Linear Algebra Exercises-n-Answers.pdf

30 Linear Algebra, by Hefferon
The first of these equations shows that c 2,1 is zero because δ 1,l1 is not zero, but since both matrices
are in echelon form, each of the entries d 2,l1 , . . . , d m,l1 , and b 2,l1 is zero. Now, with the second
equation, b 2,l2 is nonzero as it leads its row, c 2,1 is zero by the prior sentence, and each of d 3,l2 ,
. . . , d m,l2 is zero because D is in echelon form and we’ve assumed that l 2 ≤ k 2 . Thus, this second
equation shows that d 2,l2 is nonzero and so k 2 ≤ l 2 . Therefore k 2 = l 2 .
(b) For the inductive step assume that l 1 = k 1 , . . . , l j = k j (where 1 ≤ j < m); we will show that
implies l j+1 = k j+1 .
We do the l j+1 ≤ k j+1 < ∞ case here — the other cases are then easy. Consider the ρ j+1 version
of the vector equation:
( )
0 . . . 0 βj+1,lj1 . . . β j+1,n
= c j+1,1
(
0 . . . δ1,k1 . . . δ 1,kj . . . δ 1,kj+1 . . . δ 1,km . . . )
.
.
+ c j+1,j
(
0 . . . 0 . . . δj,kj . . . δ j,kj+1 . . . δ j,km . . . )
+ c j+1,j+1
(
0 . . . 0 . . . 0 . . . δj+1,kj+1 . . . δ j+1,km . . . )
.
(
+ c j+1,m 0 . . . 0 . . . 0 . . . 0 . . . δm,km . . . )
Knowing that l 1 = k 1 , . . . , l j = k j , consider the l 1 -th, . . . , l j -th component equations.
0 = c j+1,1 δ 1,k1 + c j+1,2 · 0 + · · · + c j+1,j · 0 + c j+1,j+1 · 0 + · · · + c j+1,m · 0
0 = c j+1,1 δ 1,k2 + c j+1,2 δ 2,kj · · · + c j+1,j · 0 + c j+1,j+1 · 0 + · · · + c j+1,m · 0
.
.
0 = c j+1,1 δ 1,kj + c j+1,2 δ 2,k2 · · · + c j+1,j δ j,kj + c j+1,j+1 · 0 + · · · + c j+1,m · 0
We can conclude that c j+1,1 , . . . , c j+1,j are all zero.
Now look at the l j+1 -th component equation:
β j+1,lj+1 = c j+1,j+1 δ j+1,lj+1 + c j+1,j+2 δ j+2,lj+1 + · · · + c j+1,m δ m,lj+1 .
Because D is in echelon form and because l j+1 ≤ k j+1 , each of δ j+2,lj+1 , . . . , δ m,lj+1 is zero. But
β j+1,lj+1 is nonzero since it leads its row, and so δ j+1,lj+1 is nonzero.
Conclusion: k j+1 ≤ l j+1 and so k j+1 = l j+1 .
(c) From the prior answer, we know that for any echelon form matrix, if this relationship holds
among the non-zero rows:
ρ i = c 1 ρ 1 + · · · + c i−1 ρ i−1 + c i+1 ρ i+1 + · · · + c n ρ n
(where c 1 , . . . , c n ∈ R) then c 1 ,. . . , c i−1 must all be zero (in the i = 1 case we don’t know any of the
scalars are zero).
To derive a contradiction suppose the above relationship exists and let l i be the column index
of the leading entry of ρ i . Consider the equation of l i -th components:
ρ i,li = c i+1 ρ i+1,li + · · · + c n ρ n,li
and observe that because the matrix is in echelon form each of ρ i+1,li , . . . , ρ n,li is zero. But that’s
a contradiction as ρ i,li is nonzero since it leads the i-th row.
Hence the linear relationship supposed to exist among the rows is not possible.
One.III.2.22 (a) The inductive step is to show that if the statement holds on rows 1 through r then
it also holds on row r + 1. That is, we assume that l 1 = k 1 , and l 2 = k 2 , . . . , and l r = k r , and we
will show that l r+1 = k r+1 also holds (for r in 1 .. m − 1).
(b) Lemma 2.3 gives the relationship β r+1 = s r+1,1 δ 1 + s r+2,2 δ 2 + · · · + s r+1,m δ m between rows.
Inside of those rows, consider the relationship between entries in column l 1 = k 1 . Because r + 1 > 1,
the row β r+1 has a zero in that entry (the matrix B is in echelon form), while the row δ 1 has
a nonzero entry in column k 1 (it is, by definition of k 1 , the leading entry in the first row of D).
Thus, in that column, the above relationship among rows resolves to this equation among numbers:
0 = s r+1,1 · d 1,k1 , with d 1,k1 ≠ 0. Therefore s r+1,1 = 0.
With s r+1,1 = 0, a similar argument shows that s r+1,2 = 0. With those two, another turn gives
that s r+1,3 = 0. That is, inside of the larger induction argument used to prove the entire lemma is
here an subargument by induction that shows s r+1,j = 0 for all j in 1 .. r. (We won’t write out the
details since it is just like the induction done in Exercise 21.)