Basis of a vector space and linear combinations
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Exercise:
A subset S subseteq V is a basis of V iff every vin V can be written in a unique way as a linear combination of vectors in S.
Solution:
Proof. For simplicity we will ase that |S| infty i.e. the set S is finite. The case |S|infty is similar. First we show Longleftarrow: We ase that every vin V can be written in a unique way as linear combination of elements from S. Now glqq can be writtengrqq means that SpSV. Next write Sv_...v_n. Consider the vector in V. Clearly v_+...+ v_n. Since can be written in a unique way as a linear combination of v_...v_k it follows that if a_v_+...+a_nv_n then a_...a_n Longrightarrow S is linearly indepent. This proves that S is a basis. Now we show Longrightarrow: We ase S is a basis for V. By asption SpSV so this means every vin V can be expressed as a linear combination of vectors from S. It remains to show that this is unique. Ase by contradiction that exists vin V which has two different ways to be written as a linear combination of vectors from S: va_v_+...+a_nv_n vb_v_+...+b_nv_n and a_kneq b_k for some leq kleq n. Rightarrow a_-b_v_+...+a_k-b_kv_k+...+a_n-b_nv_n and we deduce that can be written as a non-trivial linear comination of elements from S which implies that S is linearly depent.
A subset S subseteq V is a basis of V iff every vin V can be written in a unique way as a linear combination of vectors in S.
Solution:
Proof. For simplicity we will ase that |S| infty i.e. the set S is finite. The case |S|infty is similar. First we show Longleftarrow: We ase that every vin V can be written in a unique way as linear combination of elements from S. Now glqq can be writtengrqq means that SpSV. Next write Sv_...v_n. Consider the vector in V. Clearly v_+...+ v_n. Since can be written in a unique way as a linear combination of v_...v_k it follows that if a_v_+...+a_nv_n then a_...a_n Longrightarrow S is linearly indepent. This proves that S is a basis. Now we show Longrightarrow: We ase S is a basis for V. By asption SpSV so this means every vin V can be expressed as a linear combination of vectors from S. It remains to show that this is unique. Ase by contradiction that exists vin V which has two different ways to be written as a linear combination of vectors from S: va_v_+...+a_nv_n vb_v_+...+b_nv_n and a_kneq b_k for some leq kleq n. Rightarrow a_-b_v_+...+a_k-b_kv_k+...+a_n-b_nv_n and we deduce that can be written as a non-trivial linear comination of elements from S which implies that S is linearly depent.