Unboundedness in linear programming

Question

Let $K = \{ x : Ax \le b, x \ge 0 \}$, $b \ge 0$. If there are solutions $x \ge 0$ different from zero of $A x \le 0$, show that $K$ is unbounded.(A is mxn, x is 1xn b is 1xm matrix.)

Answer 1

Let $x_0 \ge 0$ be a non-zero solution of $Ax \le 0$. Then, for each $t \ge 0$, we have $tx_0 \ge 0$ and $$ A(tx_0) = tAx_0 \le 0 \le b $$ hence $tx_0 \in K$. As $$ \| tx_0 \| = t\|x_0\| \to \infty, \quad t \to \infty $$ $K$ is unbounded.