Around 33:30, how do we know that x=xp+xn gives a complete solution of Bx=b? The way I think of it is that B*xp+B*xn=b+0, so we only get one particular solution that we chose arbitrarily. Can I make linear combinations with the null space basis vectors to obtain all the other solutions or what am I missing? Thanks####[['x=xp+xn', 'Mathematical_Related_Concept', 'NEU'], ['linear combinations', 'Mathematical_Related_Concept', 'NEU']]