$ A^2(t) - 5A(t) = 6I_n$ Show there exists $P : [0,1] \to GL_n(\mathbb R), \ C^1$ s.t. $\forall t \ \ P(t)A(0)P^{-1}(t) = A(t)$ 
Let $A : [0,1] \to M_n(\mathbb R)$ s.t. $\forall \ t \in [0,1], \
A^2(t) - 5A(t) = 6I_n.$
Show there exists $P : [0,1] \to GL_n(\mathbb R), \ C^1$ s.t. $\forall
\ t \in [0,1],\ P(t)A(0)P^{-1}(t) = A(t)$

So $X^2 - 5X - 6 = \ (X-6)(X+1)$
Hence $\forall t \in [0,1] \ A(t)$ is diagonalizable with $6$ and $1$ as eigenvalues.
Then I would use the "continuity of the roots" of the carachteristic polynomial to say that the multiplicity of $6$ and $1$ the same for all $t$, so for all $t$, $A(t) \sim A(0)$.
How to proceed?
 A: We assume that $A(t)\in C^1$.
There are integers $p\geq 0,q\geq 0$ s.t. $p+q=n$ and $A(0)$ is similar to $diag(6I_p,-I_q)$. Note that $trace(A(t))$ is continuous and the eigenvalues $(\lambda_i(t)$ of $A(t)$ can be numbered in such a way that they are continuous functions of $A(t)$; consequently, for every $t$, $A(t)$ is also similar to  $diag(6I_p,-I_q)$.
More precisely $A(t)=6U(t)-V(t)$ where $U,V$ are projectors of trace $p,q$ s.t. $UV=VU=0$. From $7V=A^2-6A,42U=A^2+A$, we deduce that $U(t),V(t)$ are $C^1$. To show the existence of $P(t)$, it is enough to 
(*) find a $C^1$ parametrization of a basis of $E(t)=\ker(U(t)-I_n)$, a vector subspace of dimension $p$, that is an element of the Grassmannian $G_{p,n}$.
Since $G_{p,n}$ is compact, it suffices to prove locally (*). Let $t_0\in[0,1)$. We may assume that $U(t)-I=\begin{pmatrix}P_{p,p}&Q_{p,q}\\R_{q,p}&S_{q,q}\end{pmatrix}$ where $rank(U(t_0)-I)=rank(S(t_0))=q$. Thus, locally, $S(t)$ is invertible and $\begin{pmatrix}x_p\\y_q\end{pmatrix}\in E(t)$ iff $y=-S^{-1}Rx$, that is
$E(t)=\{\begin{pmatrix}x\\-S(t)^{-1}R(t)x\end{pmatrix};x\in \mathbb{R}^p\}\in G_{p,n}$. Finally, the required $C^1$ parametrization is locally
$t\rightarrow \{\begin{pmatrix}e_1\\-S(t)^{-1}R(t)e_1\end{pmatrix},\cdots,\begin{pmatrix}e_p\\-S(t)^{-1}R(t)e_p\end{pmatrix}\}$.
