By this point in the course, we become good at solving systems of linear equations. However, we should not use this same old problem-solving strategy over and over if a more efficient one exists. Consider the following problems:
Q1: Determine if is linearly independent in .
We can solve this linear equation to find the answer:
- where .
Yet there is a less time-consuming approach that relies on two observations:
- (1) The dimension of is so the size of a basis is also .
- (2) No linearly independent set can have more vectors than a generating set (by the Replacement Theorem).
Since a basis is a generating set and the size of is , then the Replacement Theorem tells us that cannot be linearly independent. Hence, the problem can be solved without solving any linear equations.
Q2: Determine if the polynomials generate .
Once again, we can solve a linear equation but we do not have to. Observe:
- (1) The dimension is so the size of a basis is also .
- (2) No generating set can have fewer vectors than a basis (by a Corollary to the Replacement Theorem).
Since there are only polynomials, then the Corollary tells us that it cannot generate . Once again, we used a more efficient strategy.
Extending a Linearly Independent Set to a Basis
Boris's tip (for concrete sets and vector spaces only):
If a problem requires us to extend a linearly independent set to a basis, then the easiest approach is to add vectors from the standard ordered basis. Here is an example:
Let be a linearly independent subset of . To extend to a basis, add vectors from . The only question is which vector(s) should we add?
We see that both vectors in have a as the third component so a safe choice is to add . Since has a dimension of , then is a basis of .