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I actually determined the answer shortly after posting! Sometimes just typing up the question helps as it makes you really think about it! :) Claim: {$f:V \rightarrow W$ surjective} $\iff$ {$f^*:W^* \rightarrow V^*$ injective}, where $V$ and $W$ are as in the question. Proof of Claim: "$\Rightarrow$" Suppose that $f$ is surjective. For $w \in W^*$, {$w ... 2 You have the right approach. Suppose first$f$is surjective. If$g\in W^*$and$f^*(g)=0$, then$g(f(v))=0$for all$v\in V$and since$f$is surjective this means$g(w)=0$for all$w\in W$, hence$g=0$. Thus$f^*$is injective. Now suppose$f$is not surjective, so its image is a proper linear subspace. Let$w\notin f(V)$and extend$w$and a basis for ... 1 The fact that dual to dual norm is equal to the original norm in case of finite-dimensional spaces is equivalent to the fact that the corresponding Banach space is reflexive. By James' theorem, a Banach space$B$is reflexive if and only if every continuous linear functional on$B$attains its maximum on the closed unit ball in$B$. That is surely true for ... 0 First of all, to solve this with the simplex method (tableau method) the inequalities of the contraints should be equalities. So you have to use slack variables. $$4x_1+3x_2 \leq 600$$ $$x_1+x_2 \leq 160$$ $$3x_1+7x_2 \leq 840$$ $$x_1,x_2 \geq 0$$ $$\Downarrow$$ $$4x_1+3x_2 +x_3 = 600$$ $$x_1+x_2 +x_4 =160$$ $$3x_1+7x_2 +x_5= 840$$$\$x_1,x_2,x_3, x_4, ...