Lecture 30: More on spanning and linear independence (Nicholson Section 5.2)

Alternate Video Access via MyMedia | Video Duration: 50:02

Description: Started by reminding students of definition of span and linear independence.

4:48 --- is [3;-1;2;1] in Span([1;1;0;1],[2;0;0;2],[0;2;-1;1])?

14:10 --- Does Span([1;1;2],[1;-1;-1],[2;1;1]) equal R^3? Can an arbitrary vector [a;b;c] be written as a linear combination of the three vectors?

25:45 --- Is {[1;1;2],[1;-1;-1],[2;1;1]} a basis for R^3?

29:40 --- Is a set of 4 vectors in R^3 a linearly independent set? Asked about a specific example but it should be clear to you that whenever you have 4 or more vectors in R^3 the set will be linear dependent. This is because when you set up the system of linear equations needed to address the question you’ll have more unknowns than you have equations. This means that either there’s no solution or there’re infinitely many solutions. (Having exactly one solution is not an option.) And having no solution isn’t an option (because you already know that the zero vector is a solution) so it follows that there are infinitely many solutions.

34:09 --- stated this as a general theorem.

36:40 --- Considered a specific example of 3 vectors in R^3. Are they linearly dependent or not? It turned out they aren’t.

41:00 --- Considered the same set of 3 vectors --- do they span R^3? No. What’s interesting is --- in the process of figuring out that the answer is “no” you find the scalar equation of the plane that the three vectors do span.

46:45 --- Given 3 vectors in R^3, do they form a basis for R^3? (answer maybe yes, maybe no. It depends on the specific vectors.) What about 2 vectors? (answer: never! Can’t span R^3!) What about 4 vectors (answer: never! Will always be linearly dependent!)