Finding Eigenvectors with repeated Eigenvalues

It is not a good idea to label your eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$; there are not three eigenvalues, there are only two; namely $\lambda_1=-2$ and $\lambda_2=1$.

Now for the eigenvalue $\lambda_1$, there are infinitely many eigenvectors. If you throw the zero vector into the set of all eigenvectors for $\lambda_1$, then you obtain a vector space, $E_1$, called the eigenspace of the eigenvalue $\lambda_1$. This vector space has dimension at most the multiplicity of $\lambda_1$ in the characteristic polynomial of $A$. In this case, looking at the characteristic polynomial of $A$, we see that the dimension of the eigenspace $E_1$ is at most two.

As you determined, the dimension of $E_1$ is exactly two, as you found two independent eigenvectors for $\lambda_1$. Your eigenvectors $v_1$ and $v_2$ form a basis of $E_1$. It does not matter that WA listed them in the opposite order, they are still two independent eigenvectors for $\lambda_1$; and any eigenvector for $\lambda_1$ is a linear combination of $v_1$ and $v_2$.

Now you need to find the eigenvectors for $\lambda_2$. Note the dimension of the eigenspace must be one here (since the multiplicity of $\lambda_2$ in the characteristic polynomial is 1).