# A line through fixed points, in $\mathbb{P^2}$ , is a fixed line or an invariant one?

An exercise asks me if a projectivity, represented by a matrix $A \in M(3,3, \mathbb{R})$, with parameter, has a line composed by fixed points.

Now, from the theory we know that through a fixed point, which is an eigenvector of the transformation which induces the projectivity, passes a bundle of invariant lines. To have a fixed line we need an eigenspace of dimension 2, and this projectivity $\phi_a$ doesn't for $a \neq \pm 1$.

However $\phi_a$ has two eigenvectors associated two different eigenvalues, the first question is: a line that passes through two eigenvectors, even if from two different eigenspaces, is a fixed line or an invariant one?

About the bundle of invariant lines through a fixed point (let $P$ be this). If $\mathcal{F_P}$ is this boundle centered in the $P$, its general equation is $\alpha x_1 + \beta x_2 =0$ s.t. $[\alpha, \beta] \in \mathbb{R}$, for duality it is a line in $\mathbb{(P^2)^*}$. This line is represented in the dual projective space by the restriction to $\mathcal{F_P}$ of the matrix $^tA^{-1}$.

The second question is: fixing a point of the line in the dual space, choosing a parameter for which the eigenvalues of that matrix are all multiples, non zero, of a couple of coordinates $[\alpha, \beta]$; do I fix a line or an invariant one?

For the sake of completeness, the text of this exercise is:

Consider $T_a: \mathbb{R^3} \to \mathbb{R^3}$ represented by:

$A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 1-a \\ a-1 & 0 & a \\ \end{bmatrix}$, $a \in \mathbb{R}$

When $T_a$ induces a projectivity $\phi_a: \mathbb{P^2} \to \mathbb{P^2}$. Does a fixed line, point to point, for some $a \in \mathbb{R}$, exist? If yes, give an explicit value of $a \in \mathbb{R}$ and write the equation of such line.

The characteristic polynomial is $(\lambda -1)^2(\lambda - a)$. The eigenvector for $\lambda =1$ is $v_{\lambda_1} = (0, 1, 0)$; the eigenvector for $\lambda = a$ is $v_{\lambda_2} = (0, -1, 1), \forall a \in \mathbb{R}\setminus \{\pm 1\}$. Finally $Det(A) = a$, so it is $0$ only when $a=0$ .

Thank you.

• The eigenspace of $1$ is two-dimensional when $a=\pm1$. – amd May 21 '18 at 20:52
• @amd I am feeling really stupid not having considered $\lambda =1, \lambda =-1$. Thank you. By the way the questions remain. – F.inc May 21 '18 at 21:35
• I think you can answer the first one for yourself: the line through points $\mathbf p$ and $\mathbf q$ is their join $\lambda\mathbf p+\mu\mathbf q$. What is the result of applying $T_a$ to this when $\mathbf p$ and $\mathbf q$ are eigenvectors? – amd May 22 '18 at 6:47