Question Problems Set 3.6, Question 14 (p 192) of Strang's Introduction to Linear Algebra, 4th Ed, says to find the the 4 bases of the given matrix A which has been decomposed into LU. The solution manual says the column basis is C(A) = C(U) and further says (1,0,0) (0,1,0), (0,0,1). I thought the column basis were the pivot columns of A (which we aren't supposed to calculate from the problem statement). I suppose we are supposed to see the column space is 3-dimensional and that the basic set, I, can be the column basis, but I feel I am missing something. Could someone please explain? Thx

Question

Question Problems Set 3.6, Question 14 (p 192) of Strang's Introduction to Linear Algebra, 4th Ed, says to find the the 4 bases of the given matrix A which has been decomposed into LU.  The solution manual says the column basis is C(A) = C(U) and further says (1,0,0) (0,1,0), (0,0,1). I thought the column basis were the pivot columns of A (which we aren't supposed to calculate from the problem statement).  I suppose we are supposed to see the column space is 3-dimensional and that the basic set, I, can be the column basis, but I feel I am missing something.  Could someone please explain?  Thx

anonymous · Answer

I guess we could ask ourselves a few questions:
1. Working backwards, with the proposed solution for the basis vectors, can we reproduce the corresponding columns of A? If so, the vectors are a basis for the column space of A. You're right that it states to do so without calculating anything, but in this case you may be able to simply spot the fact that the three pivot columns of L are a linear combination of e1, e2 and e3. I hope this helps.