Let {u 1 ,....,u p } be a subset of a vectorspace V and assume that b is a basis

Question

Let {u1,....,up} be a subset of a vectorspace V and assume that b is a basis for V. a)Show that if {u1,....,up} islinearly dependent in V, then{[u1]b,....,[up]b} islinearly dependent. b)Show thatif {[u1]b,....,[up]b}is linearly dependent in V, then {u1,....,up}is linearly dependent. Hint: c1u1+.....+cpup= 0(The zero vector in V) [c1u1+.....+cpup]B=[0]B(  The zero vector in Rn) Let {u1,....,up} be a subset of a vectorspace V and assume that b is a basis for V. a)Show that if {u1,....,up} islinearly dependent in V, then{[u1]b,....,[up]b} islinearly dependent. b)Show thatif {[u1]b,....,[up]b}is linearly dependent in V, then {u1,....,up}is linearly dependent. Hint: c1u1+.....+cpup= 0(The zero vector in V) [c1u1+.....+cpup]B=[0]B(  The zero vector in Rn)

Explanation / Answer

Let dim (V) =n. (n may be infinite, it doesn't matter) Note that the linear transformation given byT:V-->Rn given by T(v)=[v]b is aninjective linear transformation. I will leave its verificationfor you. Now, (a) Since  {u1,....,up} islinearly dependent in V so there exist i, not allzero, such that iui = 0. So,T(iui) = T(0) = 0 and soiT(ui) = 0, i.e. theT(ui)=[ui]b are linearlydependent. (b) The converse issimilar.T(ui)=[ui]b are linearlydependent implies there exist i, not all zero,such that iT(ui) = 0 orT(iui) = T(0). Since T isinjective so iui = 0. Hence{u1,....,up} is linearly dependent inV.

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