Exercise 3.30 (Distance from the dual plane is the reduced cost)

Proof. Let $(A_i,c_i)$, $i=1,\dots ,m+1$ be the basic points, and let $B$ denote the associated basis matrix. Fix some $1\le j\le m+1$. In order to calculate the vertical distance from $\left(A_j,c_j\right)$ to the dual plane, we find its vertical projection onto the dual plane $\left(A_j,c_j^{\text{∗}}\right)$. By the properties of the dual plane, we can find scalars ${\lambda }_1,\dots ,{\lambda }_{m+1}\in [0,1]$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{m+1}{\lambda }_i\left(A_1,c_1\right)=$ $\left(A_j,c_j^{\text{∗}}\right)$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{m+1}{\lambda }_i=1$. It is then left to solve the equation $B\lambda ={\left(A_j,1\right)}^{\prime }$ for $\lambda$. Since $B$ is invertible, we obtain $\lambda =B^{-1}{\left(A_j,1\right)}^{\prime }$. It follows that

and

In other words,

Hence, the vertical distance from the dual plane to $\left(A_j,c_j\right)$ is the reduced cost of $x_j$. □