If a matrix $A_{n\times n}$ is invertible, then $\left\|A^{-1}\right\|_2 = \dfrac{1}{\min\limits_{i} \sigma_i}$where $\sigma_i$ is the $i$-th singular value of $A$
How to prove the 2-norm of an invertible matrix is exactly the reciprocal of its minimum singular value?
matrix
Singular value decomposition of $A$ can be written as $A=USV^T$ with $S$ being a diagonal matrix of singular values of $\sigma_i$. Euclidean norm of a matrix can be written as : $||A||_2=\sigma_{max}(A)$, meaning that the norm is the maximum singular value. If $A$ is invertible, then $A^{-1}=(USV^T)^{-1}$. From here it follows:$A^{-1}=(USV^T)^{-1}\\$$      =(V^T)^{-1}S^{-1}U^{-1}\\$If $A$ is a real valued matrix, then from orthagonality:$A^{-1}=VS^{-1}U^T\\$.This new matrix $A^{-1}$ now has singular values $S^{-1}$, and its norm would be $\max(diag(S^{-1}))$, where $diag$ is the diagonal of the matrix. Because $S$ is a diagonal matrix, its inverse is computed by simply inverting every single element. Therefore,$||A^{-1}||_2=\sigma_{max}(A^{-1})=\max\limits_{i} diag(S^{-1})_i = \max\limits_{i}\frac{1}{\sigma_i}=\frac{1}{\min\limits_{i}{\sigma_i}}$ and this completes the proof.