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, \(\blacksquare\)