Suppose \(T\) is a function from \(V\) to \(W\). Let the “graph” of \(T\) be the subset of \(V \times W\) such that:
\begin{equation} graph\ T = \{(v,Tv) \in V \times W \mid v \in V\} \end{equation}
Show that \(T\) is a linear map IFF the graph of \(T\) is a subspace of \(V \times W\).
Review: A Linear Map
Recall that a function \(T: V \to W\) is called a linear map if it is a map that…
- is additive: so \(Tv + Tu = T(v+u): v,u \in V\)
- is homogeneous, so \(\lambda Tv = T\lambda v: \lambda \in \mathbb{F}, v \in V\)
Given Graph is Subspace
Given the graph of \(T\) is a subspace of \(V \times W\), we desire that the function \(T\) is a linear map and therefore additive and homogeneous.
By declaration before, \(graph\ T\) is a subspace, meaning it would be closed under adddition and scalar multiplication. We will use this fact to show that \(T\) follows the properties of a linear map.
Additivity
We first desire that \(T\) is additive, that is, for \(v,u \in V\), we desire \(Tv + Tu = T(v+u)\).
Let \(v,u \in V\), and let \(a,b \in graph\ T\) declared as follows:
\begin{equation} \begin{cases} a = (v,Tv) \in V \times W \\ b = (u,Tu) \in V \times W \end{cases} \end{equation}
We are given that \(graph\ T\) is a subspace of \(T\). As such, it is closed under addition; meaning, the sum of two elements from the space must remain in the space. Therefore:
\begin{equation} (v, Tv) + (u,Tu) = (v+u, Tv+Tu) \in graph\ T \end{equation}
And now, the latter being in \(graph\ T\) implies that \(\exists\) some \(c \in graph\ T\), \(n \in V\) such that:
\begin{equation} c := (n, Tn) = (v+u, Tv+Tu) \end{equation}
Taking the latter equivalence and solving for \(n\), we have that \(n = v+u\). And so, we have that:
\begin{equation} (v+u, T(v+u)) = (v+u, Tv+Tu) \end{equation}
Therefore, \(T(v+u) = Tv+Tu\), as desired.
Homogeneity
We now desire that \(T\) is homogeneous. That is, for \(v \in V, \lambda \in \mathbb{F}\), we desire \(\lambda Tv = T\lambda v\).
Let \(v \in V\), \(\lambda \in \mathbb{F}\), and \(a \in graph\ T\) declared as follows:
\begin{equation} a = (v, Tv) \in V \times W \end{equation}
By the same logic before, \(graph\ T\) is closed under scalar multiplication; meaning, the product of en element from the space to a scalar remain in the space. Therefore:
\begin{equation} \lambda (v,Tv) = (\lambda v, \lambda Tv) \in graph\ T \end{equation}
The latter being in \(graph\ T\) implies that \(\exists\) some \(c \in graph\ T\), \(n \in V\) such that:
\begin{equation} c :=(n,Tn) = (\lambda v, \lambda Tv) \end{equation}
Taking the latter equivalence and solving for \(n\), we have \(n = \lambda v\). And so, we have:
\begin{equation} (\lambda v, T \lambda v) = (\lambda v, \lambda Tv) \end{equation}
And therefore, \(T\lambda v = \lambda Tv\), as desired.
Having shown that \(T\) is now both additive and homogeneous, we have that \(T\) is a linear map, as desired.
Given \(T\) is a Linear Map
We will essentially prove the previous condition backwards.
We are given that the graph of \(T\) is a subset of \(V \times W\), and that \(T: V \to W\) is a linear map. We desire that the graph of \(T\) is a subspace of \(V \times W\).
Recall that to show that a subset is a subspace, on simply has to show that it has closed operations and that it contains the additive identity.
Additive Identity
Recall that the additive identity in \(V \times W\) is the tuple that’s identically \((0,0) \in V \times W\).
As \(V\) is a vector space, \(0 \in V\). Any linear map will send \(0\) to \(0\). Therefore, \(T 0 = 0\).
Therefore, construct \(a \in graph\ T\):
\begin{equation} a = (0, T 0) \in V \times W = (0, 0) \end{equation}
By construction, we have shown that the additive identity of \(V \times W\) is in \(graph\ T\).
Closure of Addition
Given WLOG \(a,b \in graph\ T\), we desire that \(a+b \in graph\ T\).
Let \(v,u \in V\), and let \(a,b \in graph\ T\) declared as follows:
\begin{equation} \begin{cases} a = (v,Tv) \in V \times W \\ b = (u,Tu) \in V \times W \end{cases} \end{equation}
Now:
\begin{equation} a+b = (v,Tv) + (u+Tu) = (v+u, Tv+Tu) \end{equation}
Given \(T\) is a linear map, we have WLOG \(Tv+Tu = T(v+u)\). And therefore:
\begin{equation} a+b = (v,Tv) + (u+Tu) = (v+u, Tv+Tu) = (v+u, T(v+u)) \in \{(v,Tv) \mid v \in V\} \end{equation}
Hence, \(graph\ T\) is closed under addition.
Closure of Scalar Multiplication
Given WLOG \(a \in graph\ T, \lambda \in \mathbb{F}\), we desire that \(\lambda a \in graph\ T\).
Let \(v \in V, \lambda \in \mathbb{F}\), and let \(a \in graph\ T\) declared as follows:
\begin{equation} a = (v,Tv) \in V \times W \end{equation}
Now:
\begin{equation} \lambda a = \lambda (v,Tv) = (\lambda v, \lambda Tv) \end{equation}
Given \(T\) is a linear map, we have WLOG \(\lambda Tv = T\lambda v\). And therefore:
\begin{equation} \lambda a = \lambda (v,Tv) = (\lambda v, \lambda Tv) = (\lambda v, T \lambda v)\in \{(v,Tv) \mid v \in V\} \end{equation}
Hence, \(graph\ T\) is closed under scalar multiplication.
Having shown \(graph\ T\) to be closed under addition and scalar multiplication, as well as containing the additive identity, we see that it is a subspace of \(V \times W\) of which it is a subset.
Having shown both directions of the proof, \(\blacksquare\)