Let $A\in \mathbb{R}^{3\times 3}$ such that $\mathrm{rank} \,A=2$ and the elements over the main diagonal are equal. Find the matrix $A$ if $\lambda=2$ is an eigenvalue and $x_1=[1,2,-1]^T$ and $x_2=[3,0,1]^T$ are the eigenvectors corresponding to $\lambda=2$.
I know I should write my attempt what I tried to do but to be honest I don't even know where to begin.
I tried writing the matrix with ordinary entries so that I can find the characteristic polynomial but I didn't get anywhere.
I would really appreciate some help. Please don't downvote for not writing my work.
Hints:
Since the matrix is not full rank (that is $\operatorname{rk}(A)\neq 3$) we have that one eigenvalue of $A$ is $0.$ Let the elements on the diagonal be equal to say $x.$ Then $$\operatorname{Trace}{A} = \lambda_1+\lambda_2+\lambda_3= 2+2 + 0 = 3x$$ and so $x= 4/3.$
Now we know that $$ Ax = \begin{bmatrix} 4/3 & a & b \\ c & 4/3 & d \\ e & f & 4/3 \\ \end{bmatrix} \begin{bmatrix} 1\\ 2\\ -1\\ \end{bmatrix} = 2\begin{bmatrix} 1\\ 2\\ -1\\ \end{bmatrix} $$ and so $$4/3 + 2a -b = 2$$ $$c + 8/3 - d = 4 $$ $$e+ 2f -4/3= -2$$ Similarily you can write a system for the eigenvector. This gives you $6$ equations for $6$ variables and so you can obtain a solution for this system (hopefully the problem is well posed).