How do we know that the eigenspaces of $T$ are the only $T$-invariant subspaces?

Question

Given $T:\mathbb{C}^2 \rightarrow \mathbb{C}^2$, if I know we have two different eigenvalues, let's call them $\lambda_1, \lambda_2$, then we can say $\mathbb{C}^2, \left\{ 0 \right\}, V_{\lambda_1} , V_{\lambda_2}$ are all invariant sub-spaces for $T$. how do I prove there aren't any more invariant sub-spaces?

Answer 1

If there was, it would have to be $1$-dimensional. In other words, it would be equal to $\Bbb Cv$, for some vector $v\ne0$. But asserting that $\Bbb Cv$ is invariant is the same thing as asserting that $v$ is an eigenvector.

Answer 2

Eigenspaces for different eigenvalues are in direct sum. As your global space is having a dimension equal to two, $V_{\lambda_1} \oplus V_{\lambda_2} = \mathbb C^2$ and it cannot exists another eigenspace.