31
Jul 19

Playing with bases




Playing with bases

Here is one last push before the main result. Exercise 1 is one of the basic facts about bases.

Exercise 1. In C^{n} any system of k<n linearly independent vectors x_{1},...,x_{k} can be completed to form a basis.

Proof. Let e_{1},...,e_{n} be a basis in C^{n}. If each of e_{1},...,e_{n} belongs to span(x_{1},...,x_{k}), then by the lemma we would have n\leq k, contradicting the assumption k<n. Hence, among e_{1},...,e_{n} there is at least on vector that does not belong to span(x_{1},...,x_{k}). We can add it to x_{1},...,x_{k} denoting it x_{k+1}.

Suppose \sum_{j=1}^{k+1}a_{j}x_{j}=0. Since x_{k+1} is independent of the other vectors, we have a_{k+1}=0 but then because of independence of x_{1},...,x_{k} all other coefficients are zero. Thus, x_{1},...,x_{k},x_{k+1} are linearly independent.

If k+1<n, we can repeat the addition process, until we obtain n linearly independent vectors x_{1},...,x_{n}. By construction, e_{1},...,e_{n} belong to span(x_{1},...,x_{n}). Since e_{1},...,e_{n} span C^{n}, x_{1},...,x_{n} do too and therefore form a basis.

Definition 1. Let L_{1}\subset L_{2} be two subspaces. Vectors x_{1},...,x_{m}\in L_{2} are called linearly independent relative to L_{1} if any nontrivial linear combination \sum a_{j}x_{j} does not belong to L_{1}. For the purposes of this definition, it is convenient to denote by \Theta a generic element of L_{1}. \Theta plays the role of zero and the definition looks similar to usual linear independence: \sum  a_{j}x_{j}\neq \Theta for any nonzero vector a. Rejecting this definition, we can say that x_{1},...,x_{m}\in L_{2} are called linearly dependent relative to L_{1} if \sum a_{j}x_{j}=\Theta for some nonzero vector a.

Definition 2. Let L_{1}\subset L_{2} be two subspaces. Vectors x_{1},...,x_{m}\in L_{2} are called a basis relative to L_{1} if they are linearly independent and can be completed by some basis from L_{1} to form a basis in L_{2}.

Exercise 2. Show existence of a relative basis in L_{2}.

Proof. Take any basis in L_{1} (say, x_{1},...,x_{k}) and, using Exercise 1, complete it by some vectors (say, x_{k+1},...,x_{n}\in L_{2}) to get a basis in L_{2}. Then, obviously, x_{k+1},...,x_{n} form a basis in L_{2} relative to L_{1}. Besides, none of x_{k+1},...,x_{n} belongs to L_{1}.

Exercise 3. Any system of vectors x_{1},...,x_{k}\in L_{2} linearly independent relative to L_{1} can be completed to form a relative basis in L_{2}.

Proof. Take a basis in L_{1} (say, x_{k+1},...,x_{l}) and add it to x_{1},...,x_{k}. The resulting system x_{1},...,x_{k},x_{k+1},...,x_{l} is linearly independent. Indeed, if \sum_{j=1}^{l}a_{j}x_{j}=0, then

\sum_{j=1}^{k}a_{j}x_{j}=-\sum_{j=k+1}^{l}a_{j}x_{j}\in L_{1}.

By assumption of relative linear independence a_{1}=...=a_{k}=0 but then the remaining coefficients are also zero.

By Exercise 1 we can find x_{l+1},...,x_{n} such that x_{1},...,x_{k},x_{k+1},...,x_{l},x_{l+1},...,x_{n} is a basis in L_{2}. Now the system x_{1},...,x_{k},x_{l+1},...,x_{n} is a relative basis because these vectors are linearly independent and together with x_{k+1},...,x_{l}\in L_{1} form a basis in L_{2}.

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