A basic example of basis for eigenspaces Hi I'm just learning how to find bases for eigenspaces and I ran into a very basic case that confused me.
So the matrix is $$A = \begin{pmatrix}2 & 1\\0 & 2\end{pmatrix}$$ and $\lambda = 2$. 
So $$A-\lambda I = \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}.$$
Here is where I'm confused... 
Solving for the null space I do $x=s, y=0$ so should my basis contain one vector $(1, 0)$? Or am I doing something wrong.
Thanks for the help.
 A: Exactly. The kernel of $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ is the subspace $\langle (1,0)\rangle$ of dimension $1$.
So the eigenspace of $A$ for the eigenvalue $2$ is $\langle (1,0)\rangle$.
A: Since eigenvalues of $A$ are degenerated, eigenspace is not much of interest. What is more valuable – generalized eigenspace based on generalized eigenvectors, i.e. one needs to find Jordan normal/canonical form.
First, find generated eigenvalue, which you've already done. It's $\lambda = 2$. Now, find first eigenvector, which is also done. It's $a = [\alpha,0]^T$. Now to find next generalized eigenvector one needs to find 
$$
\left [
\begin{array}{cc}
0 & 1 \\ 0 & 0
\end{array}
\right] \left [ 
\begin{array}{c}
\gamma \\ \delta
\end{array}
\right ] = \left [ 
\begin{array}{c}
\alpha \\ 0
\end{array}
\right ]
$$
from which $\delta = \alpha$, so eigenvector is $b = [\gamma, \alpha]^T$.
So basis is $\hat a = [1,0]^T, \hat b = [0,1]^T$. Hence one can decompose $A$ in terms of the matrix based on generalized eigenvectors and Jordan cells, each of which has the form
$$
J_\lambda = \left [
\begin{array}{ccccc}
\lambda & 1 & 0 & \ldots & 0 \\
0 & \lambda & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & 0 \\
0 & 0 & 0 & \ldots & \lambda
\end{array}
\right ]
$$
where the rank of the cell is the order of degeneration, as $A = Q^{-1}JQ$
In this case it's and Jordan matrix consists of only one cell, so
$$
J_2 = \left [
\begin{array}{cc}
2 & 1 \\
0 & 2
\end{array}
\right]
$$
So one can write
$$
A = \left [ \begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right ] \left [ \begin{array}{cc}
2 & 1 \\
0 & 2
\end{array}\right ] \left [ \begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right ]
$$
Advantage of this is one can easily find any power of $A$ as well as any function of $A$ as
$$
f(A) = Q^{-1}f(J)Q \\
A^n = Q^{-1}J^nQ
$$
since $f(J)$ is calculated by certain procedure, which is known.
You can find more information about Jordan normal form here – Jordan normal form
