Is $\bigl\|\frac{vv^T}{v^Tv}\bigr\|=1$? For any vector $v\in \mathbb{R}^{n}$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$\biggl\|\frac{vv^t}{v^tv}\biggr\|=1$$