It is true if $M$ is a diagonal matrix. Let $M=\left(\begin{array}{cc}
m_{1} & 00\\
0 & m_{2}m_{2}\\
0 & 0\end{array}\right)$. Assume that $\left|m_{1}\right|\ge\left|m_{2}\right|$, then for
any $x:=\left(\begin{array}{c}
x_{1}x_{1}\\
x_{2}\end{array}\right)\in\mathbb{R}^{2}$ with $\left\Vert x\right\Vert _{3}=1$, $\left\Vert Mx\right\Vert {3}=\left(\left|m{1}x_{1}\right|^{3}+\left|m_{2}x_{2}\right|^{3}\right)^{1/3}\geq\left(\left|m_{2}x_{1}\right|^{3}+\left|m_{2}x_{3}\right|^{3}\right)^{1/3}=\left|m_{2}\right|$.
_{3}=\left(\left|m_{1}x_{1}\right|^{3}+\left|m_{2}x_{2}\right|^{3}\right)^{1/3}\geq\left(\left|m_{2}x_{1}\right|^{3}+\left|m_{2}x_{3}\right|^{3}\right)^{1/3}=\left|m_{2}\right|$.
Take $V$ to be the space spanned by $\left(\begin{array}{c}
00\\
11\\
0\end{array}\right)$ and $\left(\begin{array}{c}
00\\
00\\
1\end{array}\right)$, then $\left\Vert M^{T}y\right\Vert {3/2}\leq\left|m{2}\right|$
_{3/2}\leq\left|m_{2}\right|$
for all $y\in V$ with $\left\Vert y\right\Vert {3/2}=1$. _{3/2}=1$. Likewise
for the case $\left|m{1}\right|\le\left|m_{2}\right|$.\left|m_{1}\right|\le\left|m_{2}\right|$.
But it is still a question for non-diagonal matrices.
