# 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 tuple 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:

 $\displaystyle \phi \left({e_1}\right)$ $=$ $\displaystyle \alpha_{11} e_1 + \alpha_{21} e_2$ $\displaystyle \phi \left({e_2}\right)$ $=$ $\displaystyle \alpha_{12} e_1 + \alpha_{22} e_2$

where $\left({e_1, e_2}\right)$ is the standard ordered basis of $\R^2$.

Then, by linearity:

 $\displaystyle \phi \left({\lambda_1, \lambda_2}\right)$ $=$ $\displaystyle \phi \left({\lambda_1 e_1 + \lambda_2 e_2}\right)$ $\displaystyle$ $=$ $\displaystyle \lambda_1 \phi \left({e_1}\right) + \lambda_2 \phi \left({e_2}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12} }\right) e_1 + \left({\lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) e_2$ $\displaystyle$ $=$ $\displaystyle \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right)$

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 $\R^2$:

 $\displaystyle b \cdot \phi \left({\lambda_1, \lambda_2}\right) + c \cdot \phi \left({\lambda_3, \lambda_4}\right)$ $=$ $\displaystyle b \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) + c \left({\lambda_3 \alpha_{11} + \lambda_4 \alpha_{12}, \lambda_3 \alpha_{21} + \lambda_4 \alpha_{22} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} }\right) + \left({c \lambda_3 \alpha_{11} + c\lambda_4 \alpha_{12}, c\lambda_3 \alpha_{21} + c\lambda_4 \alpha_{22} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12} + c \lambda_3 \alpha_{11} + c \lambda_4 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} + c \lambda_3 \alpha_{21} + c \lambda_4 \alpha_{22} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({b \lambda_1 + c \lambda_3}\right) \alpha_{11} + \left({b \lambda_2 + c \lambda_4}\right) \alpha_{12}, \left({b \lambda_1 + c \lambda_3}\right) \alpha_{21} + \left({c \lambda_2 + c \lambda_4}\right) \alpha_{22} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({b \lambda_1 + c \lambda_3, b \lambda_2 + c \lambda_4}\right)$

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 tuple:

$\left({\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22}}\right)$

of real numbers.

$\blacksquare$