Lecture 29: Linear Dependence, Linear Independence (Nicholson Section 5.2)
Sub Project: Linear Algebra Lecture Videos
Alternate Video Access via MyMedia | Video Duration: 58:21
Description: 1:30 --- introduced a set which is the set of all linear combinations of two specific vectors in R^2. Discussed the set, examples of vectors in the set, etc.
10:00 --- Theorem: the set of all linear combinations of two (fixed) vectors in R^n is a subspace of R^n. Proved the theorem.
19:00 --- Introduced Span(v1, v2 ,.. vk) and stated that it’s a subspace of R^n.
22:00 --- gave an example of a subset of R^2 which is closed under scalar multiplication but not under vector addition. Gave an example of a subset of R^2 which is closed under vector addition but not under scalar multiplication.
24:40 --- example in which I show that [7;1;13] is not in Span([2;1;4],[-1;3;1],[3;5;7]). Used high school methods to solve the problem. NOTE: there’s a mistake at 30:00. The second equation should be 6 t1 + 10 t3 = 20, not 2 t1 + 4 t3 = 6!! When plugging t1 into the third equation, one gets -2 t3 + 83/7 = 13, which has the solution t3 = -4/7. This then determines t1 = 30/7 and t2 = -1/7 and, in fact, we have that [7;1;13] is in Span([2;1;4],[-1;3;1],[3;5;7]) because (30/7)* [2;1;4]+(-1/7)*[-1;3;1]+(-4/7)*[3;5;7] = [7;1;13]. What went wrong? I made a mistake when copying from my notes. My notes had the example “[7;1;13] is not in Span([2;1;4],[-1;3;-1],[3;5;7]).” Note that the third component of the second vector in the span is -1, not +1, as written on the blackboard. If you correct that mistake then you’ll get that the rest of the approach works and that you end up with the impossibility of 44/7=6.
33:30 --- example in which I show that [7;0;13] is in Span([2;1;4],[-1;3;-1],[3;5;7]). I simply wrote down the solution without showing how to find it. Students were then asked to demonstrate that [7;0;13] is in Span([2;1;4],[-1;3;-1]) and [7;0;13] is in Span([2;1;4],[3;5;7]) and [7;0;13] is in Span([-1;3;-1],[3;5;7]).
40:00 --- proved that Span([2;1;4],[-1;3;-1],[3;5;7]) equals Span([2;1;4],[-1;3;-1]).
43:30 --- defined what it means for a set of vectors to be linearly independent.
45:00 --- defined what it means for a set of vectors to be linearly dependent. Gave an example of a linearly dependent set of four vectors in R^3.
48:06 --- Showed that {[2;1;2],[-2;2;1],[1;2;-2]} is linearly independent.