# Rows of V in reduced SVD with norm 1

Suppose, we're given the reduced/compact SVD of the rank-$$r$$ Matrix

$$A=USV^T$$

where $$U\in\mathbb{R}^{m\times r}$$, $$S\in\mathbb{R}^{r\times r}$$ and $$V\in\mathbb{R}^{n\times r}$$ and suppose the $$i$$-th column of $$A$$ ($$A_i$$) is not in the span of the other columns $$A_1,...,A_{i-1},A_{i+1},...,A_n$$ (linearly independent), then show that the $$i$$-th row of $$V$$ is of norm 1, $$||(V^T)_i||=1$$.

I came across the problem myself, while trying to implement some stuff. I'm not sure if the statement holds but I couldn't find a counterexample so far. Possibly, we could show stronger implications but this would suffice my purposes. Does anyone have an idea? Thanks!

Consider the compact SVD of the rank-$$r$$ matrix $$A=USV^T$$ with $$U\in \mathbb{R}^{m\times r}$$, $$S\in \mathbb{R}^{r\times r}$$ and $$V\in \mathbb{R}^{n\times r}$$. For the $$i$$-th column of $$A$$, $$A_i$$, and the $$i$$-th row of $$V$$, $$v_i$$, it then holds that $$A_i\notin \text{Span}(A_1,\dots,A_{i-1},A_{i+1},\dots,A_n)$$ if and only if $$\langle v_i,v_j\rangle =\delta_{ij}\enspace \forall j=1,\dots,n$$.
($$\Rightarrow$$) Consider $$k=[k_1,\dots,k_n]^T\in \ker(A)$$ for which holds \begin{align}\label{Lemmaoneweek} 0=Ak=\begin{bmatrix} a_{11}k_1+\dots+a_{1n}k_n\\\vdots \\a_{m1}k_1+\dots +a_{mn}k_n \end{bmatrix}=k_1A_1+\dots+k_iA_i+\dots+k_nA_n. \end{align} Furthermore, the following implications hold: \begin{align*} &A_i\notin \text{Span}(A_1,\dots,A_{i-1},A_{i+1},\dots,A_n)\\ \Rightarrow& A_i\neq \sum_{j\neq i}\lambda_j A_j \enspace \forall \lambda_j\in \mathbb{R}, j=1,\dots,i-1,i+1,\dots ,n\\ \Rightarrow&\left(0=\lambda A_i+\sum_{j\neq i}\lambda_jA_j,\enspace \lambda\in \mathbb{R}\enspace \Rightarrow \lambda=0\right). \end{align*}
Combining that with the kernel equation above, we obtain that if $$A_i$$ is not in the span of the other columns of $$A$$ then $$k_i$$ is zero for any arbitrary kernel element $$k$$. Thus, it also holds for any basis vector of any basis of the kernel of $$A$$. Therefore, if we were to complete $$V$$ such that it becomes an orthonormal basis of $$\mathbb{R}^n$$ the vectors to be appended all have a zero entry in the $$i$$-th entry. This means that the $$i$$-th row of $$V$$ already was of norm $$1$$ and because the rows of the completed $$V$$ are orthonormal to all other rows, the $$i$$-th row of the compact $$V$$ must already be orthogonal to all others.
($$\Leftarrow$$) Reciprocally, suppose that $$A_i\in \text{Span}(A_1,\dots,A_{i-1},A_{i+1},\dots,A_n)$$, then this implies that \begin{align*} \exists \lambda_1,\dots,\lambda_{i-1},\lambda_{i+1},\dots,\lambda_n\in \mathbb{R}\enspace :\enspace 0=-A_i+\sum_{j\neq i}\lambda_j A_j. \end{align*} Choosing $$k=[\lambda_1,\dots,\lambda_{i-1},-1,\lambda_{i+1},\dots,\lambda_n]^T$$ yields that $$k\in \ker(A)$$ with the $$i$$-th component non-zero. This means that for every basis of the kernel of $$A$$ there must exist a basis vector with the $$i$$-th component non-zero. Again, if we were to complete $$V$$ to be an orthonormal basis of $$\mathbb{R}^n$$ it would append an element to the $$i$$-th row of $$V$$ that is non zero which means that $$v_i$$ was not of norm $$1$$.