Question about Jordan base - Why is finding the kernel and finding the generelized eigenvectors the same thing? Quick question, perhaps a bit general but I don't understand something.
When finding a jordan basis for a matrix $A$, what i do is for each eigenvalue $\lambda$ I find vectors such that:
$(A-\lambda I)v_1=\lambda v_1$
$(A-\lambda I)v_2=v_1$
$(A-\lambda I)v_3=v_2$ and so on...until $(A-\lambda I)v_{r-1}=v_r$ where $r$ is the algebraic multiplicity of $\lambda$ in the characteristic polynomial.
But I've seen people do it differently. I've seen people finding jordan basis by finding:
$\operatorname{ker}(A-\lambda I), \operatorname{ker}(A-\lambda I)^2,\ldots,\operatorname{ker}(A-\lambda I)^r$, why is this the same thing? Why are both methods working?
Thank you.
 A: Recall that $r$ is the smallest natural number  $m$ such that $(A-\lambda I)^m=0_{\Bbb C^{n\times n}}$.
The following holds:
$$\begin{cases} (A-\lambda I)v_1=0_{\Bbb C^{n\times 1}}&\implies v_1\in \text{ker}(A-\lambda I) &\implies v_1\in \text{ker}\left((A-\lambda I)\right)  \\
(A-\lambda I)v_2=v_1&\implies (A-\lambda I)^2v_2=(A-\lambda I)v_1=0_{\Bbb C^{n\times n}}& \implies v_2\in \text{ker}\left((A-\lambda I)^2\right)\\
(A-\lambda I)v_3=v_2& \implies (A-\lambda I)^3v_3=(A-\lambda I)^2v_2=0_{\Bbb C^{n\times n}} & \implies v_3\in \text{ker}\left((A-\lambda I)^3\right)\\
\vdots &\implies \vdots & \implies \vdots\\
(A-\lambda I)v_{k+1}=v_{k} &\implies (A-\lambda )^{k+1}v_{k+1}=(A-\lambda I)^kv_k=0_{\Bbb C^{n\times n}} & \implies v_{k+1}\in\text{ker}\left((A-\lambda I)^k\right)\\
\vdots &\implies \vdots &\implies \vdots\\
(A-\lambda I)v_r=v_{r-1}& \implies (A-\lambda)^{r}v_r=(A-\lambda I)v_1=0_{\Bbb C^{n\times n}} & \implies v_r\in \text{ker}\left((A-\lambda I)^{r-1}\right)\end{cases}$$
More importantly, for each $k\in \{1, \ldots , r\}$, define $N_k:=\text{ker}\left((A-\lambda I)^{k}\right)\setminus \text{ker}\left((A-\lambda I)^{k-1}\right)$ and recall that $$\text{ker}\left((A-\lambda I)^0\right)\subset \text{ker}\left((A-\lambda I)^1\right)\subset \ldots \subset \text{ker}\left((A-\lambda I)^{r-1}\right)\subset \text{ker}\left((A-\lambda I)^{r}\right)=\Bbb C^{n\times n}.$$
The following holds:
$$\begin{cases} (A-\lambda I)v_1=0_{\Bbb C^{n\times 1}}&\implies v_1\in \text{ker}(A-\lambda I) &\implies v_1\in N_1  \\
(A-\lambda I)v_2=v_1&\implies (A-\lambda I)^2v_2=(A-\lambda I)v_1=0_{\Bbb C^{n\times n}}& \implies v_2\in N_2\\
(A-\lambda I)v_3=v_2& \implies (A-\lambda I)^3v_3=(A-\lambda I)^2v_2=0_{\Bbb C^{n\times n}} & \implies v_3\in N_3\\
\vdots &\implies \vdots & \implies \vdots\\
(A-\lambda I)v_{k+1}=v_{k} &\implies (A-\lambda )^{k+1}v_{k+1}=(A-\lambda I)^kv_k=0_{\Bbb C^{n\times n}} & \implies v_{k+1}\in N_{k+1}\\
\vdots &\implies \vdots &\implies \vdots\\
(A-\lambda I)v_r=v_{r-1}& \implies (A-\lambda)^{r}v_r=(A-\lambda I)v_1=0_{\Bbb C^{n\times n}} & \implies v_r\in N_{r}\end{cases}$$
This  and modus ponens assure, among other things that if $(\lambda ,v_1)$ is an eigenpair of $A$, then $\{v_1, \ldots ,v_r\}$ is linearly independent.
