Suppose I know two vectors $x$ and $b$, and we want a matrix $A$ such that $Ax=b$, we are interested in finding the $A$ with minimum 2-norm.
I know that the answer is $A=\frac{bx^*}{x^*x}$, which is rank one. I don't know how to prove it. I think we may want to use the fact that the matrix 2-norm is the largest singular value, and for rank one matrix there is only one non-zero singular value. Anyone can give a hint? Thanks!
Such a matrix $A$ will satisfy $x^\top A^\top A x = \|b\|^2$, so (by thinking about the relationship between eigenvalues of $A^\top A$ and singular values of $A$), we see that the minimum $2$-norm of $A$ cannot be smaller than $\|b\|$.
The matrix that you have given is one matrix that attains this minimum $2$-norm, since $AA^\top$ there is $bb^*$.