Question-49
Consider a matrix \(\displaystyle A\) of unit rank such that \(\displaystyle Av=12u\), where \(\displaystyle v=\frac{1}{2}( 1,1,1,1)\) and \(\displaystyle u=\frac{1}{3}( 2,2,1)\). Find the smallest possible value of the largest singular value of \(\displaystyle A\).
\(12\)
Since \(\displaystyle A\) has unit rank, it has only one non-zero singular value. Therefore, we have unit vectors \(\displaystyle u_{1} \in \mathbb{R}^{3}\) and \(\displaystyle v_{1} \in \mathbb{R}^{4}\) (not necessarily unique) such that:
\[\begin{equation*} A=\sigma _{1} u_{1} v_{1}^{T} \end{equation*}\]
We have:
\[\begin{equation*} \begin{aligned} Av & =12u \end{aligned} \Longrightarrow \begin{aligned} \sigma _{1} u_{1} v_{1}^{T} v & =12u \end{aligned} \end{equation*}\]
Taking the norm on both sides:
\[\begin{equation*} \begin{aligned} \sigma _{1} ||u_{1} ||\cdot |v_{1}^{T} v| & =12\cdot ||u||\\ |v_{1}^{T} v| & =\frac{12}{\sigma _{1}} \end{aligned} \end{equation*}\] Now applying Cauchy-Schwarz:
\[\begin{equation*} \frac{12}{\sigma _{1}} \leqslant |v_{1} |\cdot |v|=1\Longrightarrow \sigma _{1} \geqslant 12 \end{equation*}\]
The smallest value of \(\displaystyle \sigma _{1}\) is \(\displaystyle \boxed{12}\).