Throughout this discussion I am assuming we are dealing with a real vector space V. Every time I mention a collection of vectors I will assume that these vectors are in V. Vectors will usually be in bold for example w or v. Scalars will be in italics for example k or c. Remember for real vector spaces scalars are just real numbers. For complex vector spaces scalars are complex numbers. Typical vector spaces are:
Subsets are denoted by Í. Subspaces will be denoted £.
Definition: A vector w is a linear combination of the vectors v_{1}, v_{2}, … , v_{r}, if it can be expressed in the form where k_{1}, k_{2}, … , k_{r} are scalars If r = 1 then we have w= k_{1}v_{1}, and we say that w is a scalar multiple of v_{1}.
Lemma 1: If a nonzero vector w is a scalar multiple of v then v is a scalar multiple of w.
Proof:
Think 
Write 
Remember in a proof always start the proof by feeding back the given which is called the hypothesis 
Since w is a scalar multiple of w then w= kv for some scalar k. 
Check to make sure have used all of the given. We never used that w is nonzero. What does that tell us? 
Since w is nonzero then k must be nonzero because if k were zero we would have: w= kv=0v=0. 
We want to write v as a scalar multiple of w. This means we need to write v equals. 
Since k is nonzero we can write v=(1/k)w 
Make sure to state your conclusion. 
Therefore v is a scalar multiple of w. 
Ex. 1: 14+9x+4x^{2}
is a linear combination of 2+3x and 5x^{2} since
14+9x+4x^{2}=3(2+3x)+4(5x^{2}).
Ex. 2: 14+8x+4x^{2}
is not a linear combination of 2+3x and 5x^{2} since if
14+8x+4x^{2}= k_{1}(2+3x)+ k_{2}(5x^{2})
then this could be rearranged to
14+8x+4x^{2}=(2k_{1}+5k_{2})+(3k_{1})x+(k_{2})x^{2}.
Two polynomials are equal if their coefficients are equal.
So the above equation becomes 3 equations and two unknowns.
2 
k_{1} 
+ 
5 
k_{2} 
= 
14 
3 
k_{1} 




8 



1 
k_{2} 

1 
This system has no solution (you can eyeball it or use matrices). So there are scalars that work therefore 14+8x+4x^{2} is not a linear combination of 2+3x and 5x^{2}.
Since the scalars are found by solving a system of equations it is possible to have infinite solutions for the choice of scalars.
Ex 3: since w=2v_{1}+v_{2}+v_{3}= 1v_{1}+2v_{2}+2v_{3}.
When a vector can be written as two different linear combinations of a set of vectors then we say {v_{1},v_{2},v_{3}} form a dependent set. We will come back to this later.
Because saying "vector w is/is not a linear combination of the vectors v_{1}, v_{2}, … , v_{r}" is so wordy, mathematicians of course invented another word to use instead of this phrase.
Definition: Span{v_{1}, v_{2}, … , v_{r}} is the set of all linear combinations of v_{1}, v_{2}, … , v_{r}.
To save me typing let S be the set of vectors v_{1}, v_{2}, … , v_{r}. In the definition above span is used as a noun. It is a set too. Notice that v_{i}äSpan{S} for 1£ I £ r so S Í Span{S} Í V. You can also take the span of V, but Span{V}=V. Now if vector w is a linear combination of the vectors v_{1}, v_{2}, … , v_{r} we will write wÎSpan{S}. If vector w is not a linear combination of the vectors v_{1}, v_{2}, … , v_{r} we will write wÏSpan{S}.
Theorem 2: If S is the set of vectors v_{1}, v_{2}, … , v_{r} in V then,
Because of the second part of the Theorem Span{S} is said to be "the smallest subspace of V that contains S". If we let U=Span{S} then we say v_{1}, v_{2}, … , v_{r} span U. Here the first span in Span{S} is used as a noun, but the second span is used as a verb.
Proof: Part a. To prove a subset is a subspace you must only show three things.
Think 
Write 
You must show it is nonempty. Can you show zero is in Span{S} 
Since 0=0v_{1}+0v_{2}+ … +0v_{r }we have 0 is a linear combination of the vectors in S so it is in Span{S}. Thus Span{S} is nonempty. 
You need to check closure. Make sure to start with a hypothesis telling where things come from. 
Let w_{1} and w_{2} be vectors in Span{S}. Let a be a real number. 
Interpret your hypothesis. 
This means that w_{1} and w_{2} are linear combinations of the vectors in S. 
Sometimes you need to further interpret. Notice that you must use different letters for the scalars since if you used the same letters you would be implying that the scalars were equal which would mean the vectors would be equal. 
Therefore there exists scalars k_{1}, k_{2}, … , k_{r} such that w_{1}=k_{1}v_{1}+k_{2}v_{2}+ … + k_{r}v_{r} and there are scalars m_{1}, m_{2}, … , m_{r} such that w_{2}=m_{1}v_{1}+m_{2}v_{2}+ … +m_{r}v_{r}. 
Now you need to show that w_{1} +a w_{2} is in Span{S} What does this mean? It means that this new vector can be written as a linear combination of the vectors in S. This normally involves plug and chug. 
w_{1} +a w_{2}=(k_{1}v_{1}+k_{2}v_{2}+ … + k_{r}v_{r}) + a( m_{1}v_{1}+m_{2}v_{2}+ … +m_{r}v_{r})=(k_{1}+am_{1})v_{1} + (k_{2}+am_{2})v_{2} + … + (k_{r}+am_{r})v_{r} . The last equality followed because vectors satisfy all of our normal associative, commutative, and distributive rules under vector addition and scalar multiplication. 
Make a conclusion about closure. 
Thus w_{1} +a w_{2}ÎSpan{S} so Span{S} is closed under vector addition and scalar multiplication. 
Make a conclusion about the Theorem 
Since Span{S} is nonempty and closed under vector addition and scalar multiplication it follows that Span{S} is a subspace of V. 
Proof Part b.
Think 
Write 
Remember in a proof always start the proof by feeding back the given which is called the hypothesis 
Let W be a subspace of V that contains S. 
To show that W contains Span{S} means you are showing Span{S}ÍW. To do this you must start with an element in Span{S} and show it is also in W. 
Let w be a vector in Span{S}. 
Interpret 
Therefore there exists scalars k_{1}, k_{2}, … , k_{r} such that w=k_{1}v_{1}+k_{2}v_{2}+ … + k_{r}v_{r}. 
Somehow connect to given. What does the given mean? It means W is closed under vector addition and scalar multiplication. Now this was for two vectors by induction if you add up r vectors in W the resulting vector is in W. 
Since the vectors v_{i}ÎW and W is closed under vector addition and scalar multiplication it follows that w is in W. 
Conclude 
Therefore Span{S}ÍW. 
Ex 4: Describe Span{(1,2)}.
Span{(1,2)}={ a(1,2)
where a
is a scalar}. This is the straight
line through the origin with slope 2.
Ex 5: Show that Span{ (1,0,3,4),
(3,0,1,3) } Ì
Span{ (1,0,3,4), (3,0,1,3),
(2,1,5,6) }
Since (1,0,3,4) and (3,0,1,3) are in Span{ (1,0,3,4),
(3,0,1,3), (2,1,5,6) } part
b of the theorem above tells us that Span{ (1,0,3,4),
(3,0,1,3) } Í
Span{ (1,0,3,4), (3,0,1,3),
(2,1,5,6) }. We must still show that the sets cannot be equal.
To do this you must find one element of the larger set that is not in the
smaller set or you must show that the "size" of the larger set is
bigger than the "size" of the lower set.
Notice that (2,1,5,6) is not in Span{ (1,0,3,4),
(3,0,1,3) } so the sets are not equal. If the third vector in the larger
set had a 0 in the second component it would not be as easy to say it was not in
the smaller set you would have to test whether or not it was like in Ex 1 and 2.
Ex 6: Show
Span{ (1,2,3) , (5,6,1), (3,2,1) }=Span{ (5,6,1), (3,2,1) }
Like above it follows immediately from Theorem 2 that Span{ (1,2,3) , (5,6,1),
(3,2,1) } Ê
Span{ (5,6,1), (3,2,1) }. To
finish the proof you must show that Span{ (1,2,3) , (5,6,1), (3,2,1) } Í
Span{ (5,6,1), (3,2,1) } too. (Don't
forget how to show subsets!) Let wÎ
Span{ (1,2,3) , (5,6,1), (3,2,1) }. (Interpret
what does this mean?) Therefore there exists scalars k_{1}, k_{2},
k_{3} such that w=k_{1}(1,2,3)+k_{2}(5,6,1)+
k_{3}(3,2,1). We
need to show that we can find scalars that will be functions of k_{1},
k_{2}, k_{3} such that w=m_{1}(5,6,1)+m_{2}(3,2,1).
This changes to a system:
k_{1 } 
+ 
5k_{2} 
+ 
3k_{3 } 
= 
5m_{1}+3m_{2} 
2 k_{1} 
+ 
6 k_{2} 
+ 
2 k_{3} 
= 
6m_{1}+2m_{2} 
3 k_{1} 
 
k_{2} 
+ 
1 k_{3} 
= 
1m_{1}+1m_{2} 
Despite its backward form this is a system of 3 equations in two unknowns m_{1} and m_{2}. The left side of each of the above equations is just a real number; it is not unknown since the scalars are not unknown. Solving this system we get m_{1}= k_{1}+k_{2} and m_{2}= 2k_{1}+k_{3}. Since there is a solution to our system we could find the scalars thus wÎ Span{ (5,6,1), (3,2,1) }. This implies that Span{ (1,2,3) , (5,6,1), (3,2,1) } Í Span{ (5,6,1), (3,2,1) }. We therefore can conclude that Span{ (1,2,3) , (5,6,1), (3,2,1) }=Span{ (5,6,1), (3,2,1) }
When the span of a set of vectors does not change by removing one of its members we say that the larger set of vectors is said to be a dependent set or the vectors themselves are said to be dependent. So the larger set of vectors { (1,2,3) , (5,6,1), (3,2,1) } in Example 6 form a dependent set. To determine if the smaller set was a dependent set you would have to try removing a vector and see if the span changes. Be careful if you remove one and the span does not change that does not mean it is not dependent. You would have to then try removing a different one and check again. Etc. until you have tried removing everyone. If the span never changes then the set is not dependent. This is a lousy way to check for dependency.
Before I can give you a better way I need to remind you about notation. A subset T of a set S is called proper if it is nonempty and is not all of the vectors in S. If you know T is a proper subset of S you should write TÌS; if you don't know if it is proper or you don't care you write TÍS.
Definition: A set S of vectors v_{1}, v_{2}, … , v_{r} is a dependent set of vectors if there exists a choice of scalars k_{1}, k_{2}, … , k_{r} with at least one scalar not being zero such that k_{1}v_{1}+k_{2}v_{2}+ … + k_{r}v_{r}=0.
Sometimes it is easier to use the following: The set S is a dependent set if any of the following occur.
a. The Span{S}=Span{T} where T is some proper subset of S. OR
b. If one vector in S can be written as a linear combination of the others. OR
c. S contains the zero vector. OR
Choice a and b could have actually have been used for the definition; In other words the definition is equivalent to bullet a which is equivalent to bullet b. Bullet c is not equivalent to the definition. Bullet c says if you know your set contains the zero vector you know it is dependent, but if you know it does not contain the zero vector you don't know anything.
Definition: A set is called independent if it is not dependent. In other words a set is independent if
Now don't forget all of Chapter 1 and 2. If the equation in bullet c in the definition of independent translates to r equations and r unknowns it has a unique solution iff the determinant is nonzero; thus if the determinant is nonzero the set is independent and if the determinant is zero the set will be dependent. If instead it translates to m equations and r unknowns where r is bigger than m then there are infinite solutions so the set will be dependent.
Ex 7: Determine
if the set of vectors { (1,2,3), (5,6,1), (3,2,1)} is independent or dependent
a different way then in Example 6.
Assume we can find scalars k_{1}, k_{2}, k_{3}
such that k_{1}(1,2,3)+k_{2}(5,6,1)+ k_{3}(3,2,1)=0=(0,0,0).
This becomes (k_{1}+5k_{2}+3k_{3}, 2k_{1}+6k_{2}+2k_{3},
3k_{1}k_{2}+k_{3})=(0,0,0). Which translates to
three equations and three unknowns by doing first coordinate equals zero, second
coordinate equals zero, etc.
k_{1 } 
+ 
5k_{2} 
+ 
3k_{3 } 
= 
0 
2 k_{1} 
+ 
6 k_{2} 
+ 
2 k_{3} 
= 
0 
3 k_{1} 
 
k_{2} 
+ 
1 k_{3} 
= 
0 
So we can use determinants: . Thus the vectors are dependent. Much easier, right? Please notice that the vectors given appear as columns in our matrix or determinant not as rows.
Ex 8: Determine
if the set of vectors {sin^{2}x, cos^{2}x, 5} is a independent
or dependent set.
We start the same way. Assume we
can find scalars k_{1}, k_{2}, k_{3} such that k_{1}sin^{2}x+k_{2}cos^{2}x+k_{3}5=0
regardless of the value of x. You
cannot change the scalars if x changes. In other words you actually have infinite equations: k_{1}sin^{2}p+k_{2}cos^{2}p+k_{3}5=0,
k_{1}sin^{2}2p+k_{2}cos^{2}2p+k_{3}5=0,
k_{1}sin^{2}0+k_{2}cos^{2}0+k_{3}5=0
k_{1}sin^{2}(p/2)+k_{2}cos^{2}(p/2)+k_{3}5=0
etc. Even though we used four values of x we only really got two
different equations. This is not a
good approach. Maybe you remember
your trig and "see" the answer as dependent because 5sin^{2}x+5cos^{2}x+(1)5=0.
If you don't and the functions are differentiable you can try to see if
the Wronskian is helpful.
Theorem 3: Given a set of functions f_{1},f_{2},…,f_{r} in C^{(n1)} (¥,¥). If the set is dependent then
Proof: To same me typing I will assume n=2 if n is bigger the same argument works but with more typing. Since the set is dependent we can find scalars k and m at least one of which is nonzero such that kf+mg=0 The 0 vector is the function Zero(x)=0 for all real x in the interval (¥,¥). So our equation becomes kf(x)+mg(x)=0 for all real x in the interval (¥,¥). Since either k or m is nonzero the system has a nontrivial solution for every x in the interval (¥,¥). This implies that for every x in (¥,¥) the coefficient matrix is not invertible, or equivalently that its determinant the Wronskian is zero for every x in (¥,¥). Thus if the functions are linearly dependent in C^{1}(¥,¥) then W(x)=Zero(x)=0 for every x in the interval (¥,¥).
Lemma 4: If there exists an x such that W(x)¹0 then the functions are independent.
Proof: If they were dependent then W(x)=0 contradiction.
Remark: Although Theorem 3 and Lemma 4 are often helpful you must remember that the converse of Theorem 3 is not true. In other words If W(x)= 0 for all x in an interval [a,b] it does not mean the functions are dependent in C^{r}[a,b].
Ex 9: Show that if f=sin^{2}x and g=sinxsinx then W(x)=0 on the interval [1,1] but the vectors are independent on [1,1].
. Remember to think in radians.
Since on the interval [0,1] sinx and sin2x are both positive the absolute
values are not needed and then it is clear that W(x)=0.
On the interval [1,0) sinx is negative and sin2x is negative so the
expression becomes W(x)=sin^{2}x(sin2x)sinx sin2x(sinx)=0. So we have W(x)=0. Now
if we just look at kf(x)+mg(x)=0 when x=1 and when x= 1 we obtain
the two equations k+m=0 and km=0 which imply k=m=0.
Thus the functions are independent.
Ex 10: Show
that f=sin^{2}x, g=cos^{2}x, h=5 is an dependent set
and W(x)=0.
We already showed dependence in Ex 8.
Ex 11: Determine
if the vectors f=x and g=sinx form a dependent set or independent
set of vectors in C^{1}(¥,¥).
Calculate the Wronskian.
. This function has value p
when x=p
so it is not identically zero. Lemma
4 tells us that the functions must therefore be independent.