The matrix $A$ is $$\begin{bmatrix}1&0 \\ 6 & -1 \end{bmatrix}$$ For a matrix to be diagonalizable it can be rewritten as $$A=PDP^{-1}$$ where $P$'s columns are the eigenvectors of $A$ and $D$ is the diagonal matrix whose entries are the eigenvalues of $A$.
Finding the eigenvalues like so $$\begin{vmatrix}1-\lambda&0 \\ 6 & -1-\lambda \end{vmatrix}=(1-\lambda)(-1-\lambda)-0=0$$ $$\lambda^2-1=0\therefore \lambda=1$$ So $D=\begin{bmatrix}1&0 \\ 0 & 1 \end{bmatrix}$, this is where I'm lost because $D$ is just the identity matrix for $\mathbb{R}^2$ and the eigenvectors are the exact same, so how would I write $P$?
You have missed the second solution of $\lambda = - 1$ giving $\lambda^2 = 1$ so now you have the two $\lambda_1,\lambda_2$ eigenvalues you needed.