Eigenvectors of a $2 \times 2$ matrix when the eigenvalues are not integers How can I calculate the eigenvectors of the following matrix?

$$\begin{bmatrix}1&amp; 3\\3&amp; 2\end{bmatrix}$$

I calculated the eigenvalues. I got

$$\lambda_1 =  4.541381265149109$$

$$\lambda_2 = -1.5413812651491097$$

But, now I don't know how to get the eigenvectors. When I create a new matrix after I subtracted Lambda value from all the members of the matrix on the main diagonal and tried to solve the homogeneous system of equations, I get only null vector for both $\lambda_1$ and $\lambda_2$....

When I used this website for calculating eigenvalues and eigenvectors. I got these eigenvectors

$(0.6463748961301958, 0.7630199824727257)$ for $\lambda_1$  

$(-0.7630199824727257, 0.6463748961301957)$ for $\lambda_2$

.... but have no idea how to calculate them by myself...
Is it even possible? ....or it's possible to calculate it numerically?
Indeed, as Chris Eagle and Michael pointed out to you, calculators are not always your best friend.

Instead, if you do your maths with the characteristic equation, you'll find out that the eigenvalues look nicer this way:

$$
\lambda = \frac{3 \pm \sqrt{37}}{2}
$$

And it's not at all impossible to find the eigenvectors. For instance the one with the $+$ sign, you could start like this:

$$
\begin{pmatrix}
1 - \dfrac{3 + \sqrt{37}}{2}  &amp;   3                            &amp; \vert &amp; 0  \\
3                             &amp;   2 - \dfrac{3 + \sqrt{37}}{2} &amp; \vert &amp; 0 
\end{pmatrix}
$$

Hint: After some easy simplifications, you'll find out that it's very useful to multiply one of the rows by $1 - \sqrt{37}$ and that you can write the corresponding eigenvector as simple as this: $(1 - \sqrt{37}, -6)$.
By definition an eigenvector for eigenvalue $\lambda$, is a member of nullspace of $A - \lambda I$, if you can compute nullspaces, you can compute eigenvectors.