Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an exercise and it is divided into steps. The first step says:

Suppose $A\in\mathbb{R}^{m\times n}$ has rank 1. Let $u_1\in\mathbb{R}^n$ be a vector in $R(A)$ such that $\left \| u_1 \right \|_2=1$. Show that every column of A is a multiple of $u_1$. Show that A can be written in the form $A=\sigma_1u_1v_1^T$, where $v_1\in\mathbb{R}^m, \left \| v_1 \right \|_2=1$, and $\sigma_1>0$.

I'm getting confused because I keep wanting to use properties of SVD and I know I can't. I'm hoping a hint will put me on track.

share|improve this question

1 Answer 1

*(there seems to be an issue with the dimensions: the columns of $A$ are in $\mathbb{R}^m$, so $u_1\in\mathbb{R}^m$, $v_1\in\mathbb{R}^n$)*

Since $A$ has rank 1, every time you take two columns that set will be linearly dependent, i.e. one column is a multiple of the other.

I personally don't see the point, in the hint, of taking $u_1$ in $R(A)$. I would take $u_1\in\mathbb{R}^m$ to be the first column of $A$. Then we can write $$ A=\begin{bmatrix}u_1&\lambda_2 u_1&\cdots&\lambda_n u_1\end{bmatrix} =u_1\cdot\begin{bmatrix}1\\ \lambda_2\\ \vdots \\ \lambda_n\end{bmatrix}^T=u_1\cdot v_1^T. $$ Now let $\sigma_1=\|u_1\|_2\,\|v_1\|_2$. Then $$ A=\sigma_1\,\frac{u_1}{\|u_1\|_2}\,\left(\frac{v_1}{\|v_1\|_2}\right)^T. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.