Homogeneous Linear Systems

These notes closely follow the presentation of the material given in David C. Lay’s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein.

Homogeneous Linear Systems A linear system of the form a 11 x 1  a 12 x 2    a 1n x n  0

(HLS)

a 21 x 1  a 22 x 2    a 2n x n  0  a m1 x 1  a m2 x 2    a mn x n  0 (having all zeros on the right) is called a homogeneous linear system. The above system can also be written as the homogeneous vector equation x1a1  x2a2    xnan  0m or as the homogeneous matrix equation Ax  0 m with the usual interpretations that

(HVE) (HME)

a 11 a 12  a 1n A

a 21 a 22  a 2n



a1 a2  an



  

,

a m1 a m2  a mn x1 x 

x2 

,

xn and 0 0m 

0 

(the zero vector in  m ).

0 The homogeneous equation Ax  0 m always has a solution because A0 n  0 m . The solution x  0 n of the equation Ax  0 m is called the trivial solution. However, it is possible that the equation might also have non–trivial solutions. A non–trivial solution of the equation Ax  0 m is a vector x  0 n such that Ax  0 m .

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Example Determine whether or not the homogeneous linear system 3x 1  5x 2  4x 3  0  3x 1  2x 2  4x 3  0 6x 1  x 2  8x 3  0 has non–trivial solutions. If it does have non–trivial solutions, then describe its solution set.

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Example Determine whether or not the linear “system” x1  x2  x3  0 has non–trivial solutions. If it does have non–trivial solutions, then describe its solution set.

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Example Determine whether or not the homogeneous linear system x 1  9x 3  0  9x 1  4x 2  4x 3  0 2x 1  x 2  0 has non–trivial solutions. If it does have non–trivial solutions, then describe its solution set.

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Summary 1. Suppose that every column of the matrix A is a pivot column. What can we say about the solution set of the homogeneous equation Ax  0 m ?

2.

Suppose that not every column of the matrix A is a pivot column. What can we say about the solution set of the homogeneous equation Ax  0 m ?

3.

Write a single statement that summarizes the answers to questions 1 and 2 above.

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An important fact about solution sets of homogeneous equations is given in the following theorem: Theorem Any linear combination of solutions of Ax  0 is also a solution of Ax  0. Proof Suppose that A is an m  n matrix and suppose that the vectors x 1 and n x 2   are solutions of the homogeneous equation Ax  0 m . This means that Ax 1  0 m and Ax 2  0 m . Now let us take a linear combination of x 1 and x 2 , say y c 1 x 1  c 2 x 2 . By using Theorem 5 on page 45 of the textbook, we see that Ay  Ac 1 x 1  c 2 x 2   Ac 1 x 1   Ac 2 x 2   c 1 Ax 1   c 2 Ax 2   c10m  c20m  0m. This shows that y is a solution of Ax  0 m . We have proved that a linear combination of two solutions of Ax  0 is also a solution of Ax  0. However, the same type of proof would work if we were to start with three, four, or more solutions. Thus, any linear combination of solutions of Ax  0 is also a solution of Ax  0.

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The Connection Between Solutions of Ax  0 and Ax  b Let’s summarize some things that we know so far: 1. A homogeneous equation Ax  0 always has at least one solution (the trivial solution x  0). If every column of A has a pivot position, then Ax  0 has only the trivial solution. Otherwise, Ax  0 has infinitely many solutions. Furthermore, any linear combination of solutions of Ax  0 is also a solution of Ax  0. 2. A non–homogeneous equation, Ax  b (where b  0 m ) may or may not have a solution. If every row of A has a pivot position, then Ax  b has at least one solution no matter what b   m is. Otherwise, whether or not Ax  b has a solution depends on what b is. Assuming that Ax  b does have a solution, this solution is unique if and only if every column of A has a pivot position.

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The following theorem relates the solution set of Ax  b to the solution set of Ax  0: Theorem Suppose that A is an m  n matrix and suppose that b   m . Also, suppose that y is a solution of Ax  0 m and suppose that z is a solution of Ax  b. Then w  z  y is also a solution of Ax  b. Conversely, every solution, w, of the equation Ax  b can be written as w  z  y where y is a solution of Ax  0 m and z is a solution of Ax  b. Proof Since y is a solution of Ax  0 m , we know that Ay  0 m . Since z is a solution of Ax  b, we know that Az  b. Now let w  z  y. Then Aw  Az  y  Az  Ay  b  0m  b, which shows that w is a solution of Ax  b. The proof of the converse statement is easy: If w is a solution of Ax  b, then we simply observe that w  w  0 n , and that w is a solution of Ax b, and that 0 n is a solution of Ax  0 m .

Remark Another way to state the above theorem is to say that the solution set of Ax  b consists of all vectors of the form w  z  y where z is some particular solution of Ax  b, and y can be any solution of Ax  0 m .

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Visualizing the Solution Sets of Ax  0 and Ax  b in  2 The solution sets of Ax  0 and Ax  b are translations of each other. This is rather easy to visualize in  2 and in  3 . For instance, the solution set of Ax  0 might just consist of the zero vector, and in this case the solution set (if it is not empty) of Ax  b (where b  0) just consists of a single non–zero vector. Another possibility is that the solution set of Ax  0 is a line passing through the origin, in which case the solution set of Ax  b (if not empty) is a line not passing through the origin. These ideas are illustrated in the two examples that follow. Example For the homogeneous system x 1  3x 2  0

(H1)

 4x 1  4x 2  0, we have 1

3 0

4 4 0

~

1 0 0 0 1 0

showing that the only solution of system (H1) is the trivial solution x  0 2 .

For the non-homogeneous system x 1  3x 2  8

(NH1)

 4x 1  4x 2  0, we have 1

3 8

4 4 0

~

1 0 4 0 1 4

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showing that the only solution of system (NH1) is x

4 4

.

Example For the homogeneous “system” x 1  3x 2  0, we have solution x 1  3t

(H2)

x 2  t (a free variable) which can be written as x

3t t

or as x t

3 1

,

so the solution set of (H2) is a line passing through the origin in  2 .

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For the non–homogeneous “system” x 1  3x 2  3, we have solution x 1  3  3t

(NH2)

x 2  t (a free variable) which can be written as x

3  3t t

or as x

x

3 0 3 0



t

3t t 3 1

,

so the solution set of (NH2) is a line not passing through the origin in  2 .

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A possibility that can occur in  3 (although we will not attempt to draw a picture) is that the solution set of Ax  0 is a plane passing through the origin in  3 and that the solution set of Ax  b (for some b  0) is a plane not passing through the origin in  3 .

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