3.E Problem 1

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