I am trying to find out if the following function is a linear transformation from $\mathbb{R^2}$ $\to$ $\mathbb{R^3}$.
T: $\mathbb{R^2}$ $\to$ $\mathbb{R^3}$, T $\left(
\begin{bmatrix}
1 \\
1 \\
\end{bmatrix}
\right) =$
$\begin{bmatrix}
2 \\
0 \\
2 \\
\end{bmatrix}$
I know for a function to be a linear transformation the following two conditons must hold:
Given u,v $\in M_{2,1}$ (The 1st vector space, an analog for $\mathbb{R^2}$) & c$\in \mathbb{R}$ then:
- T(u + v) = T(u) + T(v)
- T(cu) = cT(u)
I claimed the function was indeed a linear transformation and denoted u,v as:
u =
$\begin{bmatrix}
u_1 \\
u_2 \\
\end{bmatrix}$
and v =
$\begin{bmatrix}
v_1 \\
v_2 \\
\end{bmatrix}$
I then showed T(u + v) = T(u) + T(v)
$\to T(u + v) = T\left(
\begin{bmatrix}
u_1 + v_1 \\
u_2 + v_2 \\
\end{bmatrix}\right)
= \begin{bmatrix}
2(u_1+v_1) \\
0 \\
2(u_2+v_2) \\
\end{bmatrix}$
$\to T(u) = T\left(
\begin{bmatrix}
u_1 \\
u_2 \\
\end{bmatrix} \right)
=
\begin{bmatrix}
2(u_1) \\
0 \\
2(u_2) \\
\end{bmatrix},
T(v) = \left(
\begin{bmatrix}
v_1 \\
v_2 \\
\end{bmatrix} \right)
=
\begin{bmatrix}
2(v_1) \\
0 \\
2(v_2) \\
\end{bmatrix}$
By inspection $T(u,v) = T(u) +T(v) \implies$ Condition 1 holds
Next I showed T(cu) = cT(u)
$\to T\left( c
\begin{bmatrix}
u_1 \\
u_2 \\
\end{bmatrix} \right) =
T\left(
\begin{bmatrix}
c(u_1) \\
c(u_2) \\
\end{bmatrix} \right) =
\begin{bmatrix}
2c(u_1) \\
0 \\
2c(u_2) \\
\end{bmatrix}$
$ \to cT \left(
\begin{bmatrix}
u_1 \\
u_2 \\
\end{bmatrix} \right) =
c
\begin{bmatrix}
2(u_1) \\
0 \\
2(u_2) \\
\end{bmatrix}
=
\begin{bmatrix}
2c(u_1) \\
0 \\
2c(u_2) \\
\end{bmatrix}$
Thus T(cu) = cT(u) $\implies$ Condition 2 holds
I think this is right, but I'd like to have confirmation regarding whether the transformation is indeed linear and if the way in which I proved it was sufficiently rigorous.
Best Answer
You can't define a linear transformation from $\mathbb{R}^2$ to another space by defining only the image of one vector. You have to define it on a basis of the vector space.
For sure you can define a linear transformation $T$ from $\mathbb{R}^2$ to $\mathbb{R}^3$ such that $T\bigl([1,1]\bigr)=[2,0,2]$. It is sufficient to complete $[1,1]$ to a basis of $\mathbb{R}^2$ with another vector $v$, linearly independent from $[1,1]$ (for instance with the vector $[1,0]$) and define the image of $v$. In whatever way you define the image of the second vector of the basis you obtain a linear function.
Of course the way you define $T$ on another basis' vector will change the function, but for sure it will be a linear function such that $T\bigl([1,1]\bigr)=[2,0,2]$.
What you did while trying to verifying the linearity of your function is forcing other properties to the function: why should $T([u_1+v_1, u_2+v_2])=[2(u_1+v_1), 0, 2(u_2+v_2)]$? By the only property you stated at the beginning you don't have this equality. A function which maps $[1,1]$ to $[2,0,2]$ is not forced to be linear. Think for instance to the constant function $T(v)=[2,0,2] \ \forall \ v \in \mathbb{R}^2$.
An example of a linear function $T$ which satisfies your property is the following:
Take the standard basis of $\mathbb{R}^2$, let's say $B=\bigl\{[1,0],[0,1]\bigr\}$. If you define $T$ by $T\bigl([1,0]\bigr)=[2,0,0]$ and $T\bigl([0,1]\bigr)=[0,0,2]$, then, by linearity of $T$ you have: $$T\bigl([1,1]\bigr) \ = \ T\bigl([1,0]\bigr)+T\bigl([0,1]\bigr) \ = \ [2,0,0]+[0,0,2]=[2,0,2] \ .$$
This function is represented by the following matrix (the columns are the images of the vectors of the standard basis): $$\left[\begin{matrix} 2 & 0 \\ 0 & 0 \\ 0 & 2 \end{matrix} \right] \ .$$
You can also swap the images of the two basis vectors and you obtain the same result but a different function.