If $L \in GL_n(\mathbb{Q})$ such that $L^{-1} = L + L^2$ then $3\mid n$ Suppose $L \in GL_n(\mathbb{Q})$, i.e, a linear invertible map on $\mathbb{Q}^n$. Then  I want to see that if $L^{-1} = L + L^2$ then $3\mid n$.
I have no idea how to start this.
 A: It's easy to see that the minimal polynomial of such a linear transformation is $X^3+X^2-1$, which is irreducible over $\mathbb{Q}$.  Since the minimal and characteristic polynomials have the same set of roots, the characteristic polynomial must be $(X^3+X^2-1)^k$ for some $k$, so its degree is $n=3k$.
A: Let $f = x^3 + x^2 - {1}$. The polynomial $f$ is irreducible over $\mathbb{Q}$, and it has three roots - one real and two complex conjugate.
All the eigenvalues of $L$ are exactly the roots of $f$.
Let $\sigma$ be an element of the Galois group of the splitting field of $f$ (which is isomorphic to $S_3$), we extend it's action to matrices naturally. Certainly $\sigma(L)=L$.
If $D$ is any diagonalisation of $L$, then $\sigma(D)$ also is one:
$$
D = PLP^{-1} \implies \sigma(D) = \sigma(P) L \sigma(P)^{-1}.
$$
Now let $\lambda_0, \lambda_1, \lambda_2$ be the roots of $f$, and let $a_i$ count the number of eigenvalues of $D$ of each type, i.e.
$$
a^D_i = \text{multiplicity of $\lambda_i$ as eigenvalue of $D$ }.
$$
We call $\sigma(i)$ the index of $\sigma(\lambda_i)$, i.e. $\sigma(\lambda_i) = \lambda_{\sigma(i)}$. Then
$$
a^D_i = a^{\sigma(D)}_{\sigma(i)}
$$
However, the values $a^D_i$ are independent of $D$, they only depend on $L$. Therefore we have $a_i = a_{\sigma{(i)}}$ for all $i=0,1,2$.
Since the Galois group is transitive, we must have $a_i = a_j$ for all $i,j\in\{0,1,2\}$.
Finally, since we have $n = a_0 + a_1 + a_2$, we must have that $3 \mid n$.
A: A vector space $V$ over $\mathbb{Q}$ with a linear operator $L$ is a $\mathbb{Q}[X]$ module, with $X \cdot v = L v$. Now, if the operator $L$ satisfies $P(L)= 0$, then our $\mathbb{Q}$-vector space is a module over $\mathbb{Q}[X]/(P)$. If $P$ is irreducible over $\mathbb{Q}$, then $\mathbb{Q}[X]/(P)$ is a field $K$, with $[K\colon \mathbb{Q}] = \deg P$. We have
$$\dim_\mathbb{Q}V = \dim_K V \cdot [K\colon \mathbb{Q}] = \dim_K V \cdot \deg P$$
Now apply this to the irreducible polynomial $X^3 + X^2 -1 \in \mathbb{Q}[X]$.
