14240/TutorialOctober28

Contents 
Boris
Be Efficient
By this point in the course, we become good at solving systems of linear equations. However, we should not use this same old problemsolving 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 timeconsuming 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 .