Is $\bigl\|\frac{vv^T}{v^Tv}\bigr\|=1$? For any vector $v\in \mathbb{R}^{n}$ 
I am stuck while showing that
  $$\biggl\|\frac{vv^T}{v^Tv}\biggr\|=1,$$ where $v\in \mathbb{R}^n$, and $\|.\|$ is a matrix norm.


Here is my steps:
I used Frobenius norm: A Frobenius matrix norm for any matrix $A$ is defined by 
\begin{equation*}
\begin{split}
||A\|_F & = \biggl( \sum_{i=1}^{m}\sum_{j=1}^{n}|a_{ij}|^2\biggr)^\frac{1}{2}\\
& = \biggl(tr(A^TA)\biggr)^\frac{1}{2}
\end{split}
\end{equation*}
Now
\begin{equation*}
\begin{split}
 \biggl\|\frac{vv^T}{v^Tv}\biggr\|_F& = \biggl (tr\biggl(\frac{vv^T}{v^Tv}\biggr)^T\biggl(\frac{vv^T}{v^Tv}\biggr)\biggl)^{\frac{1}{2}} \\
& = \biggl (tr\biggl(\frac{vv^T}{v^Tv}\biggr)^2\biggr)^{\frac{1}{2}}\\
%& =\frac{1}{\|v\|_{F}}\biggr(tr\bigl(vv^T\bigr)^2\biggl)^{\frac{1}{2}}\\
\end{split}
\end{equation*} 

Then how can I continue from here?
 A: First notice that $v^Tv$ is a scalar. Moreover
$$\operatorname{tr}((vv^T)^Tvv^T)=\operatorname{tr}(vv^Tvv^T)=v^Tv\operatorname{tr}(vv^T)$$
and using the fact that
$$\operatorname{tr}(AB)=\operatorname{tr}(BA)$$
we get that
$$\operatorname{tr}((vv^T)^Tvv^T)=(v^Tv)^2$$
so we deduce the desired result easily.
A: Let $Q = vv^t$. And note that $v^t v$ is just a (nonnegative) number, so 
$$
\biggl\| \frac{Q}{v^t v} \biggr\| = \frac{\| Q \|}{v^t v}
$$
so that all you need to prove is that 
$$
\| Q \| = v^t v. 
$$ 
Now, let $Q=[q]_{ij}$, then
\begin{align}
tr(Q) &= \sum q_{kk}\\
 &= \sum v_k v_k
\end{align}
On the other hand, 
\begin{align}
v^t v   &= \sum_i v_i v_i \ge 0.
\end{align}
Hence these are equal. 
An example helps a lot here. Pick $v = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Then 
\begin{align}
v^t v 
&= \begin{bmatrix} 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \end{bmatrix}\\
&= 1 \cdot 1 + 3 \cdot 3, \text{, while} \\
v v^t 
&= \begin{bmatrix} 1\cdot 1 & 1\cdot 3 \\
3\cdot 1 & 3\cdot 3 \end{bmatrix} \text{, so} \\
tr(v v^t) 
&= 1\cdot 1 + 3\cdot 3
\end{align}
and it's pretty easy to see where the corresponding terms come from. 
A: For Frobenius norm, simply use the property  trace(AB)=trace(BA), or in our case $\operatorname{tr}(uv^t)=\sum(uv^t)_{ii}=\sum u_iv_i=u^tv$, so we even have a more general
$$
\biggl\|\frac{uv^t}{u^tv}\biggr\|=1.
$$
If we choose instead the operator 2-norm,
$$
\|A\|_2 = \sup_{\|x\|_2 = 1} \|Ax\|_2
$$
and write
$\|(vv^t)x\|=\|vv^tx\|=\underbrace{\|v\|\color{red}{|v^tx|}\leq\|v\|\color{red}{\|v\|\|x\|}}_{\text{Cauchy-Schwarz}}=\|v\|^2\|x\|$, where the equality, the maximum of the operator norm, is attained for $x$ proportional to $v$, while $\|x\|=1$; then dividing both sides by the scalar $\|v\|^2=v^tv$ we get 
$$\biggl\|\frac{vv^t}{v^tv}\biggr\|=1$$
