show lambda is an eigenvalue of matrix A and find one eigenvector x
Andrew Henderson 
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
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 3
