Linear Operator on the Plane

Theorem
Let $\phi$ be a linear operator on the real vector space of two dimensions $\R^2$.

Then $\phi$ is completely determined by an ordered sequence of $4$ real numbers.

Proof

 * Let $\phi$ be a linear operator on $\R^2$.

Let $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ be the real numbers which satisfy the eqns:

where $\left({e_1, e_2}\right)$ is the Standard Ordered Basis of $\R^2$.

Then, by linearity:


 * Conversely, if $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ are any real numbers, then we can define the mapping $\phi$ as:
 * $\phi \left({\lambda_1, \lambda_2}\right) = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22}}\right)$

which is easily verified as being a linear operator on $\R^2$:

Thus, by Condition for Linear Transformation, $\phi$ is a linear operator on $\R^2$.

Thus each linear operator on $\R^2$ is completely determined by the ordered sequence $\left({\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22}}\right)$ of real numbers.