If eigenspaces $V_{\lambda} (\phi )= V_{\lambda} (\phi ‘)$ then $ \phi = \phi ‘$

Question

Let $\phi, \phi ‘$ be diagonalisable elements of $\text{End}_K(V)$ where $V$ is finite dimensional over $K$.

Setting $V_\lambda (\phi )=\ker (\phi - \lambda \text{Id}_V )$ if we have $V_\lambda (\phi )= V_\lambda (\phi ‘ ) $ for all $\lambda \in \mathbb{C}$ then how can we show that $\phi = \phi ‘ $?

Answer 1

In the comments, you have stated that you know $$V = \sum_{\lambda\in\mathbb{C}} V_\lambda(\phi),$$ so I'm going to assume that below.

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For $x \in V$, this tells you that there is some finite set $F \subset \mathbb{C}$ and some $x_\lambda \in V_\lambda(\phi)=V_\lambda(\phi')$ for each $\lambda \in F$ such that $$x = \sum_{\lambda\in F} x_\lambda.$$

Applying $\phi$ and $\phi'$ shows $$\phi(x) = \sum_{\lambda\in F} \phi(x_\lambda) = \sum_{\lambda\in F} \lambda x_\lambda = \sum_{\lambda\in F} \phi'(x_\lambda) = \phi'(x)$$ so $\phi=\phi'$.