Prove or disprove that the statement P implies Q (and conversely) for T: $\mathbb R^2 \to\mathbb R^2$ be a linear transformation. Question: Let $T:\mathbb R^2\to \mathbb R^2$ be a linear transformation and let A be the matrix representation of T with respect to the standard basis of $\mathbb R^2$.Consider two statements:

(P): There are exactly two distinct lines $L_1 ,L_2 \in \mathbb R^2$ passing through the origin that are mapped onto themselves: $T(L_1)=L1$ and $T(L_2)=L_2$.


(Q):The matrix A has two distinct nonzero real eigenvalues.

My Attempt:
Given that $T:\mathbb R^2 \to \mathbb R^2$ is a linear transformation and A is its matrix representation then the lines passing through origin will look like $y=mx$, when it comes to "mapping onto itself", I thought about $T(x)=x$ and $T(-x)=-x$.
Clearly we have that $T^n(x)=x$ and $T^n(-x)=-x$ for all n.Thus I see only these two distinct lines mapping onto "themselves" and they have $1$ and $-1$ eigenvalues respectively (So (P) implies (Q)) . Can we have a counter for Q implies P? I didn't got that. Any hint for constructing the theoretical proofs?Thanks.
 A: Suppose that $T$ has two distinct real eigenvalues, $\alpha$ and $\beta$. Let $v$ be an eigenvector corresponding to $\alpha$ and let $w$ be en eigenvalue corresponding to $\beta$. Now, let $L_1=\{\lambda v\mid\lambda\in\Bbb R\}$ and let $L_2=\{\lambda w\mid\lambda\in\Bbb R\}$. Then $L_1$ and $L_2$ are two distinct lines passing through the origin, $T(L_1)\subset L_1$ and $T(L_2)\subset L_2$.
Now, suppos that there are two distinct lines passing through the origin such that $T(L_1)\subset L_1$ and $T(L_2)\subset L_2$, and that these are the only lines passing throu the origin with this property. If $v\in L_1\setminus\{0\}$, then $T(v)\in L_1$, which means that $T(v)=\alpha v$, for some real number $\alpha$. So, $\alpha$ is an eigenvalue of $T$. And if $w\in L_1\setminus\{0\}$, then $T(w)\in L_2$, which means that $T(v)=\beta v$, for some real number $\beta$. So, $\beta$ is an eigenvalue of $T$. If $\alpha=\beta$, then $T=\alpha\operatorname{Id}$, and therefore, for every line $L$ passing through the origin, $T(L)=L$. But we are assuming that no such line exists other than $L_1$ and $L_2$. So, $\alpha\ne\beta$.
