Let $A = \left[\begin{matrix} a & b \\ 0 & c \end{matrix} \right] \in \mathbb C^{2\times 2}$. Find a unitary $U$ such that $U^HAU = \left[\begin{matrix} c & b \\ 0 & a \end{matrix}\right]$.
NOTE: we have dealt with the cases when $b=0$ or $a = c$. Now we solve for the case when $b \ne 0 \wedge a \ne c$.
Let $U= (u_c, u_a)$ where the $u_{c,a}$'s are the unit length right and left eigenvectors of $A$, respectively. Then $u_c \perp a_a$ and $U$ is unitary. We have $$ U^HAU = \left[\begin{matrix} u_c^HAu_c & u_c^HAu_a \\ u_a^HAu_c & u_a^HAu_a\end{matrix}\right] = \left[\begin{matrix} cu_c^Hu_c & u_c^HAu_a \\ au_a^Hu_c & au_a^Hu_a\end{matrix}\right] = \left[\begin{matrix} c & b \\ 0 & a\end{matrix}\right]. $$ How does one then find that $$ U = \frac 1{\sqrt{1 + \left| \frac{a-c}a\right|^2}}\left[\begin{matrix} 1 & \frac{a-c}b \\ -\frac{a-c}b & 1\end{matrix}\right] $$
Any help is appreciated! Thanks!
You're on the right track. To find $u_c$, note that $$ (A - c I)u_c = 0 \implies \pmatrix{a-c&b\\0&0}u_c = 0. $$ With that, we can conclude that $u_c$ a multiple (in particular a unit vector in the direction of) $(b,-(a-c))$. From there, note that $u_a$ must be a unit vector that is orthogonal to $u_c$. Putting it all together, we find that $$ u_c = \frac{e^{i\theta_c}}{\sqrt{1 + |\frac{a-c}{a}|^2}}\left(1, -\frac{a-c}{b}\right), \quad u_a = \frac{e^{i\theta_a}}{\sqrt{1 + |\frac{a-c}{a}|^2}}\left(\frac{\bar a-\bar c}{\bar b},1\right) $$ for some real numbers $\theta_a,\theta_c$. Choosing $\theta_a = \theta_c = 0$ leads to a solution close to that which you presented.