# SVD for augmented matrix

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?

## 1 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 $\sqrt{\lambda_i + 1} = \sqrt{\sigma_i^2 + 1}$.

• Why is $A^TA$ positive semi-definite? Does that hold for any choice of $A$? – Jeff Koff Feb 16 '16 at 10:43
• Yes. The matrix $A^TA$ is symmetric and $\left< A^TAv, v \right> = (A^T Av)^T v = v^T A^T A v = (Av)^T Av = \left< Av, Av \right> \geq 0$. – levap Feb 16 '16 at 10:45