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Answer to Question #23762 in Linear Algebra for Matthew Lind

Question #23762
Show that:
a) If U = {u^1, u^2,...,u^N} is a spanning set for W, then U is a basis for W iff, after the removal of any one member, U is no longer a spanning set for W; and

b) If U = {u^1, u^2,...,u^N} is a linearly independent set in W, then U is a basis for W iff, after including in U any new member from W, U is no longer linearly independent.
Expert's answer
a) if all u_i are basis thenthey are linearly independent (LI), and thus each u_j cannot be experessed by
linear combination of other u_i, so W is not a Span of {u_i, 1<=i<=n}.
Conversely, suppose that {u_i, 1<=i<=n} are LD (linearly
dependent). Then some u_j can be expressed as linear combination of other
elements {u_i, 1<=i<=n, i <> j}. So, we can exclude u_j from
U and still U is spanning set of W, that contradicts assumption.

b) Suppose U is basis, then U is LI, and moreover it is maximal LI set in W.
So, including any new member to U cannot preserve its linear independence.
Conversely, if we include some v to U and they are now LD then there is nonzero
set of scalars a,b,c,...,d that

au1+bu2+...+cun+dv=0

If d=0, then a*u1+b*u2+...+c*un=0 and then a=b=...=c=0, since U is LI. So
{a,b,c,...,d}={0} a contradiction.
Thus d is nonzero, and so d = (-a/d)u1+(-b/d)u2+...+(-c/d)un belongs to
Span(U).

Thus as any v belongs to Span(U) then W is in Span(U) and obviously then
W=Span(U) and U is basis.

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