# Singular Value Decomposition of Rank 1 matrix

I am trying to understand singular value decomposition. I get the general definition and how to solve for the singular values of form the SVD of a given matrix however, I came across the following problem and realized that I did not fully understand how SVD works:

Let $0\ne u\in \mathbb{R}^{m}$. Determine an SVD for the matrix $uu^{*}$.

I understand that $uu^{*}$ has rank 1 and thus only has one singular value i.e.

$$\Sigma = \begin{pmatrix} \sigma_1 & \ldots & 0\\ 0& \ldots & 0 \end{pmatrix} \in\mathbb{R}^{m\times m}$$

and I realize since $uu^{*}\in\mathbb{R}^{m\times m}$ then for $uu^{*}=U\sum V^{*}$ that $U,\Sigma,V\in\mathbb{R}^{m\times m}$. Additionally, I realize that the columns of $U$ and $V$ are orthonormal.

I guess my question is how do you determine U and V from $uu^{*}$?

The short SVD of $uu^*$ is $v\sigma v^*$ with $v=\frac{u}{\|u\|}$ and $σ=\|u\|^2$.
If you compute the bisector $w=u+\frac{u_1+0}{|u_1|+0}\|u\|e_1$ of $u$ and $e_1$, then you can get a full unitary matrix $V$ as the reflection matrix $I-2\frac{ww^*}{\|w\|^2}$, so that $uu^*=V\Sigma V^*$
• How did you get that $\sigma=\|u\|^{2}$? – user133541 Mar 8 '14 at 15:48
• Because $v$ has norm $1$ and $uu^*=vσv^*$ has to be an identity. Note that this works in general for $ab^*=uσv^*$ with $u=\frac{a}{\|a\|}$, $v=\frac{b}{\|b\|}$ and $σ=\|a\|\,\|b\|$. – LutzL Mar 8 '14 at 15:52
• So, this implies that the singular value for a rank 1 matrix given by $ab^{*}$ is always $\sigma =\|a\|\|b\|$? – user133541 Mar 8 '14 at 15:56
• Yes, there is only one positive singular value and that has to coincide with the operator norm $\|ab^*\|_{op}=\sup_{\|x\|=1}\|ab^*x\|_2=\sup_{\|x\|=1}\sqrt{x^*ba^*ab^*x}=\|a\| \ \sup_{\|x\|=1}|b^*x|$... – LutzL Mar 8 '14 at 16:00