I'm doing the following problem to prepare for my exam tomorrow:
Suppose that $A$ is a $n \times m$ matrix with $n \geq m$ and that the singular values of $A$ are $\sigma_1, \sigma_2, \dots, \sigma_m$. Now consider the augmented matrix $$B = \left[ \begin{array}{cc} A \\ I \end{array} \right],$$ where $I$ is the $m \times m$ identity matrix. We want to show that the singular values of $B$ are $\tilde{\sigma}_i = \sqrt{1 + \sigma_i^2}$ for each $i$.
I'm currently lost on where to begin. I know that $A = U\Sigma V^T$ for orthogonal matrices $U$ and $V$ with $\Sigma$ containing the singular values on the diagonal, by definition. However, I'm not seeing how to extend this to show the desired result for $B$. Can someone lend a hand?
Best Answer
Let $\lambda_1 \geq \dots \geq \lambda_m \geq 0$ be the eigenvalues of the positive semidefinite matrix $A^TA$ so $\sigma_i = \sqrt{\lambda_i}$. Consider
$$ B^T B = \left[ \begin{matrix} A^T & I_m^T \end{matrix} \right] \left[ \begin{matrix} A \\ I_m \end{matrix} \right] = A^TA + I_m. $$
Thus, the eigenvalues of $A^TA + I_m$ are $\lambda_i + 1$ and so the singular values of $B$ are $\sqrt{\lambda_i + 1} = \sqrt{\sigma_i^2 + 1}$.