The problem is $$\begin{aligned} \min ~&\frac{1}{2}\|x\|_2^2\\ \text{s.t.}~ &a^Tx=b\\ &x\ge 0 \end{aligned}$$ How can we find the optimal solution with KKT condition?
My thinking:
Firstly, it's trival to write the KKT condition as follows: $$L(x,w,v)=\frac{1}{2}\|x\|_2^2+w(a^Tx-b)-v^Tx$$ $$\begin{aligned} \nabla_x L(x,w,v)=x+wa-v&=0\\ v&\ge 0\\ a^Tx&=b\\ x&\ge 0\\ v_ix_i&=0 \end{aligned}$$ But how can we solve $x$ with expression of $a,b$? Could anyone help to give the detailed steps? Thanks.