Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf 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.)
Answers to Exercises 31 (c) First, l r+1 < k r+1 is impossible. In the columns of D to the left of column k r+1 the entries are are all zeroes as d r+1,kr+1 leads the row k + 1) and so if l k+1 < k k+1 then the equation of entries from column l k+1 would be b r+1,lr+1 = s r+1,1 · 0 + · · · + s r+1,m · 0, but b r+1,lr+1 isn’t zero since it leads its row. A symmetric argument shows that k r+1 < l r+1 also is impossible. One.III.2.23 The zero rows could have nonzero coefficients, and so the statement would not be true. One.III.2.24 We know that 4s + c + 10d = 8.45 and that 3s + c + 7d = 6.30, and we’d like to know what s + c + d is. Fortunately, s + c + d is a linear combination of 4s + c + 10d and 3s + c + 7d. Calling the unknown price p, we have this reduction. ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 4 1 10 8.45 4 1 10 8.45 4 1 10 8.45 ⎝3 1 7 6.30⎠ −(3/4)ρ 1+ρ 2 −→ ⎝0 1/4 −1/2 −0.037 5 ⎠ −3ρ 2+ρ 3 −→ ⎝0 1/4 −1/2 −0.037 5⎠ −(1/4)ρ 1 1 1 p 1 +ρ 3 0 3/4 −3/2 p − 2.112 5 0 0 0 p − 2.00 The price paid is $2.00. One.III.2.25 If multiplication of a row by zero were allowed then Lemma 2.6 would not hold. That is, where ( ) ( ) 1 3 0ρ 2 1 3 −→ 2 1 0 0 all the rows of the second matrix can be expressed as linear combinations of the rows of the first, but the converse does not hold. The second row of the first matrix is not a linear combination of the rows of the second matrix. One.III.2.26 (1) An easy answer is this: 0 = 3. For a less wise-guy-ish answer, solve the system: ( ) ( ) 3 −1 8 −(2/3)ρ 1 +ρ 2 3 −1 8 −→ 2 1 3 0 5/3 −7/3 gives y = −7/5 and x = 11/5. 5x + 5y = 3. Now any equation not satisfied by (−7/5, 11/5) will do, e.g., (2) Every equation can be derived from an inconsistent system. For instance, here is how to derive “3x + 2y = 4” from “0 = 5”. First, 0 = 5 (3/5)ρ 1 −→ 0 = 3 xρ 1 −→ 0 = 3x (validity of the x = 0 case is separate but clear). Similarly, 0 = 2y. Ditto for 0 = 4. But now, 0 + 0 = 0 gives 3x + 2y = 4. One.III.2.27 Define linear systems to be equivalent if their augmented matrices are row equivalent. The proof that equivalent systems have the same solution set is easy. One.III.2.28 (a) The three possible row swaps are easy, as are the three possible rescalings. One of the six possible pivots is kρ 1 + ρ 2 : ⎛ ⎝ 1 2 3 ⎞ k · 1 + 3 k · 2 + 0 k · 3 + 3⎠ 1 4 5 and again the first and second columns add to the third. The other five pivots are similar. (b) The obvious conjecture is that row operations do not change linear relationships among columns. (c) A case-by-case proof follows the sketch given in the first item. Topic: Computer Algebra Systems 1 (a) The commands > A:=array( [[40,15], [-50,25]] ); > u:=array([100,50]); > linsolve(A,u); yield the answer [1, 4]. (b) Here there is a free variable:
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<strong>Answers</strong> to <strong>Exercises</strong> 31<br />
(c) First, l r+1 < k r+1 is impossible. In the columns of D to the left of column k r+1 the entries are<br />
are all zeroes as d r+1,kr+1 leads the row k + 1) and so if l k+1 < k k+1 then the equation of entries<br />
from column l k+1 would be b r+1,lr+1 = s r+1,1 · 0 + · · · + s r+1,m · 0, but b r+1,lr+1 isn’t zero since it<br />
leads its row. A symmetric argument shows that k r+1 < l r+1 also is impossible.<br />
One.III.2.23<br />
The zero rows could have nonzero coefficients, and so the statement would not be true.<br />
One.III.2.24 We know that 4s + c + 10d = 8.45 and that 3s + c + 7d = 6.30, and we’d like to know<br />
what s + c + d is. Fortunately, s + c + d is a linear combination of 4s + c + 10d and 3s + c + 7d. Calling<br />
the unknown price p, we have this reduction.<br />
⎛<br />
⎞<br />
⎛<br />
⎞ ⎛<br />
⎞<br />
4 1 10 8.45<br />
4 1 10 8.45<br />
4 1 10 8.45<br />
⎝3 1 7 6.30⎠ −(3/4)ρ 1+ρ 2<br />
−→ ⎝0 1/4 −1/2 −0.037 5 ⎠ −3ρ 2+ρ 3<br />
−→ ⎝0 1/4 −1/2 −0.037 5⎠<br />
−(1/4)ρ<br />
1 1 1 p<br />
1 +ρ 3<br />
0 3/4 −3/2 p − 2.112 5<br />
0 0 0 p − 2.00<br />
The price paid is $2.00.<br />
One.III.2.25 If multiplication of a row by zero were allowed then Lemma 2.6 would not hold. That<br />
is, where ( ) ( )<br />
1 3 0ρ 2 1 3 −→<br />
2 1 0 0<br />
all the rows of the second matrix can be expressed as linear combinations of the rows of the first, but<br />
the converse does not hold. The second row of the first matrix is not a linear combination of the rows<br />
of the second matrix.<br />
One.III.2.26 (1) An easy answer is this:<br />
0 = 3.<br />
For a less wise-guy-ish answer, solve the system:<br />
( )<br />
( )<br />
3 −1 8 −(2/3)ρ 1 +ρ 2 3 −1 8<br />
−→<br />
2 1 3<br />
0 5/3 −7/3<br />
gives y = −7/5 and x = 11/5.<br />
5x + 5y = 3.<br />
Now any equation not satisfied by (−7/5, 11/5) will do, e.g.,<br />
(2) Every equation can be derived from an inconsistent system. For instance, here is how to derive<br />
“3x + 2y = 4” from “0 = 5”. First,<br />
0 = 5 (3/5)ρ 1<br />
−→ 0 = 3 xρ 1<br />
−→ 0 = 3x<br />
(validity of the x = 0 case is separate but clear). Similarly, 0 = 2y. Ditto for 0 = 4. But now,<br />
0 + 0 = 0 gives 3x + 2y = 4.<br />
One.III.2.27 Define linear systems to be equivalent if their augmented matrices are row equivalent.<br />
The proof that equivalent systems have the same solution set is easy.<br />
One.III.2.28 (a) The three possible row swaps are easy, as are the three possible rescalings. One of<br />
the six possible pivots is kρ 1 + ρ 2 : ⎛<br />
⎝ 1 2 3<br />
⎞<br />
k · 1 + 3 k · 2 + 0 k · 3 + 3⎠<br />
1 4 5<br />
and again the first and second columns add to the third. The other five pivots are similar.<br />
(b) The obvious conjecture is that row operations do not change linear relationships among columns.<br />
(c) A case-by-case proof follows the sketch given in the first item.<br />
Topic: Computer <strong>Algebra</strong> Systems<br />
1 (a) The commands<br />
> A:=array( [[40,15],<br />
[-50,25]] );<br />
> u:=array([100,50]);<br />
> linsolve(A,u);<br />
yield the answer [1, 4].<br />
(b) Here there is a free variable:
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