Exercise 1.5.12. Could be factored into the product where is upper triangular and is lower triangular, instead of being factored into the product ? If so, how could this other factorization be carried out? Would and be the same in both cases?
Answer: The matrix can be factored into by doing elimination in reverse, starting at row of and continuing through rows , , and so on, up to row 1, and trying to produce zeros starting from column on down. For example, in the first steps of such elimination a multiple of row would be subtracted from row in order to produce a zero in position ; a multiple of row would then be subtracted from row in order to produce a zero in position ; and so on. This produces a lower triangular matrix.
The multipliers from the reverse elimination steps are used to create an upper triangular matrix, in analogous manner to regular elimination: When subtracting a multiple of row from row we put the multiplier in position of the lower triangular matrix; when subtracting a multiple of row from row we put the multiplier in position of the lower triangular matrix; and so on. The upper triangular matrix has ones on the diagonal.
As an example, consider the matrix defined at the beginning of section 15:
The reverse elimination steps would go as follows. At each step we construct the upper triangular matrix using the multipliers used in the corresponding elimination step.
The first step is null because the entry in the 2,3 position of is already zero. The intermediate matrices are as follows, with the upper triangular matrix having a zero in the 2,3 position and the other entries being yet undetermined:
For the second step we subtract times row 3 from row 1, and put the multiplier in the 1,3 position of the upper triangular matrix:
For the third step we subtract times row 2 from row 1, and put the multiplier in the 1,2 position of the upper triangular matrix. This completes elimination and produces the final lower triangular and upper triangular matrices and respectively:
We now compute the product :
So we have found and such that . Note that if we reverse the order of the factors we have
So if it does not necessarily follow that , or vice versa—and indeed we would not expect this in general, since matrix multiplication is not commutative.
Extra credit: The matrices and found via reverse elimination are not exactly in the form we have otherwise been using, since does not have diagonal entries equal to 1. We can transform to the proper form by factoring it into a diagonal matrix times a new lower triangular matrix. We can then multiply the matrix by to obtain a new upper triangular matrix.
In our example we have
Thus is another factoring of , this time into an upper triangular matrix and a lower triangular matrix with ones on the diagonal. Again if we reverse the order of multiplication then the product is no longer a factoring:
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.