Show that matrix is unitary Suppose $0\neq v \in \mathbb{C}^n$. Show that the matrix
$$U=I - 2\frac{vv^H}{\|v\|^2}$$
is unitary (with $^H$ being the conjugate transpose and I the identity matrix).
What I've tried:
A matrix U is unitary if $U^HU$ is equal to the identity matrix. So I've tried to plug U into this term but unfortunately I really don't know how to calculate the conjugate transpose of this long string.
Another tip: The script hints to Schur decomposition, which may be useful, but I don't know how I could apply it to this problem.
 A: Let $A = \frac{vv^H}{\|v\|^2}$, I interpret this as the matrix with coefficients $A_{ij} = v_i \overline{v_j}$.
Then $(A^H)_{ij} = \overline{A_{ji}} = \overline{v_j \overline{v_i}}= \overline{v_j} v_i = A_{ij}$, so $A^H = A$.
Similarly we can show $A^HA = A^H = A$, the result then follows from a direct calculation of $U^HU$.
A: No decomposition necessary. The straight-forward calculation workds:
\begin{align}
U^HU
&= \left(I-2\frac{vv^H}{|v|^2}\right)^H \left(I-2\frac{vv^H}{|v|^2}\right) \\
&= \left(I^H-2\frac{\left(vv^H\right)^H}{|v|^2}\right) \left(I-2\frac{vv^H}{|v|^2}\right) \\
&= \left(I-2\frac{vv^H}{|v|^2}\right) \left(I-2\frac{vv^H}{|v|^2}\right) \\
&= \left(I -4\frac{vv^H}{|v|^2} + 4\frac{(vv^H) (vv^H)}{|v|^4}\right) \\
&= \left(I -4\frac{vv^H}{|v|^2} + 4\frac{v(v^H v)v^H}{|v|^4}\right) \\
&= \left(I -4\frac{vv^H}{|v|^2} + 4\frac{vv^H}{|v|^2}\right) \\
&= I
\end{align}
hope this helps.
In particular we used the formulas:
\begin{align}
(vv^H)^H &= {v^H}^H v^H = vv^H \\
v^Hv &=|v|^2
\end{align}
