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 equations:

$$ $$

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.