How do I find the basis of this eigen space? Given $\lambda$ is -2 and after calculating $A-\lambda I$,
$$\left(\begin{array}{rrr}
3 & 3 & 3\\
-3 & -3 & -3\\
3 & 3 & 3
\end{array}\right).$$
The reduced row echelon form is:
$$\left(\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right).$$
The solution is actually:
$$\left(\begin{array}{aaa}
-1\\
1\\
0
\end{array}\right).$$
and
$$\left(\begin{array}{bbb}
-1\\
0\\
1
\end{array}\right).$$
Can someone explain how the book arrived at this?
 A: Your reduced row echelon form is
$$\left(\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right).$$
You have one leading variable (the first term), and two free variables. The system is equivalent to the single equation
$$x+y+z=0.$$
That means that you have two degrees of freedom, $y$ and $z$. You can express $x$ in terms of $y$ and $z$ and $x=-y-z$. So the solutions are given by:
$$\begin{align*}
x &= -s - t\\
y & = s\\
z &= t
\end{align*}\qquad\qquad s,t\in\mathbb{R}.$$
You get a basis for the space of solutions by taking the parameters (in this case, $s$ and $t$), and putting one of them equal to $1$ and the rest to $0$, one at a time. Setting $s=1$ and $t=0$, we get $x=-1$, $y=1$, $z=0$, leading to the vector $(-1,1,0)$; setting $s=0$ and $t=1$ we get $x=-1$, $y=0$, $z=1$, leading to the vector $(-1,0,1)$. 
Note: What you claim is "the solution" is just one possible basis for the eigenspace. The eigenspace is equivalent to the nullspace of $A-\lambda I$; which is equivalent to the nullspace of the reduced row echelon form of $A-\lambda I$. So this question really comes down to:

Do you know how to find a basis for the nullspace of a matrix $A$?

Equivalently:

Do you know how to express the set of solutions to a homogeneous system of equations in vector form?

Above, we end up with the single-equation "system" given by
$$x+y+z=0.$$
Do you know how to write down/describe the set of all solutions to this equation in terms of vectors?
