I am trying to solve the following problem:
Let z $\in \mathbb{R}^{n}$, A $\in \mathbb{R}^{m \times n}$, $m \geq n$ and let $B = \begin{pmatrix} A \\ z^T \end{pmatrix}$. Let us call $\sigma_1(C)$ to the first singular value of any $C$ matrix. Show that $$\sigma_1(A) \leq \sigma_1(B) \leq \sqrt{\sigma_1(A)^2 + {||z||_2}^2}$$
What I've tried to do is to take the eigenvector associated with $A^tA$ and I got:
$$(B^tB)v=$$ $$(A^tA + z^tz)v=$$ $${\sigma_1(A)^2}v + (z^tz)v$$
At this point I got stuck, I don't know how to arrive to the inequalities from here. I would really appreciate some help with this. Thanks in advance
Let $v$ be a unit singular vector of $A$ corresponding to the singular value $\sigma_1(A)$. Using the definition/characterisation $\sigma_1(M)=\max_{\|x\|_2=1}\|Mx\|_2$ and Cauchy-Schwarz inequality $(z^Tx)^2\le\|z\|_2^2\|x\|_2^2$, we get
$$ \sigma_1(A)=\|Av\|_2\le\|Bv\|_2\le\sigma_1(B) =\max_{\|x\|_2=1}\|Bx\|_2 =\max_{\|x\|_2=1}\sqrt{\|Ax\|_2^2+(z^Tx)^2} \le\sqrt{\sigma_1(A)^2+\|z\|^2}. $$