Norm of matrix $M= u \otimes v^*$ if $u$ and $v$ are unit norm vectors

Question

Let $u$ and $v$ be an $n \times 1$ and $m \times 1$ unit norm (L-2 norm) vectors, respectively.
Let us define matrix $M$ (of dimension $n \times m$) as the kronecker product of $u$ and $v^*$ $$M= u \otimes v^*,$$ where $v^*$ denotes the conjugate transpose of $v$.

My question: can we claim that the resulting matrix $M$ is of unit norm ?

Update:
P.S.: I saw this result in a paper where they don't precise which norm is used for the matrix.

Answer 1

\begin{align} ||u\otimes v^*||_{op} &= \sup_{||x||_2\leq 1} ||(u\otimes v^*)(x)||_2\\ &= \sup_{||x||_2 \leq 1} ||\langle x, v \rangle u||_{2} \\ &= \sup_{||x||_2 \leq 1} |\langle x, v \rangle |. ||u||_{2}\\ &= ||u||_2. \sup_{||x||_2 \leq 1} |\langle x, v \rangle |\\ &= ||u||_2 .||v||_2 \;\;\;\;\text{(By Cauchy Schwarz)} \end{align}

Answer 2

The matrix $M$ is given by

$$ M = u\otimes v^* = uv^* = \left[ \begin{array}{c} u_1v^*\\ \vdots\\ u_n v^* \end{array} \right] $$

The goal is to find the maximum eigenvalue of $M^*M$. Since $\|u\|_2 = 1$, we have

$$ M^*M = \sum_{i=1}^n |u_i|^2 vv^* = vv^* $$

Observing that $(vv^*)v = v(v^*v) = v$, it follows that $\lambda = 1$ is the only nonzero eigenvalue of $M^*M$ since $vv^*$ is rank $1$. Hence $\|M\|_2 = \sqrt{\lambda_{\max}(M^*M)} = 1$.

Answer 3

$$\eqalign{ \|u\otimes v^*\|_F^2 &= (u\otimes v^*)^*:(u\otimes v^*) \cr &= (u^*\otimes v):(u\otimes v^*) \cr &= (u^*:u)\otimes(v:v^*) \cr &= \|u\|_F^2 \otimes \|v\|_F^2 \cr &= 1 \otimes 1 \cr &= 1 \cr }$$