’ve recently came upon an interesting problem in Axler’s Linear Algebra Done Right:
Given \(\phi \in \mathcal{L}(V, \mathbb{F})\) and \(u \in V\) not in \(\text{null }\phi\) show that \(V = \text{null }\phi \oplus \{au | a \in \mathbb{F}\}\).
The proof is quite straightforward. Since \(\phi u \neq 0\), \(\phi u = c \in \mathbb{F}\) is a basis of \(\mathbb{F}\) (notice that now we’re talking about \(\mathbb{F}\) as vector space, not a field). So, for any \(v \in V\), we have that \(\phi v = bc = b \phi u = \phi (bu)\) for some \(b \in \mathbb{F}\). Thus, \(0 = \phi v - \phi (bu) = \phi(b - bu)\) which means that \(v - bu = v_{\text{null}} \in \text{null } \phi\).
This proves that \(v \in \text{null }\phi + \{au | a \in \mathbb{F}\}\). Moreover, if \(v_\text{null} + au = 0\) for some \(v_\text{null} \in \text{null }\phi\), then \(\phi(v_\text{null} + au) = a\phi u = ac = 0\), so \(a = 0\) and \(v_\text{null} = 0\), meaning that we have a direct sum.
It’s a known fact that for any given subspace \(U\) of a finite-dimensional space \(V\), it is always possible to find a subspace \(W\) such that \(V = U \oplus W\) (in some sense this subspace is complementary to \(U\)). Hence, it must be possible to find a subspace complementary to \(\text{null } T\) for any linear map. This proof above can be generalized to any one-dimensional space, i.e. to a map in \(\mathcal{L}(V, W)\), where \(\text{dim } W = 1\). It is not clear however how to complement the null space in other cases. So here I’ll show one way of doing that. The decomposition we get as a result, can be applied to another problem which I’ll also discuss.
The decomposition
Assume that \(W\) is finite-dimensional, and we have a linear map \(T \in \mathcal{L}(V, W)\). Since \(W\) is finite-dimensional, \(\text{range } T\) is also finite-dimensional. Let \(w_1, \dots, w_n\) be the basis of \(\text{range }T\). There are also some \(v_i \in V\) such that \(Tv_i = w_i\).
With this setup out of the way, we can state the fact that’s the highlight of this post: \[V = \text{null }T \oplus \text{span}(v_1, \dots, v_n).\] Let’s look at the proof (in essence very similar to the one we’ve seen before).
For any \(v \in V\), we have that [\(i\) is the summation index] \[Tv = \sum a_i w_i = \sum a_i T v_i = T\left( \sum a_i v_i \right).\] This implies that \[0 = Tv - T\left( \sum a_iv_i \right) = T \left( v - \sum a_i v_i \right).\] Hence, \[v - \sum a_i v_i \in \text{null }T,\] so \(v = v_\text{null} + \sum a_i v_i\), for some \(v_\text{null} \in \text{null }T\); in other words, \(v \in \text{null }T + \text{span}(v_1, \dots, v_n)\).
To show that this is a direct sum, suppose that \(v_\text{null} + \sum a_i v_i = 0\). Applying \(T\) to both sides gives \[T(v_\text{null} + \sum a_i v_i) = 0,\] which implies (by linearlity of \(T\)) that \[\sum a_i T v_i = \sum a_i w_i = 0.\] Since \(w_1, \dots w_n\) is linearly independent, for all \(i\), \(a_i = 0\). This means that the “span element” is \(0\) and so is \(v_\text{null}\), so this is indeed a direct sum.
Application
This decomposition can be used to prove the following statement:
Assume that \(W\) is finite dimensional, and \(T_1, T_2 \in \mathcal{L}(V, W)\), then \(\text{null }T_1 = \text{null }T_2\) if and only if there exists an invertible \(S \in \mathcal{L}(W)\) such that \(T_1 = ST_2\).
The “if” part is very easy to prove: let \(v \in \text{null }T_1\), then \(0 = T_1v = (ST_2)v = S(T_2v)\). Since \(S\) is injective, \(\text{null }S = \{0\}\), so \(T_2v = 0\), showing that \(v \in \text{null }T_2\). The proof for the inclusion \(\text{null }T_2 \subseteq \text{null }T_1\) is completely analogous.
Now for the juicy part.
Just as in the proof for the direct sum decomposition, we let \(w_1, \dots, w_n\) be the basis of \(\text{range }T_1\) and \(v_i \in V\) be such that \(T_1v_i = w_i\). It’s not hard to prove that \(T_2v_1, \dots, T_2v_n\) is linearly independent. Indeed, from the assumption that there are some \(a_i\)’s such that \(\sum a_i T_2 v_i = T_2 \left( \sum a_i v_i \right) = 0\), we get that \(\sum a_i v_i \in \text{null }T_2 = \text{null }T_1\), so \[\sum a_i v_i \in \text{null } T_1 \cap \text{span}(v_1, \dots, v_n) = \{0\}.\] Hence, \(\sum a_i v_i = 0\) which implies that \(T_1 \left( \sum a_i v_i \right) = \sum a_i w_i = 0\). Now, linear independence of \(w_i\)’s forces each \(a_i\) to be \(0\) (by the way, as a byproduct we get that \(v_i\)’s are also linearly independent).
Since \(T_2v_1, \dots, T_2v_n\) is linearly independent in \(W\), it can be extended to a basis of \(W\) as \(T_2v_1, \dots, T_2v_n, e_1, \dots, e_k\). As another basis of \(W\) we have \(w_1, \dots, w_n, f_1, \dots, f_k\). We can define \(S \in \mathcal{L}(W)\) in the following way: \[S(T_2v_i) = w_i \text{ and } Se_i = f_i.\] Now, for any \(v \in V\), using our decomposition we have that \(v = v_\text{null} + \sum a_i v_i\), and \[(ST_2)v = S\left(T_2\left( v_\text{null} + \sum a_i v_i\right)\right) = S \left( \sum a_i T_2 v_i \right) = \sum a_i S(T_2 v_i) = \sum a_i S(T_2 v_i).\] Using the definition of \(S\), the sum above is equal to \[\sum a_i w_i = \sum a_i T_1 v_i = T_1 \left( \sum a_i v_i \right) = T_1\left(v_\text{null} + \sum a_i v_i \right) = T_1v.\]
This proves that \(ST_2 = T_1\). All we have to do now is show that \(S\) is invertible.
For that we need a short lemma:
If \(V = \text{span}(v_1, \dots, v_n), T \in \mathcal{L}(V, W)\), then \(\text{span}(Tv_1, \dots, Tv_n) = \text{range }T\).
For \(w \in \text{span}(Tv_1, \dots, Tv_n)\), \(w = \sum a_i Tv_i = T\left(\sum a_i v_i\right)\), so \(w \in \text{range }T\).
For \(w \in \text{range }T\), there exists \(v \in V\) such that \(w = Tv\), and \(v = \sum a_i v_i\) for some \(a_i\)’s. So \(w = T\left( \sum a_i v_i \right) = \sum a_i T v_i \in \text{span}(Tv_1, \dots, Tv_n)\).
Back to the invertibility argument. Since \(W = \text{span}(T_2v_1, \dots, T_2v_n, e_1, \dots e_k)\), using the lemma above we get \[\text{span}(S(T_2v_1), \dots S(T_2v_n), Se_1, \dots, Se_k) = \text{span}(w_1, \dots, w_n, f_1, \dots, f_k) = \text{range }S.\] This proves that \(W = \text{range }S\), i.e. that \(S\) is surjective, and since \(S\) is an operator, this equivalent to \(S\) being invertible.