I have two eigenspaces $E_1$ and $E_{-1}$ (where $1$ and $-1$ are not necessarily distinct) of a matrix $A$ where I have found their intersection to be $0$, so they form a direct sum of $\Bbb{R}^n$. Is this enough to show that $A$ is diagonalizable?
If not, then should I show that the algebraic multiplicity equals the geometric multiplicity somehow? I do not have the specific values of the matrices, but I do know that $\forall v \in R^{n}$, $$(A + I)\vec{v} \in \ker(A - I)$$ and $$(A - I)\vec{v} \in \ker(A + I).$$
I haven't learned minimal polynomials in my course yet, so I cannot use them.
Please correct me if I am misunderstanding anything. I am new to linear algebra.