# Calculating the Dual Problem to a Minimisation Problem - confusion

Let $A$ be an $m$x$n$ real matrix. Let QP denote the problem: minimise $f(x,y) = \frac12 x^Tx$ such that $x \ge 0, y\ge 0$, and $Ax-y=b$.

I want to prove that the dual of this problem QP* is: maximise $f^*(\lambda, \mu, x) = \lambda^Tb - \frac{1}{2} x^Tx$ such that $\lambda\ge 0, \mu \ge 0$ and $x = A^T \lambda + \mu$. I am having some trouble however and I was hoping somebody would point me in the right direction.

I started by defining the lagrange $L(x,y,\lambda) = \frac12 x^Tx - \lambda^T (Ax - y -b)$, and my goal is to find $g(\lambda):= \displaystyle\inf_{x,y}L(x,y,\lambda)$, and then maximise over such $\lambda$ where this $g(\lambda)$ has a finite minimum. I observed that since $L(x,y,\lambda) = \frac12 x^Tx - \lambda^T Ax + \lambda^T y +\lambda^T b$, then this only has a finite minimum if we ensure $\lambda^T y \ge 0$, i.e. that $\lambda \ge 0$. So to minimise the Lagrange, it makes sense to set $y=0$.

By partially differentiating the Lagrange, I get that the vector $(x,y) = (A^T \lambda, 0)$ (thinking of vectors within a vector here) minimises L, so I have $$g(\lambda) = -\frac12(A^T \lambda)^T (A^T \lambda) + \lambda^T b.$$

Is this analysis correct? How do I write it in the given form QP*? I am very confused about where the $x$ and the $\mu$ come from, as well as the constraints. Any help would be greatly appreciated with this!

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