Consider the real matrix $A: m \times n$, rank $(A)=n$, your SVD decomposition $A=U \Sigma V^t$. Consider now $z: m \times 1$ and $B = [A \;\;z]$. Show that

(a) $\sigma_{n}(B) \geq \sigma_{n}(A)$

(b) $\sigma_1(B) \geq \sigma_1(A).$

Tip: analyze $Bw$ and $By$ and corresponding norm 2, for $ w = (x \;\;\;0)^t$ and $y = (0 \;\;\;\alpha)^t$, $x \in \mathbb{R}^n$ and $\alpha \in \mathbb{R}$.


I could solve the item (b). I know that $\sigma_{1}(B) = \max_{x \neq 0}\dfrac{||Bx||_{2}}{||x||_{2}} \geq \max_{x \neq 0, x_{n+1}=0}\dfrac{||Bx||_{2}}{||x||_{2}} = \sigma_1(A)$. But I can't to solve the item (a).


1 Answer 1


Here is a roundabout way of doing it:

Note that $\sigma_n(A)$ is the largest $\mu$ such that $\|Ax\| \ge \mu \|x\|$ for all $x$.

Note that $\sigma_n(B)=\sigma_n(B^T)$.

Note that $\|B^Tx\| =\|\begin{bmatrix} A^T x \\ z^T x \end{bmatrix}\| = \sqrt{\|Ax\|^2+(z^Tx)^2 } \ge \|Ax\| \ge \sigma_n(A)\|x\|$ for all $x$. Hence $\sigma_n(B) \ge \sigma_n(A)$.


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