How to find the third eigenvector of a 3x3 matrix given the other two eigenvectors? In the following question, I believe that there is a way to find the third eigenvector without computing everything explicitly i.e. a way to find the third eigenvector given two eigenvectors. I am not sure how to find this eigenvector without tedious calculation?
In my textbook I have read something about Perron-Frobenius theorem but not sure if this is relevant to my question. 
Any help would be much appreciated!

 A: The mutation probabilities are symmetric, which makes $M$ symmetric as well. This means that it can be orthogonally diagonalized, i.e., there’s an orthonormal basis that consists of eigenvectors of $M$. The two eigenvectors that you’ve been given are obviously orthogonal (check this by computing their dot product), so their cross product $(1,1,1)^T\times(1,0,-1)^T=(-1,2,-1)^T$, which is orthogonal to both of them, must also be an eigenvector of $M$.  
As for the eigenvalues, recall the basic definition of an eigenvector: $\mathbf v\ne0$ is an eigenvector of $M$ iff $M\mathbf v=\lambda\mathbf v$ for some scalar $\lambda$. So, multiply the eigenvectors that you have by $M$ and extract the corresponding eigenvector with a simple division. In fact, in this problem you only have to go through the computation for $M(1,0,-1)^T$. $M$’s columns must all sum to $1$ and since it’s symmetric, so do its rows. Multiplying by $(1,1,1)^T$ is the same a summing rows, therefore you know its corresponding eigenvalue is $1$ without doing any computations. Then, once you have the second eigenvalue you can find the last one “for free” because the sum of the eigenvalues, taking multiplicities into account, is equal to the trace of the matrix.
A: The third eigenvector is $v_3 =(1, -2, 1)^{\rm T}$.
Some general insight. The first eigenvector is $v_1=(1, 1, 1)^{\rm T}$, so all others eigenvectors must be such that their entries add up to zero. The second eigenvector is $v_2=(1, 0, -1)^{\rm T}$, which complies. To generate the third eigenvector, take into acount that its entries must add up to zero, and also has to be perpendicular to $v_2$. You can exploit the form of $v_2$ to start with $(1, ?, 1)^{\rm T}$. Now what is left is to choose the $?$ so that $1+?+1=0$. Thus, $?=-2$.
A naive guess is $(1, -1, 0)^{\rm T}$, albeit it is not perpendicular to $v_2$. Still, it spans $\mathbb{R}^3$ altogether with $v_1$ and $v_2$.
Now, for the eigenvalues, I can't be of that much help. If the $3\times3$ matrix is given, just compute the characteristic polynomial. If the matrix AND two of its eigenvalues are given, just sum the diagonal entries (the trace) and substract the two known eigenvalues.
