Let $$J=\oplus_{i=1}^{k} J_{n_i}(\lambda)$$
where $J_{n_i}(\lambda)$ is a jordan block of size $n_i$ with $\lambda $ on its diagonal, and $\sum_{i=1}^{k}n_i = n $, so $J$ is $n\times n$ matrix,
then is it correct to write the following :
$$(1)\;\;\;\;\;\;\;J^{r}=\oplus_{i=1}^{k} [J_{n_i}(\lambda)]^r\;\; \forall r>0\;\; ?$$
$$(2)\;\;\;\;\;\;\; (J-\lambda I_n)^r = \oplus_{i=1}^{k} [J_{n_i} (\lambda) -\lambda I_{n_i}]^r =\oplus_{i=1}^{k} [J_{n_i} (0)]^r \;\; \forall r>0\;\; ?$$
$$(3)\;\;\;\;\;\;\;\text{rank}(J^{r})=\sum_{i=1}^{k} \text{rank}\big([J_{n_i}(\lambda)]^r\big)\;\; \forall r>0\;\; ?$$
and hence,
$$(4)\;\;\;\;\;\;\; \text{rank}((J-\lambda I_n)^r) = \sum_{i=1}^{k} \text{rank}\big([J_{n_i} (0)]^r\big) \;\; \forall r>0\;\; ?$$
Also please tell me, how can I show that $J$ has $k$ linearly independent eigenvectors corresponding to $\lambda$ if I know that Geometric multiplicity of $\lambda$ in $J_{n_i}(\lambda)$ equals $1$? Can I say that, I will get an eigenvector $x_i \in C^{n_i}$ for $\lambda$ from each of $J_{n_i}(\lambda)$ and then make this I can extend this vector by add $0s$ at the right positions (where their jordan blocks lie) so that $x_i$ will now lie in $C^n$. But how do I show that they will be linearly independent?
Having a direct sum of linear transformations presumes a direct sum of the ambient vector spaces.
Here we have $\Bbb C^n=\oplus_i\Bbb C^{n_i}$ with the embeddings $\Bbb C^{n_i}\hookrightarrow \Bbb C^n$ which extend the vectors by $0$ coordinates out of the range of indices $\sum_{j<i}n_j+1,\dots, \sum_{j\le i}n_j$.
So, any vectors belonging to distinct $i$'s are linearly independent.
Each $\lambda$-block has a unique eigenvector, and since these are in different subspaces of the direct sum, they are linearly independent.