I am trying to show that for $A \in \mathbb{R}^{m\times n}$, where $m > n$, $\sigma_{min}(A) \leq \sigma_{min}([A | x])$ for some $x \in \mathbb{R}^m$. My approach is using the two norm, but am not sure if that is correct. If this is correct, I believe it involving showing that $||A||_2 \leq ||[A | x]||_2$. How can this be shown?
Would anyone be able to point me in the right direction? Thanks.
It's the other way round, $\sigma_{min}([A|x])\le\sigma_{min}(A)$.
Let $B:=[A|x]$. Since $B^TB$ is positive semi-definite, for any vector $v\in\mathbb{R}^{n+1}$, $$v^TB^TBv\ge\sigma^2_{min}(B)|v|^2.$$ In particular pick $v=(w,0)$ where $w$ is a unit singular vector satisfying $Aw=\sigma_{min}(A)w'$, then $$\sigma^2_{min}(B)\le v^TB^TBv=(w,0)\begin{pmatrix}A^TA&A^Tx\\x^TA&x^Tx\end{pmatrix}\begin{pmatrix}w\\0\end{pmatrix}=w^TA^TAw=\sigma_{min}^2(A)$$