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Section 2.9 Homogeneous Linear Systems (VS9)

Subsection 2.9.1 Class Activities

Definition 2.9.1.

A homogeneous system of linear equations is one of the form:

a11x1+a12x2++a1nxn=0a21x1+a22x2++a2nxn=0am1x1+am2x2++amnxn=0

This system is equivalent to the vector equation:

x1v1++xnvn=0

and the augmented matrix:

[a11a12a1n0a21a22a2n0am1am2amn0]

Activity 2.9.2.

Note that if [a1an] and [b1bn] are solutions to x1v1++xnvn=0 so is [a1+b1an+bn], since

a1v1++anvn=0 and b1v1++bnvn=0

implies

(a1+b1)v1++(an+bn)vn=0.

Similarly, if cR, [ca1can] is a solution. Thus the solution set of a homogeneous system is...

  1. A basis for Rn.

  2. A subspace of Rn.

  3. The empty set.

Activity 2.9.3.

Consider the homogeneous system of equations

x1+2x2+x4=02x1+4x2x32x4=03x1+6x2x3x4=0
(a)

Find its solution set (a subspace of R4).

(b)

Rewrite this solution space in the form

{a[????]+b[????]|a,bR}.
(c)

Rewrite this solution space in the form

span{[????],[????]}.

Activity 2.9.5.

Consider the homogeneous system of equations

2x1+4x2+2x34x4=02x14x2+x3+x4=03x1+6x2x34x4=0

Find a basis for its solution space.

Activity 2.9.6.

Consider the homogeneous vector equation

x1[223]+x2[446]+x3[211]+x4[414]=[000]

Find a basis for its solution space.

Activity 2.9.7.

Consider the homogeneous system of equations

x13x2+2x3=02x1+6x2+4x3=0x1+6x24x3=0

Find a basis for its solution space.

Observation 2.9.8.

The basis of the trivial vector space is the empty set. You can denote this as either or {}.

Thus, if 0 is the only solution of a homogeneous system, the basis of the solution space is .

Subsection 2.9.2 Videos

Figure 22. Video: Polynomial and matrix calculations

Subsection 2.9.3 Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/VS9.slides.html.

Exercises 2.9.4 Exercises

Exercises available at https://teambasedinquirylearning.github.io/linear-algebra/2022/exercises/#/bank/VS9/.

Subsection 2.9.5 Mathematical Writing Explorations

Exploration 2.9.9.

An n×n matrix M is non-singular if the associated homogeneous system with coefficient matrix M is consistent with one solution. Assume the matrices in the writing explorations in this section are all non-singular.

  • Prove that the reduced row echelon form of M is the identity matrix.

  • Prove that, for any column vector b=[b1b2bn], the system of equations given by [Mb] has a unique solution.

  • Prove that the columns of M form a basis for Rn.

  • Prove that the rank of M is n.

Subsection 2.9.6 Sample Problem and Solution

Sample problem Example B.1.13.

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