So I'm given a matrix $A$
$$A=\begin{pmatrix}0.4 & 0.3& 0.3\\ 0.3 & 0.5 & 0.2 \\ 0.3& 0.2 & 0.5 \end{pmatrix}$$
Find the eigenvector corresponding to the eigenvalue $\lambda_1=1$ then proceed to find the remaining eigenvalues using a built in command in MatLab
So far I've found the eigenvector for $\lambda_1=1$ which is $v_1=\begin{pmatrix}1\\1\\1\end{pmatrix}$ and after i've used the eig command I found the remaining eigenvalues to be $\lambda_2=0.3,\lambda_3=0.1$ now it wants me to literary "guess" the corresponding eigenvectors.And this is where i'm stuck. I have read lots of guides and stuff about how to find eigenvectors by inspection but that was for 2x2 matrices and for 3x3 matrices they just mentioned that they need a 3rd fact which is a bit more complicated. Besides I can't see any obvious eigenvectors.....
Not really "guesses", but...
Note that $$A - 0.3 I = \pmatrix{0.1 & 0.3 & 0.3\cr 0.3 & 0.2 & 0.2\cr 0.3 & 0.2 & 0.2\cr} $$ has its second and third columns equal. That says that $\pmatrix{0\cr 1 \cr -1\cr}$ will be in its null space, i.e. an eigenvector of $A$ for eigenvalue $0.3$.
Now $A$ being a symmetric matrix, its eigenvectors are orthogonal, so the eigenvector for the remaining eigenvalue $0.1$ has sum $0$ and its second and third entries equal. Thus it should be (a multiple of) $\pmatrix{2 \cr -1\cr -1\cr}$.