MAT 2401 Discovery Lab 4.4 Names___________________________________
Objectives
 To explore the concept of linear independence.
 To develop procedures to determine whether a set of vectors is linearly independent.
Instructions
 Do not look up any references including the textbook and internet.
 Use correct notations and do not skip steps.
 Two persons per group. Do not communicate with other groups.
Common Vector Spaces
We will use the following vector spaces in this lab.
1. R n   x1 , x2 , , xn  | x1 , x2 , , xn  R Set of all n-tuples.
2. C 2  ,   Set of all functions defined on the number line and have continuous second
derivatives.
We are going to describe how the concept of linear independence arises from differential
equations. For this introductory part, we are going to use the vector space C 2  ,   .
Afterwards, the examples will be concentrated on Rn for which we are more familiar
with.
1
Differential Equations (DE)
DE are used to model physical phenomenon. So they have huge amounts of applications in
sciences and engineering. You have learned the elements of DE in calculus II. However,
you are not expected to remember a lot of DE from calculus II. You will be fine as long as
you remember how to take derivatives.
Assume y is a function in x . A DE is an equation involves x, y, y, y,... For example,
y  2 y  3 y  0 .
(1)
A solution of a given DE is a function that satisfy the DE.
1. (a) Show that the function y  e3x is a solution of DE (1).
y 
and
y 
First and second derivatives
So,
y  2 y  3 y 
We calculate the left side of the DE.
Thus, y  e3x is a solution of DE (1).
 y  e3x is a solution of the DE if
this equals to 0, the value on the right
side.
(b) Show that the function y  e x is a solution of DE (1).
y 
and
y 
First and second derivatives
So,
y  2 y  3 y 
We calculate the left side of the DE.
Thus, y  e x is a solution of DE (1).
 y  e x is a solution of the DE if
this equals to 0, the value on the right
side.
Note that both e 3 x and e  x are twice differentiable functions and therefore are vectors of
the vector space C 2  ,   .
2
(c) Show that the function y  2e3 x is a solution of DE (1).
y 
and
y 
So,
y  2 y  3 y 
Thus, y  2e3 x is a solution of DE (1).
(d) Show that the function y  7e3 x is a solution of DE (1).
y 
and
y 
So,
y  2 y  3 y 
Thus, y  7e3 x is a solution of DE (1).
(e) What conclusion/speculation can you make based on part (a), (c), (d)?
Same or Different?
In practice, we consider y  e3x and y  e x as two “different” solutions of DE (1).
However, it is not desirable to consider y  e3x , y  2e3 x , and y  7e3 x as different
solutions because they all “depend” on each other. We can always get one function from
another one with the multiplication of a constant.
For classification purpose, we have this new notion of “difference” termed linear
independence as described below.
3
Linearly Independent Functions
Version 1
Two functions y1 and y2 are linearly independent if they are not constant multiple of
each other. Otherwise, they are linearly dependent.
Version 1 is equivalent to version 2 below which can be generalized to more than two
“vectors”.
Version 2
Two functions y1 and y2 are linearly independent if the only time that
c1 y1  c2 y2  0
is when the constants c1  c2  0 .
2. Use version 2 to demonstrate that y1  e3 x and y2  e x are linearly independent.
Suppose c1 y1  c2 y2  0 for constants c1 , c2 .
c1 y1  c2 y2  0
c1 y1  c2 y2
The last expression cannot be correct because
______________________________________.
We want to determine c1 , c2 and
see that they must be zeros.
 Substitute y1  e3 x and y2  e x
Combine the exponential functions
and put it on the left. Combine the
constants and put them on the right.
Make an observation why the last
expression does not make sense.
So the last step above is invalid.
In the last step, we made an assumption that c1  0
and therefore we divided both side by c1 . This step
is invalid because this assumption is incorrect.
Thus c1 must equal to zero.
Now explain why c2 must equal to zero.
Starts with c2 y2  c1 y1 and
substitute c1  0 .
Since c1  c2  0 , y1  e3 x and y2  e x are linearly independent.
4
Linear Dependence in R2
It is rather straight forward to see that the two vectors in S  1,1 ,  2, 2 are
“dependent” on each other because one is a constant multiple of the other,
1
 2, 2  2 1,1 or 1,1   2, 2  .
2
When a set has three vectors, we can look at whether one vector “depends” on the other two
vectors. For example, let
S   2,3 , 1,0 ,  0,1 .
Here, a quick look can discover the following dependence.
3. (a) Write  2,3 as a linear combination of 1,0  and  0,1 .
 2,3 
1, 0  
 0,1
(b) Write 1,0  as a linear combination of  0,1 and  2,3 .
1, 0  
 0,1 
 2,3
(c) Write  0,1 as a linear combination of 1,0  and  2,3 .
 0,1 
1, 0  
 2,3
You can see that if one vector depends on the other two (as in (a)), then any one vector
depends on the rest of the vectors as shown in (b) and (c) above.
Having three dependence equations is somewhat redundant. In fact, the three dependence
equations above can be lumped into one single equation
 1 2,3  2 1,0  3 0,1  0 .
This explains the general definition below.
5
General Definition
In other words, the collection is linearly independent when...
The only time a linear combination of vectors is the zero vector is when the
coefficients are all zeros.
Linear Dependence in R3
3. Show that the set S  1, 0, 0  ,  0, 0, 2  , 1, 0,1 is linearly dependent.
c1 1,0,0  c2  0,0, 2  c3 1,0,1   0,0,0 
Solve for the constants c1 , c2 , and c3
such that
c1 1,0,0  c2  0,0, 2  c3 1,0,1   0,0,0  .
Combine the 3 vectors on the left and
setup a system of 3 linear equations in 3
unknowns c1 , c2 , and c3 . Note that one of
the equation is degenerated, i.e. 0  0 .
Solve the system with your favorite
method.
Notice in this case, the system has infinite
number of solution. You just need one set of
values for c1 , c2 , and c3 that are not all
zeros.
Since c1 , c2 , and c3 are not all zeros, S
is linearly dependent.
Must have a conclusion here with the
appropriate reasoning.
We can solve the system in the example above by an augmented matrix and GJE. The steps
are formally summarized in the following test.
6
Test
4. Determine if S  1,1, 4  , 1, 0,3 , 1, 1, 0  is linearly independent or linearly
dependent.
Step 1
c1 1,1, 4  c2 1,0,3  c3 1, 1,0    0,0,0 
Write down a system of
equations in c1 , c2 , and c3 .
7
Step 2
Peform the GJE. Be sure to put
down the necessary details.
Step 3
 Make a conclusion here with
the appropriate reasoning.
8