How to find eigenvectors of a matrix given only eigenvalues? I'm trying to solve this problem:

Given a $3\times 3$ matrix with eigen values $3,0,-1$, are their associated
eigen vectors $(v_1,v_2,v_3)$ orthogonal to each other?

My thought process is that we need to find three possible eigen vectors, then test whether they're orthogonal. However how would I go about reverse engineering the eigen vectors? Or am I going about the problem wrong? For some reason I'm having a hard time understand how I would do that without being given a matrix.
 A: Fix any non-orthogonal basis $\{v_1,v_2,v_3\}$, such has, say $\{(1,0,0),(1,1,0),(0,0,1)\}$. Now, take the matrix $A$ of the linear map from $F^3$ into itself (where $F$ is the field with which you are working) which maps $v_1$ into $3v_1$, $v_2$ into $0$, and $v_3$ into $-v_3$. And you're done: the eigenvalues of $A$ are $3$, $0$, and $-1$ and its eigenvectors are not orthogonal.
In this case, you have$$A=\begin{bmatrix}3 & -3 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}.$$
A: With any set of real eigenvalues, the eigenvectors are orthogonal iff the matrix is Hermitian -- meaning, each $(i,j)$ element is the complex conjugate of the $(j,i)$ element.
Thus the matrix
$A=\begin{bmatrix}1 & 0 & \color{blue}{1+i\sqrt3} \\ 0 & 0 & 0 \\ \color{brown}{1-i\sqrt3} & 0 & 1 \end{bmatrix}$
with eigenvalues $3,0,-1$ will have orthogonal rigenvectors because the $\{(i,j),(j,i)\}$ pairs (like the blue and brown elements above) are all complex conjugates. By contrast the matrix
$B=\begin{bmatrix}1 & 0 & \color{blue}{1} \\ 0 & 0 & 0 \\ \color{brown}{4} & 0 & 1 \end{bmatrix}$
whose eigenvalues are also $3,0,-1$ does not have orthogonal eigenvectors; the blue and brown elements are not complex conjugates.
