# Proving the SVD Theorem by induction on the rank of $A$

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.

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*(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 $$A=\sigma_1\,\frac{u_1}{\|u_1\|_2}\,\left(\frac{v_1}{\|v_1\|_2}\right)^T.$$