Originally Posted by

**scorpion007** For

$\displaystyle

\underbrace{\left[ \begin{array}{cccc} 1 & 1 & 1 \\ 1 & 1& 1 \\ 1 & 1 & 1 \end{array}\right]}_{A} \begin{bmatrix} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \end{bmatrix}=0

$

Finding the RREF(A) is simply:

$\displaystyle \begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}$

and so

$\displaystyle \epsilon_1 + \epsilon_2 + \epsilon_3 = 0$

$\displaystyle \Rightarrow \epsilon_1 = -\epsilon_2 - \epsilon_3,\quad \epsilon_2,\epsilon_3 \in \mathbb{R}$.

So the solution set $\displaystyle S = \text{Ker}(A) = \text{span}\lbrace \begin{bmatrix} -1\\1\\0\end{bmatrix}, \begin{bmatrix} -1\\0\\1\end{bmatrix} \rbrace$ and so those are your two general e-vectors for $\displaystyle \lambda_1 = \lambda_2$.

(This is a linear algebra question and really belongs in that forum)