Trouble determining basis of eigenspace?

Question

I'm supposed to find the basis for eigenspace of a particular $\lambda$.

Solving the three by three matrix $(3*3)\begin{pmatrix} v_1 \\ v_2 \\ v_3\end{pmatrix} = 0$. I get: $$6v_1 + v_2 + 2v_3 = 0$$ $$6v_2 -2v_3 = 0$$

I express everything in the free variable $v_3$ because my last row has only $0 's$ (no pivot). $$v_1 = {7\over 18 }v_3$$ $$v_2 = {1\over 3 }v_3$$

Can someone help me in the final step to represent the basis for eigenspace, I can't get the grip of it..

Answer 1

Because you derived only one free variable, say $x_3$, the eigenspace $E$ is of dimension $1$ and is of the form \begin{align*} E=\text{span}\left(\left\{\begin{pmatrix}\frac{7}{18}\\\frac{1}{3}\\1 \end{pmatrix}\right\}\right). \end{align*}