Raising the bar

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

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

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

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

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

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

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

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

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

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

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

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

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

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