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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.

illuminatitruthseeker
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2 Answers2

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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 $U^HU$.

jdoe
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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}

Simon
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