Proof for: the norm of the product between two vectors is equal to the product of their norms

Question

I need to prove the following $\lVert \pmb{a}\pmb{b}^T\rVert_2 = \lVert \pmb{a}\rVert_2\cdot\lVert \pmb{b}\rVert_2$ where $\pmb{a}$ and $\pmb{b}$ are vectors and $\pmb{a}\pmb{b}^T$ is a matrix. This is one of the properties for vector norms.

All help is really appreciated. I have a suspicion that the following equations might be helpful in the derivations, considering they were given alongside the question:

$\lVert A\rVert_2 = \sqrt{\rho(A^TA)}$ where $\rho(A)$ is the spectral radius of $A$.

Answer 1

Every where I use the $2$-norm

Observe that we can write $$ \pmb{a}\pmb{b}^T = \lVert \pmb{a} \rVert \cdot \lVert \pmb{b} \rVert \cdot\pmb{n}_{\pmb{a}}\pmb{n}_{\pmb{b}}^T $$

Where $\pmb{n}_{\pmb{u}} = \frac{1}{\lVert \pmb{u} \rVert} \pmb{u}$ is the normalization of $\pmb{u}$. Now we have \begin{align*} (\pmb{n}_{\pmb{a}}\pmb{n}_{\pmb{b}}^T)^T (\pmb{n}_{\pmb{a}}\pmb{n}_{\pmb{b}}^T)&= \pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{a}}^T \pmb{n}_{\pmb{a}}\pmb{n}_{\pmb{b}}^T\\ &=\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T \end{align*} So that \begin{align*} \lVert \pmb{a}\pmb{b}^T\rVert &= \lVert \pmb{a} \rVert \cdot \lVert \pmb{b} \rVert \cdot \sqrt{\rho(\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T)} \end{align*}

so we need to find the largest eigenvalue of $\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T$ but observe that it is a projection matrix :

$$(\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T)^2=\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T \pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T = \pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T$$

if we write $\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T = D\Sigma D^T$ the eigen decomposition, then we have $D\Sigma D^T = D\Sigma D^T D \Sigma D^T = D\Sigma^2 D^T$ which means by uniqueness of the eigen decomposition that $\Sigma^2=\Sigma$. Since it is a diagonal matrix we have that every eigen values satisfies $\sigma^2=\sigma$ so that they take value in $\lbrace 0 , 1 \rbrace$.

Now it only remains to show that there is a eigen value that is $1$. But $\pmb{n}_{\pmb{b}}$ is an eigenvector associated to eigenvalue $1$, indeed $$(\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T)\pmb{n}_{\pmb{b}} = \pmb{n}_{\pmb{b}}$$

and so $\rho(\pmb{n}_{\pmb{b}}\pmb{n}_{\pmb{b}}^T) = 1$ which in turn means that $$\lVert \pmb{a}\pmb{b}^T\rVert = \lVert \pmb{a} \rVert \cdot \lVert \pmb{b} \rVert $$