Consider the matrix $$ A = \begin{bmatrix} 1 & \alpha_1 \\ 1 & \alpha_2 \\ \vdots & \vdots \\ 1 & \alpha_n\end{bmatrix} $$ where $\alpha_1, \dots, \alpha_n$ are real numbers. I was trying to find a sleek way to write down the smallest singular value of this matrix. But the only way that I know is to compute $A^t A$ and solve for the eigenvalues and then pick one of them. My computations did not turn out nice.
Is there perhaps a nice argument that I am overlooking? Thanks.
$$ A^TA=\pmatrix{n&\sum_i\alpha_i\\ \sum_i\alpha_i&\|\alpha\|^2} \Rightarrow \sigma_\min(A)=\sqrt{\frac{n+\|\alpha\|^2-\sqrt{(n-\|\alpha\|^2)^2 + 4\left(\sum_i\alpha_i\right)^2}}2}. $$ I am not sure what you are expecting, but I think this is nice enough.