Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf Linear Algebra Exercises-n-Answers.pdf
116 Linear Algebra, by Hefferon Three.IV.3.44 This is how the answer was given in the cited source. No, it does not. Let A and B represent, with respect to the standard bases, these transformations of R 3 . ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ Observe that ⎝ x y z ⎛ ⎞ ⎠ a ↦−→ ⎛ ⎝ x y 0 ⎞ ⎝ x y⎠ ↦−→ abab ⎝ 0 0⎠ z 0 ⎠ but ⎝ x y z ⎛ ⎠ a ↦−→ ⎞ ⎝ 0 x y ⎛ ⎠ ⎞ ⎝ x y⎠ ↦−→ baba ⎝ 0 0⎠ . z x Three.IV.3.45 This is how the answer was given in the cited source. (a) Obvious. (b) If A trans A⃗x = ⃗0 then ⃗y · ⃗y = 0 where ⃗y = A⃗x. Hence ⃗y = ⃗0 by (a). The converse is obvious. (c) By (b), A⃗x 1 ,. . . ,A⃗x n are linearly independent iff A trans A⃗x 1 ,. . . , A trans A⃗v n are linearly independent. (d) We have col rank(A) = col rank(A trans A) = dim {A trans (A⃗x) ∣ all ⃗x} ≤ dim {A trans ⃗y ∣ all ⃗y} = col rank(A trans ). Thus also col rank(A trans ) ≤ col rank(A transtrans ) and so we have col rank(A) = col rank(A trans ) = row rank(A). Three.IV.3.46 This is how the answer was given in the cited source. Let 〈⃗z 1 , . . . , ⃗z k 〉 be a basis for R(A) ∩ N (A) (k might be 0). Let ⃗x 1 , . . . , ⃗x k ∈ V be such that A⃗x i = ⃗z i . Note {A⃗x 1 , . . . , A⃗x k } is linearly independent, and extend to a basis for R(A): A⃗x 1 , . . . , A⃗x k , A⃗x k+1 , . . . , A⃗x r1 where r 1 = dim(R(A)). Now take ⃗x ∈ V . Write A⃗x = a 1 (A⃗x 1 ) + · · · + a r1 (A⃗x r1 ) and so A 2 ⃗x = a 1 (A 2 ⃗x 1 ) + · · · + a r1 (A 2 ⃗x r1 ). But A⃗x 1 , . . . , A⃗x k ∈ N (A), so A 2 ⃗x 1 = ⃗0, . . . , A 2 ⃗x k = ⃗0 and we now know A 2 ⃗x k+1 , . . . , A 2 ⃗x r1 spans R(A 2 ). To see {A 2 ⃗x k+1 , . . . , A 2 ⃗x r1 } is linearly independent, write b k+1 A 2 ⃗x k+1 + · · · + b r1 A 2 ⃗x r1 = ⃗0 A[b k+1 A⃗x k+1 + · · · + b r1 A⃗x r1 ] = ⃗0 and, since b k+1 A⃗x k+1 + · · · + b r1 A⃗x r1 ∈ N (A) we get a contradiction unless it is ⃗0 (clearly it is in R(A), but A⃗x 1 , . . . , A⃗x k is a basis for R(A) ∩ N (A)). Hence dim(R(A 2 )) = r 1 − k = dim(R(A)) − dim(R(A) ∩ N (A)). Subsection Three.IV.4: Inverses Three.IV.4.13 Here is one way to proceed. ⎛ ρ 1 ↔ρ 2 −→ ⎝ 1 0 1 0 1 0 ⎞ ⎛ 0 3 −1 1 0 0⎠ −ρ 1+ρ 3 −→ ⎝ 1 0 1 0 1 0 ⎞ 0 3 −1 1 0 0⎠ 1 −1 0 0 0 1 0 −1 −1 0 −1 1 ⎛ (1/3)ρ 2 +ρ 3 −→ ⎝ 1 0 1 0 1 0 ⎞ ⎛ 0 3 −1 1 0 0⎠ (1/3)ρ 2 −→ ⎝ 1 0 1 0 1 0 ⎞ 0 1 −1/3 1/3 0 0 ⎠ −(3/4)ρ 0 0 −4/3 1/3 −1 1 3 0 0 1 −1/4 3/4 −3/4 ⎛ ⎞ 1 0 0 1/4 1/4 3/4 (1/3)ρ 3 +ρ 2 −→ ⎝0 1 0 1/4 1/4 −1/4⎠ −ρ 3 +ρ 1 0 0 1 −1/4 3/4 −3/4
Answers to Exercises 117 Three.IV.4.14 (a) Yes, it has an inverse: ad − bc = 2 · 1 − 1 · (−1) ≠ 0. (b) Yes. (c) No. ( ) 1 1 −1 Three.IV.4.15 (a) 2 · 1 − 1 · (−1) · = 1 ( ) ( ) 1 −1 1/3 −1/3 1 2 3 · = ( ) ( ) 1 2 1/3 2/3 1 −3 −4 3/4 1 (b) 0 · (−3) − 4 · 1 · = −1 0 1/4 0 (c) The prior question shows that no inverse exists. Three.IV.4.16 ( (a) The ) reduction ( is routine. ) ( ) 3 1 1 0 (1/3)ρ 1 1 1/3 1/3 0 −(1/3)ρ 2 +ρ 1 1 0 1/3 −1/6 −→ −→ 0 2 0 1 (1/2)ρ 2 0 1 0 1/2 0 1 0 1/2 This answer agrees with the answer from the check. ( ) −1 ( ) 3 1 1 2 −1 = 0 2 3 · 2 − 0 · 1 · = 1 ( ) 2 −1 0 3 6 · 0 3 (b) This reduction is easy. ( ) ( ) 2 1/2 1 0 −(3/2)ρ 1 +ρ 2 2 1/2 1 0 −→ 3 1 0 1 0 1/4 −3/2 1 ( (1/2)ρ 1 1 1/4 1/2 0 −→ 4ρ2 0 1 −6 4 ) ( ) −(1/4)ρ 2 +ρ 1 1 0 2 −1 −→ 0 1 −6 4 The check agrees. ( ) ( ) 1 1 −1/2 1 −1/2 2 · 1 − 3 · (1/2) · = 2 · −3 2 −3 2 (c) Trying the Gauss-Jordan ( reduction ) ( ) 2 −4 1 0 (1/2)ρ 1 +ρ 2 2 −4 1 0 −→ −1 2 0 1 0 0 1/2 1 shows that the left side won’t reduce to the identity, so no inverse exists. The check ad − bc = 2 · 2 − (−4) · (−1) = 0 agrees. (d) This produces an inverse. ⎛ ⎝ 1 1 3 1 0 0 ⎞ 0 2 4 0 1 0 −1 1 0 0 0 1 (1/2)ρ 2 −→ −ρ3 ⎛ −→ ⎝ 1 1 3 1 0 0 ⎞ 0 2 4 0 1 0 0 2 3 1 0 1 ⎛ ⎝ 1 1 3 1 0 0 ⎞ 0 1 2 0 1/2 0 0 0 1 −1 1 −1 ⎠ ρ 1+ρ 3 ⎠ −ρ 2+ρ 3 −→ ⎠ −2ρ 3+ρ 2 −→ −3ρ 3 +ρ 1 ⎛ ⎝ 1 1 3 1 0 0 ⎞ 0 2 4 0 1 0⎠ 0 0 −1 1 −1 1 ⎛ ⎝ 1 1 0 4 −3 3 0 1 0 2 −3/2 2 0 0 1 −1 1 −1 ⎛ −ρ 2 +ρ 1 −→ ⎝ 1 0 0 2 −3/2 1 ⎞ 0 1 0 2 −3/2 2 ⎠ 0 0 1 −1 1 −1 (e) This is one way to do the reduction. ⎛ ⎝ 0 1 5 1 0 0 ⎞ ⎛ 0 −2 4 0 1 0⎠ ρ 3↔ρ 1 −→ ⎝ 2 3 −2 0 0 1 ⎞ 0 −2 4 0 1 0⎠ 2 3 −2 0 0 1 0 1 5 1 0 0 ⎛ ⎞ ⎛ ⎞ 2 3 −2 0 0 1 1 3/2 −1 0 0 1/2 (1/2)ρ 2 +ρ 3 −→ ⎝0 −2 4 0 1 0⎠ (1/2)ρ 1 −→ ⎝0 1 −2 0 −1/2 0 ⎠ −(1/2)ρ 0 0 7 1 1/2 0 2 (1/7)ρ 3 0 0 1 1/7 1/14 0 ⎛ ⎞ ⎛ ⎞ 1 3/2 0 1/7 1/14 1/2 1 0 0 −2/7 17/28 1/2 2ρ 3+ρ 2 −→ ⎝0 1 0 2/7 −5/14 0 ⎠ −(3/2)ρ2+ρ1 −→ ⎝0 1 0 2/7 −5/14 0 ⎠ ρ 3+ρ 1 0 0 1 1/7 1/14 0 0 0 1 1/7 1/14 0 (f) There is no ⎛ inverse. ⎞ ⎛ ⎞ 2 2 3 1 0 0 2 2 3 1 0 0 ⎝1 −2 −3 0 1 0⎠ −(1/2)ρ1+ρ2 −→ ⎝0 −3 −9/2 −1/2 1 0⎠ −2ρ 4 −2 −3 0 0 1 1 +ρ 3 0 −6 −9 −2 0 1 ⎛ −2ρ 2 +ρ 3 −→ ⎝ 2 2 3 1 0 0 ⎞ 0 −3 −9/2 −1/2 1 0⎠ 0 0 0 −1 −2 1 ⎞ ⎠
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<strong>Answers</strong> to <strong>Exercises</strong> 117<br />
Three.IV.4.14 (a) Yes, it has an inverse: ad − bc = 2 · 1 − 1 · (−1) ≠ 0. (b) Yes.<br />
(c) No.<br />
( )<br />
1 1 −1<br />
Three.IV.4.15 (a)<br />
2 · 1 − 1 · (−1) · = 1 ( ) ( )<br />
1 −1 1/3 −1/3<br />
1 2 3 · =<br />
( ) ( )<br />
1 2 1/3 2/3<br />
1 −3 −4 3/4 1<br />
(b)<br />
0 · (−3) − 4 · 1 ·<br />
=<br />
−1 0 1/4 0<br />
(c) The prior question shows that no inverse exists.<br />
Three.IV.4.16<br />
(<br />
(a) The<br />
)<br />
reduction<br />
(<br />
is routine.<br />
)<br />
( )<br />
3 1 1 0 (1/3)ρ 1 1 1/3 1/3 0 −(1/3)ρ 2 +ρ 1 1 0 1/3 −1/6<br />
−→<br />
−→<br />
0 2 0 1 (1/2)ρ 2<br />
0 1 0 1/2<br />
0 1 0 1/2<br />
This answer agrees with the answer from the check.<br />
( ) −1 ( )<br />
3 1<br />
1 2 −1<br />
=<br />
0 2 3 · 2 − 0 · 1 · = 1 ( ) 2 −1<br />
0 3 6 · 0 3<br />
(b) This reduction is easy.<br />
( )<br />
( )<br />
2 1/2 1 0 −(3/2)ρ 1 +ρ 2 2 1/2 1 0<br />
−→<br />
3 1 0 1<br />
0 1/4 −3/2 1<br />
(<br />
(1/2)ρ 1 1 1/4 1/2 0<br />
−→<br />
4ρ2 0 1 −6 4<br />
)<br />
( )<br />
−(1/4)ρ 2 +ρ 1 1 0 2 −1<br />
−→<br />
0 1 −6 4<br />
The check agrees.<br />
( ) ( )<br />
1 1 −1/2 1 −1/2<br />
2 · 1 − 3 · (1/2) ·<br />
= 2 ·<br />
−3 2 −3 2<br />
(c) Trying the Gauss-Jordan<br />
(<br />
reduction<br />
)<br />
( )<br />
2 −4 1 0 (1/2)ρ 1 +ρ 2 2 −4 1 0<br />
−→<br />
−1 2 0 1<br />
0 0 1/2 1<br />
shows that the left side won’t reduce to the identity, so no inverse exists. The check ad − bc =<br />
2 · 2 − (−4) · (−1) = 0 agrees.<br />
(d) This produces an inverse.<br />
⎛<br />
⎝ 1 1 3 1 0 0 ⎞<br />
0 2 4 0 1 0<br />
−1 1 0 0 0 1<br />
(1/2)ρ 2<br />
−→<br />
−ρ3<br />
⎛<br />
−→ ⎝ 1 1 3 1 0 0 ⎞<br />
0 2 4 0 1 0<br />
0 2 3 1 0 1<br />
⎛<br />
⎝ 1 1 3 1 0 0 ⎞<br />
0 1 2 0 1/2 0<br />
0 0 1 −1 1 −1<br />
⎠ ρ 1+ρ 3<br />
⎠ −ρ 2+ρ 3<br />
−→<br />
⎠ −2ρ 3+ρ 2<br />
−→<br />
−3ρ 3 +ρ 1<br />
⎛<br />
⎝ 1 1 3 1 0 0 ⎞<br />
0 2 4 0 1 0⎠<br />
0 0 −1 1 −1 1<br />
⎛<br />
⎝ 1 1 0 4 −3 3<br />
0 1 0 2 −3/2 2<br />
0 0 1 −1 1 −1<br />
⎛<br />
−ρ 2 +ρ 1<br />
−→ ⎝ 1 0 0 2 −3/2 1 ⎞<br />
0 1 0 2 −3/2 2 ⎠<br />
0 0 1 −1 1 −1<br />
(e) This is one way to do the reduction.<br />
⎛<br />
⎝ 0 1 5 1 0 0<br />
⎞ ⎛<br />
0 −2 4 0 1 0⎠ ρ 3↔ρ 1<br />
−→ ⎝ 2 3 −2 0 0 1<br />
⎞<br />
0 −2 4 0 1 0⎠<br />
2 3 −2 0 0 1<br />
0 1 5 1 0 0<br />
⎛<br />
⎞ ⎛<br />
⎞<br />
2 3 −2 0 0 1<br />
1 3/2 −1 0 0 1/2<br />
(1/2)ρ 2 +ρ 3<br />
−→ ⎝0 −2 4 0 1 0⎠ (1/2)ρ 1<br />
−→ ⎝0 1 −2 0 −1/2 0 ⎠<br />
−(1/2)ρ<br />
0 0 7 1 1/2 0<br />
2<br />
(1/7)ρ 3<br />
0 0 1 1/7 1/14 0<br />
⎛<br />
⎞<br />
⎛<br />
⎞<br />
1 3/2 0 1/7 1/14 1/2<br />
1 0 0 −2/7 17/28 1/2<br />
2ρ 3+ρ 2<br />
−→ ⎝0 1 0 2/7 −5/14 0 ⎠ −(3/2)ρ2+ρ1<br />
−→ ⎝0 1 0 2/7 −5/14 0 ⎠<br />
ρ 3+ρ 1<br />
0 0 1 1/7 1/14 0<br />
0 0 1 1/7 1/14 0<br />
(f) There is no<br />
⎛<br />
inverse.<br />
⎞<br />
⎛<br />
⎞<br />
2 2 3 1 0 0<br />
2 2 3 1 0 0<br />
⎝1 −2 −3 0 1 0⎠ −(1/2)ρ1+ρ2<br />
−→ ⎝0 −3 −9/2 −1/2 1 0⎠<br />
−2ρ<br />
4 −2 −3 0 0 1<br />
1 +ρ 3<br />
0 −6 −9 −2 0 1<br />
⎛<br />
−2ρ 2 +ρ 3<br />
−→ ⎝ 2 2 3 1 0 0<br />
⎞<br />
0 −3 −9/2 −1/2 1 0⎠<br />
0 0 0 −1 −2 1<br />
⎞<br />
⎠
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