I'm looking at a $2\times 2$ real symmetric matrix
$$\begin{pmatrix} a & b \\ b & d \\ \end{pmatrix} $$
For the eigenvalues I have
$$\lambda_{1,2} = \frac{a+d}{2} \pm \sqrt{\frac{(a+d)^{2}}{4}-ad+b^{2}}$$
Now, I'm trying to find an eigenvector $(u_{1}, u_{2})^{T}$ corresponding to $\lambda_{1}$.
To do so, I start with
$$(a-\lambda)u_{1}+bu_{2} = 0$$
which comes from the formulation of the eigenproblem. I know that in order to solve this I have to choose $u_{1}$ or $u_{2}$ arbitrarily and substitute in $\lambda_{1}$. Doing so yields:
$$u_{1} = - \frac{bu_{2}}{a-\lambda_{1}}$$. Choosing $u_{2} = 1$, I get $(-b, a-\lambda_{1})^{T}$. Comparing this with the output from wolfram alpha, it seems my coordinates are reversed. This is the output I'm referring to: http://www.wolframalpha.com/input/?i=eigenvectors%20%7B%7Ba%2C%20b%7D%2C%20%7Bc%2C%20d%7D%7D&lk=2
Any help would be appreciated!
Some ideas:
$$p_A(t)=\det (tI-A)=\begin{vmatrix}t-a&-b\\-b&t-d\end{vmatrix}=t^2-(a+d)t+ad-b^2$$
The quadratic's discriminant is
$$\Delta=(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2\ge 0\stackrel{\text{the root(s)}}\implies\;\lambda_{1,2}=\frac{(a+d)\pm\sqrt\Delta}{2}$$
Forming the homogeneous system to find eigenvectors:
$$(\lambda_1-a)x-by=0\implies \begin{cases}y=\frac{\lambda_1-a}bx\;,\;\;b\neq 0\implies\binom b{\lambda_1-a}\;\text{is an eigenvector}\\{}\\x=0\;,\;\;y\in\Bbb F,\;\;b=0\,,\,\,\lambda_1\neq a\implies\binom01\;\text{is an eigenvector}\\{}\\x,y\in\Bbb F\;,\;\;b=\lambda_1-a=0\implies\binom10\,,\,\,\binom01\;\;\text{are eigenvectors}\end{cases}$$