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\begin{frame}{One direction}
\large
For a rank-one matrix $A=uv^{\mathsf T}$,
\[\|A\|_{\mathrm{op}}=\|u\|_2\,\|v\|_2.\]
\medskip
The geometry tells us where to look for equality.
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\begin{frame}{A proof in three steps}
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\begin{enumerate}
\item Write $Ax=u(v^{\mathsf T}x)$.
\item<2-> Bound the scalar product by Cauchy--Schwarz.
\item<3-> Take $x=v/\|v\|_2$ when $v\ne0$.
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\begin{frame}{Beyond one component}
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For $A=\sum_{j=1}^r u_jv_j^{\mathsf T}$,
\[\|A\|_{\mathrm{op}}\leq\sum_{j=1}^r\|u_j\|_2\,\|v_j\|_2.\]
\medskip
What changes when the directions are orthogonal?
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