Uniqueness of a linear transformation?

Question

I was requested to decide whether there is some linear transformation $f: \mathbb{R}^2 \rightarrow \mathbb{R}^4$ satisfying

\begin{equation*} \begin{cases} f(0, 1) &= (1, 2, 0, 0) \\ f(1, 0) &= (1, 1, 0, 0) \end{cases} \end{equation*}

and, if it exists, whether it is unique or not.

It seems perfectly possible that such linear transformation exists. If we assume $f$ is a linear transformation satisfying the aforementioned properties, then it is such that any $f(\textbf{x}) \in \text{Im}(f)$ is of the form

\begin{align*} f(\textbf{x}) &= f\big(x_1(0, 1) + x_2(1, 0)\big) \\ &= x_1f(0, 1) + x_2f(1, 0) \\ &= x_1(1, 2, 0, 0) + x_2(1, 1, 0, 0) \end{align*}

From this would follow $\text{Im}(f) = \{x_1(1, 2, 0, 0) + x_2(1, 1, 0, 0) \mid x_i \in \mathbb{R}\}$, which on its turn implies $\text{Im}(f)$ is isomorphic to $\mathbb{R}^2$. In short, the transformation turns out to be

\begin{align*} f(\textbf{x}) &= (x_1, 2x_2, 0, 0) + (x_2, x_2, 0, 0) \\ &= (x_1 + x_2, 2x_2 + x_2, 0, 0) \end{align*}

We have shown that a linear transformation $f$ satisfies the given equalities if and only if $f(\textbf{x}) = (x_1 + x_2, 2x_2 + x_2, 0, 0)$. Hence $f$ is unique.

I : Is this solution correct? II : What would be an example of non-uniqueness? Restraining ourselves to the same spaces of the problem: what $f_1, f_2$ mapping $\mathbb{R}^2 \rightarrow \mathbb{R}^4$ are such that $f_1(\textbf{x}_1) = f_2(\textbf{x}_1), ..., f_1(\textbf{x}_n) = f_2(\textbf{x}_n)$ for some set of $n$ vectors $\mathbf{x} \in \mathbb{R}^2$? Are such mappings special in any relevant sense, or studied for some particular purpose?

Answer 1

Generally speaking, let $U$ and $V$ be two linear spaces and $\{e_1,\text{ }\cdots,\text{ }e_n\}$ a basis of $U$. For any mapping $f:\{e_1,\text{ }\cdots,\text{ }e_n\}\rightarrow V$, $\exists!$ a linear mapping $\bar{f}:U\rightarrow V$ s.t. $\bar{f}(e_i)=f(e_i)$, $i\in\{1,\text{ }\cdots\, \text{ }n\}$

For the existence we directly construct a $\bar{f}$. Define the image of $x\in U$ as follows : if $x=x_1e_1+\text{ }\cdots+\text{ }x_ne_n$, then $\bar{f}(x)=x_1f(e_1)+\cdots+x_nf(e_n)$. You can check $\bar{f}$ is linear and coincides with $f$ on the basis vectors.

For the uniqueness, if there exists $g: U\rightarrow V$ which also coincides with $f$ on the basis vectors, then it suffices to show $\psi(x)=\bar{f}(x)-g(x)$ is always $0$ for all $x\in U$. In fact $\psi(e_i)=\bar{f}(e_i)-g(e_i)=0$ $\forall i$, hence $\forall x\in U$, where $x=x_1e_1+\text{ }\cdots+\text{ }x_ne_n$, $\psi(x)=x_1\psi(e_1)+\cdots+x_n\psi(e_n)=0$ as desired.