Invariant subspaces of a linear transformation with different eigenvalue multiplicity Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation such that all it's eigenvalues are $\lambda_1$ and $\lambda_2$ and the eigenvectors are $v_1, v_2$ (corresponding with $\lambda_1$) and $v_3$ (corresponding with $\lambda_2$) .
Find all the T invariant subspaces of $\mathbb{R}^3$.
This question helped me to understand invariant subspace if every eigenvalue are associated with only one eigenvector. But if one eigenvalue are related with two vectors? 
I think that the answer is related with my other question: Are eigenvectors related with same eigenvalue always linearly independent?
 A: Luckily, here we have as many eigenvectors as the dimension of the space. (I presume that, as is common, when someone says that "the eigenvectors are $v_1$, $v_2$", they mean that these are linearly independent eigenvectors.) Since we have enough eigenvectors, they form a basis for the entire space. And in particular, each invariant subspace is the span of the corresponding eigenvectors.
As to your second question, the answer is NO. But… As I pointed out above, people often slightly abuse the language.


*

*If $v$ is an eigenvector, then e.g. $2v$ is an eigenvector corresponding to the same eigenvalue, and so is any scalar multiple of $v$. So if you know that several vectors are eigenvectors corresponding to the same eigenvalue, you can't conclude that they are independent.

*But when describing eigenvalues and eigenvectors, it is a common figure of speach to just say something like "Let $v_1,v_2$ be the eigenvectors corresponding to an eigenvalue $\lambda$." This is an inaccurate (but quite common!) statement, where it is silently assumed that these eigenvectors are linearly independent. The purpose of such a phrase is literally to state that the invariant subspace is two dimensional, and these two vectors form one of its possible bases.
