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Solutions To Selected Exercises Part 1

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Learning objectives​

  • Explain the main ideas and vocabulary in Solutions To Selected Exercises Part 1.
  • Work through the source examples for Solutions To Selected Exercises Part 1 without depending on raw chunk order.
  • Use Solutions To Selected Exercises Part 1 as selective reference when learner modules point back to Linear Algebra And Its Applications.

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  • module-04-linear-algebra

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This unit is anchored to Linear Algebra And Its Applications and the source chapter "Solutions To Selected Exercises Part 1". Use external resources only to clarify, extend, or modernize details without replacing the chapter's conceptual spine.

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Source provenance​

  • Primary source: Linear Algebra And Its Applications
  • Source chapter: Solutions To Selected Exercises Part 1
  • Raw source file: 165-solutions-to-selected-exercises-part-1.md

Merged source​

Solutions To Selected Exercises Part 1​

id: '165-solutions-to-selected-exercises-part-1' book: "Linear Algebra and Its Applications" source_file: "../Gilbert_Strang_Linear_Algebra_and_Its_Applicatio_230928_225121.pdf" chunk_index: 165 chunk_total: 187 title: "Solutions to Selected Exercises (Part 1)" part: "Back Matter" chapter: "Solutions to Selected Exercises" pages: start: 497 end: 498 words: 633​

Solutions to Selected Exercises (Part 1)

1 1 1

  1. 1 == 1 1 + 1 +

1 1 1 Check the(1, 1) entry, then(3, 2),then(3, 3),then(1, 2) to show that those five P 's are independent. Four conditions on the nine entries make all row sums and column

sums equal: row sum1 ==row sum2==row sum3 ==column sum1 ==column sum

2(== column sum3is automatic because- sum of all rows== sum of all columns).

  1. If the5by5matrix[A b]is invertible, bis not a combination of the columns ofA.

If [A b] is singular, and Ahas independent columns,bis a combination of those columns.

Problem Set2.lt page110

  1. False, we only know dimensions are equal. Left nUllspace has smaller dim== m-r.

  2. C(A): r == 2, (1, 0, 1) (0, 1, 0); N(A):n-r == 2, (2, - 1, 1, 0), (-1, 0, 0, 1);

C(AT):r == 2, (1, 2, 0, 1), (0, 1, 1, 0); N(AT):m-r == 1, (-1, 0, 1);

C(U): (1, 0, 0), (0, 1, 0); N(U): (2, -1, 1, 0), (- 1, 0, 0, 0);

C(UT): (1, 2, 0, 1), (0, 1, 1, 0); N(AT): (0, 0, 1).

  1. Atimes every column ofBis zero, so C(B)is contained in the nUllspaceN(A).

  2. FromAx== 0,the row space and the nullspace must be orthogonal. See Chapter3.

2

  1. [1 2 4]; 2 4 8 has the same nUllspace.

6

  1. IfAx == 0has a nonzero solution, then r < n andC(AT) is smaller than Rn. So

ATY == fis not solvable for somef.Example:A== [1 1] and f== (1, 2).

  1. d== be/a;the only pivot isa.

n

'i5. With independent columns: rankn;nullpace== {OJ;row space is R; left inverse.

  1. A== [1 1 0]; B== [0 0 1].

Soluti o n s to Sele cted Exe rcises

  1. No-for example, all invertiblenbynmatrices have the same four subspaces.

  2. (a) 1 °. (b) Impossible: dimensions 1 +1 =j:. 3. (c) [1 1].

-3

(d) l'] (e) Impossible: Row space == column space requires m== n.

[-93

Thenm-r == n-r.

  1. Invertible A: row space basis== column space basis== (1, 0, 0), (0, 1, 0), (0, 0, 1);

nullspace and left nullspace bases are empty. B: row space basis (1, 0, 0, 1, 0, 0), (0,"'1, 0; 0, 1, 0),and(0, 0, 1, 0, 0, 1); column space basis(1, 0, 0), (0, 1, 0), (0, 0, 1); nullspace basis (-1, 0, 0, 1, 0, 0), (0, -1, 0, 0, 1, 0), and (0, 0, -1, 0, 0, 1); left nUllspace basis is empty..

  1. (a) Same row space and nUllspace. Therefore rank (dimension of row space) is

the same. (b) Same column space and left nullspace. Same rank (dimension of column space).

  1. (a) No solution means thatr <m.Alw aysr <n. Can't comparemandn.

(b) Ifm-r >0,the nullspace ofA1.: contains a nonzero vector.

  1. Row space basis (1, 2, 3, 4), (0, 1, 2; 3), (0, 0, 1, 2); nullspace basis (0, 1, -2, 1);

column space basis (1, 0, 0),. (0, 1, 0), (0,0, 1); left nullspace has empty basis.

  1. IfAv== °andvis a row ofAthenv. v == 0. Onlyv == °is in both spaces.

  2. Row3 -2(row2) +row1== zero row, so the vectors e(l, -2, 1) are in the left

nUllspace. The same vectors happen to be in the nullspace.

  1. (a) uand wspanC(A). (b) vandz spanC(AT). (c) rank<2ifuand ware

dependent orvandz are dependent. (d) The rank ofuvT+wzTis 2.

  1. (a) True (same rank). (b) False (A == [1 0]). (c) False (A can be invertible

and also unsymmetric). (d) True.

  1. a11 == 1, a12 == 0,a13 == 1,a22 == 0,a32 == 1, a31 == 0,a23 == 1, a33 == 0,a21 ==

(not unique).

  1. Rankr == nmeans nullspace== zero vector andXn == 0.