If $b^*\cdot a=1$, then $\|a.b^*\|=?$

Question

Let $a$ and $b$ two column-vectors $\in C^n$. If we know that the inner product $b^* \cdot a=1$, (where $b^*$ is conjugate transpose of $b$), what follows for the norm of the matrix $P=a \cdot b^*$ $$ \|P\|_2=? $$ is it equal to $1$?

Answer 1

\begin{eqnarray} \|P\|_2 &=& \sup_\limits{\|z\|=1}\|Pz\|_2\\ &=& \sup_\limits{\|z\|=1} \sqrt{z^*P^*Pz}\\ &=& \sup_\limits{\|z\|=1} \sqrt{z^*ba^*ab^*z}\\ &=& \|a\|_2\sup_\limits{\|z\|=1}\sqrt{z^*bb^*z}\\ &=& \|a\|_2\|b\|_2\\ &\ge& \vert b^*a\vert = 1 \end{eqnarray} by the Cauchy-Schwarz inequality. By making $\ a\ $ and $\ b\ $ very nearly orthogonal, you can make the product $\ \|a\|_2\|b\|_2\ $ arbitrarily large while still keeping the inner product $\ b^*a=1\ $, so I don't think there's much more you can say about the size of $\ \|P\|_2\ $.