Linear Operator on the Plane
From ProofWiki
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 eqns:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \phi \left({e_1}\right)\) | \(=\) | \(\displaystyle \alpha_{11} e_1 + \alpha_{21} e_2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \phi \left({e_2}\right)\) | \(=\) | \(\displaystyle \alpha_{12} e_1 + \alpha_{22} e_2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
where $\left({e_1, e_2}\right)$ is the Standard Ordered Basis of $\R^2$.
Then, by linearity:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \phi \left({\lambda_1, \lambda_2}\right)\) | \(=\) | \(\displaystyle \phi \left({\lambda_1 e_1 + \lambda_2 e_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lambda_1 \phi \left({e_1}\right) + \lambda_2 \phi \left({e_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12} }\right) e_1 + \left({\lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) e_2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
- 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 $\R^2$:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle b \cdot \phi \left({\lambda_1, \lambda_2}\right) + c \cdot \phi \left({\lambda_3, \lambda_4}\right)\) | \(=\) | \(\displaystyle b \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) + c \left({\lambda_3 \alpha_{11} + \lambda_4 \alpha_{12}, \lambda_3 \alpha_{21} + \lambda_4 \alpha_{22} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} }\right) + \left({c \lambda_3 \alpha_{11} + c\lambda_4 \alpha_{12}, c\lambda_3 \alpha_{21} + c\lambda_4 \alpha_{22} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12} + c \lambda_3 \alpha_{11} + c \lambda_4 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} + c \lambda_3 \alpha_{21} + c \lambda_4 \alpha_{22} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({\left({b \lambda_1 + c \lambda_3}\right) \alpha_{11} + \left({b \lambda_2 + c \lambda_4}\right) \alpha_{12}, \left({b \lambda_1 + c \lambda_3}\right) \alpha_{21} + \left({c \lambda_2 + c \lambda_4}\right) \alpha_{22} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({b \lambda_1 + c \lambda_3, b \lambda_2 + c \lambda_4}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965): $\S 28$: Example $28.1$