Linear Algebra: Basis Reduction and Extension Theorems

The problem asks us to prove two statements related to bases in a vector space .
Let be a generating set (spanning set) for .
Let be a linearly independent set of vectors in .

We need to prove two things:

Let’s address each part separately.

Given that is a spanning set for , this means that every vector in can be written as a linear combination of vectors in .
A basis for is a linearly independent spanning set for .
If is already linearly independent, then itself is a basis, and we don’t need to remove any vectors.

If is linearly dependent, then there exists at least one vector in that can be written as a linear combination of the other vectors in .
Let’s use the following procedure:
Start with .
Consider the vectors one by one.
If , remove it. The span remains the same.
If , keep it.
Consider . If is a linear combination of , remove . The span remains the same.
If is not a linear combination of , keep .
Continue this process. For each vector , check if is a linear combination of the previously kept vectors. If it is, remove . If it is not, keep .

Let be the set of vectors kept after this process.
By construction, is linearly independent. This is because if was a linear combination of , it would have been removed.
Also, the span of is the same as the span of . This is because whenever we remove a vector , it means is a linear combination of the vectors (or more precisely, the vectors from that were kept before ). Therefore, removing does not change the span.
Since spans , also spans .
Since is linearly independent and spans , is a basis for .
This procedure shows that we can reduce to a basis for by removing some vectors.

Given that is a linearly independent set of vectors in .
If already spans , then is a basis, and we don’t need to add any vectors.

If does not span , then there exists at least one vector in that is not in the span of .
Let be a spanning set for . (The problem statement implies such an exists).
Consider the set . This set spans because spans .
Now, we can apply the reduction process from Part 1 to the set .
However, we need to ensure that the vectors in are kept in the resulting basis.
We can modify the reduction process as follows:
Start with the set . Since is linearly independent, all its vectors are kept.
Now, iterate through the vectors in .
For each :
Check if is a linear combination of the vectors currently in .
If is a linear combination of the vectors in , then .
If is not a linear combination of the vectors in , then .
Continue this process until all vectors in have been considered.
Let the final set be .

By construction, contains and is linearly independent. This is because we only add a vector if it is not in the span of the previously selected vectors.
Also, spans . This is because spans , and we only removed vectors that were linearly dependent on the preceding ones. Since spans , and , the span of must be .
Therefore, is a basis for that contains . This means we have extended to a basis for by adding some vectors from .

Part 1: Reduction of a spanning set to a basis.
Let be a spanning set for .
Consider the vectors in one by one.
Let .
For :
If , then set .
If , then set .
Let .
By construction, is linearly independent.
Also, for every , . Therefore, .
Since spans , also spans .
Thus, is a basis for , and . This means can be reduced to a basis by removing some vectors.

Part 2: Extension of a linearly independent set to a basis.
Let be a linearly independent set in .
Let be a spanning set for .
Consider the set . This set spans .
Now, we apply the reduction process from Part 1 to , but with a slight modification to ensure is included.
Let .
For :
If , then set .
If , then set .
Let .
By construction, is linearly independent because is linearly independent, and we only add if it is not in the span of the previous vectors.
Also, . Since spans , spans . Therefore, spans .
Thus, is a basis for , and . This means can be extended to a basis by adding some vectors (from ).

These two results are fundamental theorems in linear algebra, often referred to as the “Basis Reduction Theorem” and the “Basis Extension Theorem” (or “Steinitz Exchange Lemma” for the extension part).

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