Are Jordan chains always linearly independent? Firstly, I know this question might seem slightly vague, but I can't find any examples or specific questions to specify it. 
My question is in regards to finding the Jordan Canonical Form of a matrix, more specifically when you have two Jordan chain of length greater than $1$ for a particular eigenvalue. After you've found the eigenvectors, are these eigenvectors always linearly independent? Also, if they are, is it possible to find eigenvectors which form Jordan chains which are 'wrong' because they aren't linear independent? 
 A: 
After you've found the eigenvectors, are these eigenvectors always linearly independent?

Depends on what you mean by "find the eigenvectors." Remember that if $\mathbf{v}$ is an eigenvector, and $\alpha\neq 0$ is a scalar, then $\alpha\mathbf{v}$ is also an eigenvector (associated to the same eigenvalue), and $\mathbf{v},\alpha\mathbf{v}$ are certainly not linearly independent.
Rather: what we do is find a basis for the eigenspace, which, being a basis, guarantees that the eigenvectors we found are linearly independent. 
More generally, to deal with the Jordan canonical form, we prove two things:


*

*If $\gamma$ is a Jordan chain, then $\gamma$ is linearly independent.

*If $\gamma_1,\ldots,\gamma_k$ are Jordan chains, and their initial vectors are linearly independent, then $\gamma_1\cup\cdots\cup\gamma_k$ is linearly independent.
(If the eigenvalue is $\lambda$, then the Jordan chain is of length $k$ is of the form $$\gamma=[(T-\lambda I)^{k-1}\mathbf{v},(T-\lambda I)^{k-2}\mathbf{v},\ldots,(T-\lambda I)\mathbf{v},\mathbf{v}]$$
where $(T-\lambda I)^{k-1}\mathbf{v}$ is an eigenvector corresponding to $\lambda$; the "initial vector" of the cycle is $(T-\lambda I)^{k-1}\mathbf{v}$.)
There are two ways to proceed to construct the Jordan chains; both methods guarantee that you will end up with a linearly independent set in the end.
You can start with the eigenspace: find a basis for the eigenspace; then you find appropriate vectors in $\mathrm{ker}(T-\lambda I)^2$ that map to (some of) the eigenvectors you found; then in $\mathrm{ker}(T-\lambda I)^3$, etc. Because the initial vectors of the chains we are constructing are linearly independent (being a basis of the eigenspace) we are guaranteed these sets are linearly independent.
The other method is to first determine the lengths of the chains (there are formulas you can use to find them). If the longest chains have length $k$, then you find linearly independent vectors in $\mathrm{ker}(T-\lambda I)^{k-1}$ that are not in $\mathrm{ker}(T-\lambda I)^{k-2}$. Then you apply $(T-\lambda I)$ to them, and if necessary find linearly independent vectors from the ones you have so far that are in $\mathrm{ker}(T-\lambda I)^{k-2}$ but not in $\mathrm{ker}(T-\lambda I)^{k-3}$. You continue this way, always selecting linearly independent vectors, until you are down to finding enough linearly independent vectors in $\mathrm{ker}(T-\lambda I)$ besides the ones you already have; you will be guaranteed they are linearly independent because at each stage we are always selecting vectors that are linearly independent.
In summary: so long as you don't purposefully pick a vector that is linearly dependent from what you already have, you will always be able to continue and obtain a Jordan canonical basis: you will not run into a "blind alley" due to poor choice of vectors. 
