upper bound for the norm of vector belonging to the column space of a matrix

Question

If $u$ is a vector that belong to the column space of a matrix $A$ and $\sigma_{\min}(A)$ is the smallest non zero eigenvalue of $A^T A$, then I read in a paper that we can write $$ \sigma_{\min}(A)^{\frac{1}{2}}\|u \| \leq \|A u \|$$ Any hints on how to prove this inequality.

Answer 1

The standard definition would be that $\sigma_\min(A)$ is the square root of the smallest nonzero eigenvalue of $A^T A$. Anyway, since $A^T A$ is symmetric there exists $m$ orthonormal eigenvectors of $A^T A$ that spans $\operatorname{Im}(A^T A)$, where $m=\operatorname{rank}(A^TA)=\operatorname{rank}(A)$. This means $u$ can be uniquely written as $u=\sum_{i=1}^m \alpha_i x_i$ for some scalars $\alpha_i$ where $x_i$ is the eigenvector corresponding to eigenvalue $\lambda_i$, i.e. $A^TA x_i = \lambda_i x_i$ and $\lVert x_i \rVert=1$. Note that also $A^TAu=\sum_{i=1}^m \alpha_i \lambda_i x_i$ and consequently $u^TA^TAu=\sum_{i=1}^m \lambda_i \alpha_i^2$. Therefore, $$\lambda_\min \lVert u \rVert^2 = \lambda_\min \sum_{i=1}^m \alpha_i^2 \leq \sum_{i=1}^m \lambda_i \alpha_i^2 = u^TA^TAu = \lVert Au \rVert^2$$