# 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.

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.

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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