Lecture 31: Bases, finding bases for R^n (Nicholson, Section 5.2)

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Description: Started with a review of the definitions of linear independence and basis. 

1:50 --- Need to consider the one bizarre subspace of R^n. The subset S = {0} is a subspace of R^n. But it has no basis because no set of vectors in S will be a linearly independent set.  This discussion is the reason why theorems in the book (like Theorem 5.2.6) assume that the subspace in question is not {0}.

5:30 --- Is a given set of 4 vectors in R^3 a basis for R^3?  No: demonstrated that the vectors are linearly dependent. 

12:00 --- Discussed theorem that says that if you have k vectors in R^n and you build an n x k matrix A by putting the vectors into the columns of A then: the k vectors are linearly independent if and only if rank(A) = k.  Discussed the implications of the theorem. 

17:20 --- Given any m x n matrix A, if you find its RREF then rank(A) = number of leading 1s in RREF and so rank(A) ≤ number of columns of A.  But we also know rank(A) = number of nonzero rows of the RREF and so rank(A) ≤ number of rows of A.  So rank(A) ≤ min{m,n}.  This is a super-important fact that we use all the time. 

23:15 --- Two vectors in R^3.  Can they be a basis for R^3?  No: demonstrated that they cannot span R^3. 

34:00 --- Discussed theorem that says: given k vectors in R^n, if k < n then the vectors cannot span R^n. 

37:00 --- Is a given set of 3 vectors in R^3 a basis for R^3?  Yes: it turns out to be a basis.  Given n vectors in R^n, then figuring out whether or not it’s a basis takes work --- there’s no fast answer. 

42:34 --- I wrote "k vectors in R^n form a basis for R^n if and only if the rank of the coefficient matrix equals k." This is ridiculously wrong. If I have two vectors {<1,0,0>,<0,1,0>} in R^3 then clearly they aren't a basis for R^3. But for this example, k=2 and the rank of the coefficient matrix is 2. So by the (wrong) theorem I wrote up, those two vectors form a basis for R^3.

43:00 --- Stated and discussed theorem that says that k vectors will span R^n if rank(A)=n where A is the n x k matrix whose columns are the vectors in question. 

46:45 --- Up to this point in the lecture, all I was working on was whether or not a set of vectors was a basis for R^n.  At this point, I turned to a proper subspace, S, of R^4 and sought a basis for the subspace S.  (“Proper subspace of R^4” means a subspace which is smaller than R^4.) 

52:25 --- corrected mistake from last class. 

55:25 --- In general, given k vectors in R^4, can they be a basis for R^4?  Discussed this for k<4, k=4, and k>4.  Make sure that you understand this argument and that it has nothing to do with R^4 --- if the vectors were in R^n you’d be looking at three cases k<n, k=n, and k>n.