# Why are the ordered singular values of $A^{−1}$ given by $\sigma _n^{ - 1} \ge … \ge \sigma _1^{ - 1}$?

Let $A\in{M_n}$ be nonsingular, with ordered singular values $\sigma _n^{} \le ..... \le \sigma _1^{}$. Is it true that the ordered singular values of $A^{−1}$ are $\sigma _n^{ - 1} \ge ..... \ge \sigma _1^{ - 1}$?

• If $A=USV^*$, $A^{-1}=\ldots$ – Algebraic Pavel Oct 22 '15 at 16:36

I'm following the idea of Algebraic Pavel that gave you the hint. There's a result from linear Algebra (called singular value decomposition) that tells you that you can decompose any non-singular matrix $A$ into a product $$A = U\Sigma V,$$ where $U$ and $V$ are unitary, and $\Sigma$ is a diagonal matrix, with entries exactly the singular values $\sigma_i$, $i=1\dots n$.
Now clearly $A^{-1}=V^{-1}\Sigma^{-1}U^{-1}$, since $$U\Sigma VV^{-1}\Sigma^{-1}U^{-1}=I,$$ where $I$ is the identity matrix.
Your result follows from the properties of a diagonal matrix and the uniqueness of the SVD. Indeed, $\Sigma^{-1}$ is again diagonal, with diagonal values $\sigma_i^{-1}$, $i=1,\dots n$ (an easy to show property of diagonal matrices).