# Are linear operator just projections in projective geometry?

If we view (linear) operators as nothing than multipliers multiply their eigenvectors with numbers

$$Tu=λu$$

then aren't they the same with projections from a finite point in projective geometry?

There are projection operators in functional analysis projecting the whole space into its subspace, but they're not linear, while this projection is analogous to the definition of linear operator and it stretches the whole space. The Möbius transform is linear transform on homogeneous coordinates in $\Bbb C^2$.

Can I see linear operators as protections in projective geometry without trouble?

*What is the significance of differential operators over other operators in group theory?

• I don't think that this goes anywhere, frankly – Giuseppe Negro Nov 25 '17 at 10:39
• Can you elaborate more? Thanks – Ooker Nov 25 '17 at 11:50
• No, in projective geometry an eigenvector is a fixed point. You can still have interesting dynamics. – Max Nov 28 '17 at 4:09
• @Max What interesting dynamics are you talking with? – Ooker Nov 28 '17 at 4:31

Let's consider a simple example. A generic $2 \times 2$ matrix $A$ has $2$ distinct eigenvalues, so let's say $Av=\lambda v$ and $Aw=\sigma w$ with $\lambda \ne \sigma$, hence $v \ne w$. That means that $A$ dilates the eigenlines, that is, $A$ scales the span of $v$ by $\lambda$ and it scales the span of $w$ by $\sigma$. In projective space $\mathbb{P}^1$, these lines are now two fixed points. If $|\lambda|>|\sigma|$ then the span of $v$ is an attracting fixed point, while if $|\lambda|<|\sigma|$ the span of $v$ is a repelling fixed point.