show lambda is an eigenvalue of matrix A and find one eigenvector x Hello Lovely people of the Overflow :)
I am working on a homework assigntment for my linear algebra class and i am stumped on this pesky question which is as follows:
Show that λ is an eigenvalue of A and find one eigenvector, x, corresponding to this eigenvalue.
$$
A=\begin{bmatrix}6 & 6\\6 & -3\end{bmatrix},\qquad \lambda=-6.
$$
In my attempts I:
a) tired to find $A-6I$ (I being the identity matrix for $2\times2$ matrix)
b) The result of the above gave me the matrix :
$$
\begin{bmatrix}12 & 6\\6 & 3\end{bmatrix}
$$
From which i said that since Column 2 is 2x column 1 it is linear independent which implies null space is non zero. Now i am lost and do not know what to do. More so im not sure what to do next. My textbook does an example similar to this but i do not understand what steps it takes after this. Any suggestions , hints and helpful input is greatly appreciated :)
Thankyou

The characteristic polynomial is given by $|A - \lambda I| = 0$, hence:
$$\lambda ^2-3 \lambda -54 = 0 \implies (\lambda +6)(\lambda -9) = 0 \implies\lambda_1 = -6, ~ \lambda_2 = 9$$
The eigenvectors are found by $[A - \lambda I]v_i = 0$. For $\lambda_1 = -6$, we have 
$$\begin{bmatrix} 12 &\ 6\\ 6 & 3\\ \end{bmatrix}v_1 = 0$$
The rref of this is:
$$\begin{bmatrix} 1&\dfrac{1}{2}\\0&0\\ \end{bmatrix}v_1 = 0$$
This gives us an eigenvector of:
$$v_1 = (-1, 2)$$
Of course, there are other possible choices for the eigenvector.
