Let E1 = (1, 0, ... ,0), E2 = (0, 1, 0, ... ,0), ... , En = (0, ... ,0, 1)
be the standard unit vectors of Rn. Let x1 ... ,xn be numbers. Show that if
x1E1+...+xnEn=0 then xi=0 for all i.
The Attempt at a Solution
Proof By contradiction
Assume to the contrary that x1E1+...+xnEn=0 then xi≠0 for some i. We also assume that x1...xi-1 and xi+1...xn are zero. Rewriting the equation we get
x1E1+.xpEp+...+xnEn=0 where xpEp is a nonzero scalar. xpEp=-x1E1-...-xpEp-1-xpEp+1-..-xnEn. But this leads to a contradiction since we assumed earlier that x1...xi-1 and xi+1...xn are zero. Thus x1E1+...+xnEn=0 xi=0 for all i.
Let me know where my proof begins to fall apart? And how do I go about it?