I am considering the following questions: If $V$ is a finite dimensional vector space and $E_1$ and $E_2$ are projections such that $R(E_1) = R(E_2)$, does this imply that $\ker E_1 = \ker E_2$?
The best I seem to come up with is that the kernels must be isomorphic. This is a bit trivial. Since $\dim \ker E_1 = \dim \ker E_2$, then they are isomorphic by the map that takes one basis to the other.
So I am wondering, is the answer to the questions, "no", the kernels of $E_1$ and $E_2$ do not have the be the same. Suppose $V = \mathbb{R}^2$. Let $V = (1,0) \oplus (1/2,1/2)$. Then for each $x \in \mathbb{R}^2$, we can represent $x$ uniquely as $x=u+v$ for $u$ in the span of $(1,0)$ and $v$ in the span of $(1/2,1/2)$. Does this induce the operator $E : R^2 \to R^2$ by $E(x) = E(u + v) = u$ so that $E^2 (x) = E(u) = u$.
It is clear that we could define the kernel as the span of a different vector, say $(0,1)$. Call the projection induced by this decomposition of $\mathbb{R}^2$ as $F$. Then it is clear that the kernels of $E$ and $F$ are not the same.
So my question: is this analysis fundamentally flawed, or is all of this true?