Do I find a (of many) solution with the gradient descent, if
$$a^Tx=b,$$ $a=(-1,1)$ and $b = 2$
For example: $Ax=b \rightarrow (-1 \; 1)\cdot (x_1 \; x_2)^T = 2$
I set $x_1$ to -1 and $x_2$ to 1:
$-1 + x_2 = 2 \rightarrow x_2 = 3$
$-x_1 + 1 = 2 \rightarrow x_1 = 1$
Then I get a possible solution $ x = (3 \; 1)^T $, but I think that it is not a minimum one.