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

Is this your matrix?
$$
\begin{bmatrix}
6 & 6 \\
6 & -3
\end{bmatrix}
$$
The definition of the eigenvalue problem is as follows:
$$
Ax = \lambda x \rightarrow \left(A - \lambda \mathbb{I}\right)x = 0
$$
For that to have non-trivial solutions, the determinant of $A - \lambda \mathbb{I}$ must be zero--which basically just means subtracting $\lambda$ from each of the diagonals and setting the determinant to zero:
$$
det\left(
\begin{bmatrix}
6 - \lambda & 6 \\
6 & -3 - \lambda
\end{bmatrix}\right) = (6 - \lambda)(-3 - \lambda) - 36 = 0 \\
-18 - 6\lambda + 3\lambda + \lambda^2 - 36 = 0 \\
\lambda^2 - 3\lambda - 54 = (\lambda - 9)(\lambda + 6) = 0 \\
\lambda = 9, -6
$$
Now you go back to find the eigenvectors by solving each problem individually.
$\lambda = -6$:
$$
\begin{bmatrix}
12 & 6 \\
6 & 3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} = 0 \\
12x + 6y = 0 \rightarrow y = -2x
$$
All this means is that the eignenvector is any vector such that the "y" component is the negative of twice the x, such as:
$$
\begin{bmatrix}
1 \\
-2
\end{bmatrix}
$$
It's particularly easy to find the eigenvector because it's a 2$\times$2 matrix.  Had it been a larger matrix, you would have to actually go through the steps of getting the matrix into reduced-echelon form (except for one row--usually).  The only thing that could make these problems more complicated is when you have a multiplicity (or degeneracy) in the eigenvalues which means there are multiple distinct eigenvectors with the same eigenvalue.
