The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector

If $V=0$, then The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| gives $\parallel T\parallel =0$. Assume now that $V\mathrm{\ne }0$, and choose an SVD from [L1] with orthonormal bases $(e_1,\dots ,e_n)$ of $V$ and $(f_1,\dots ,f_m)$ of $W$, where $n\ge 1$. Write its positive singular values as ${\sigma }_1\ge \cdots \ge {\sigma }_r>0$ and set ${\sigma }_j=0$ for $r<j\le n$. The SVD formula gives $T{e}_j={\sigma }_j{f}_j$ only for $j\le r$, while $T{e}_j=0$ for $j>r$. In particular, when $r=0$, the formula is the empty sum, $T=0$, and the largest singular value on the nonzero domain is ${\sigma }_1=0$.

If $r>0$, write a unit vector as $v={\sum }_{j=1}^n{c}_j{e}_j$. Orthonormality of the input and output bases gives $\parallel Tv{\parallel }^2={\sum }_{j=1}^r{\sigma }_j^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le {\sigma }_1^2{\sum }_{j=1}^n\mathrm{\mid }{c}_j{\mathrm{\mid }}^2={\sigma }_1^2\mathrm{.}$ Thus every unit vector satisfies $\parallel Tv\parallel \le {\sigma }_1$. If $r=0$, then $Tv=0$ for every $v$, and the same bound holds with ${\sigma }_1=0$.

Since $n\ge 1$, the unit right-singular vector $e_1$ exists. If $r>0$, then $f_1$ exists and $\parallel T{e}_1\parallel =\parallel {\sigma }_1{f}_1\parallel ={\sigma }_1$. If $r=0$, then $T{e}_1=0$ and $\parallel T{e}_1\parallel =0={\sigma }_1$, without requiring any vector of $W$. Together with step 2.1, this proves that the maximum defining $\parallel T\parallel$ equals the largest singular value and is attained at $e_1$ whenever $V\mathrm{\ne }0$, including $W=0$.