<s>[INST] Finding rank-$1$ matrix 
Let $$S = \frac{1}{12} \begin{pmatrix} 1 & 10 & 1 \\ 5 & 2 & 5\\ 1 & 2 & 9\end{pmatrix}$$ Find a rank-$1$ matrix $R$ so that $$ M = S + R $$ will have the same eigenvalues as $S$ and all the diagonal elements equal.

I have found the eigenvalues of $S$ to be: $$\{ 1, \frac{\sqrt{2}}{3},\frac{-\sqrt{2}}{3} \}$$
So I now have 
$$M = \frac{1}{12} \begin{pmatrix}
 1 & 10 & 1 \\
5 & 2 & 5\\
1&2&9\\ 
\end{pmatrix} + R $$
and $M$ has eigenvalues $$\{ 1, \frac{\sqrt{2}}{3},\frac{-\sqrt{2}}{3}\}$$
How do I find the rank-$1$ matrix $R$ with diagonal elements equal?
 [/INST] I'm assuming you meant the three diagonal elements of $S+R$ are equal.
Let $R = u v^\top$.  You have $5$ equations to solve for $6$ variables $u_1,u_2,u_3,v_1,v_2,v_3$ (of course there is redundancy here): the coefficients of $\lambda^0$ to $\lambda^2$ in $\det(S+R-\lambda I) - \det(S-\lambda I)$ are $0$, $S_{11} + R_{11}-S_{22} - R_{22}=0$ and $S_{11}+R_{11}-S_{33} - R_{33} = 0$.
One of the solutions is
$$ u_1 = \frac{1}{4},\; u_2 = \frac{5}{48},\; u_3 = \frac{-5}{36}, \;v_1 = 1, \;v_2 = \frac{8}{5}, \;v_3 = 3 $$
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