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Vector Spaces: Linear Independence Notes for CSci 124/
Poorvi L. Vora
We reproduce Example 4 from the previous set of notes: Example 4: Determine if (^)
is a linear combination of
v 1 =
(^) , v 2 =
(^) , v 3 =
If so, what are the coefficients? Again, in matrix form, the question is, find a solution to:
a 1 a 2 a 3
if it exists. Gaussian elimination gives:
a 1 a 2 a 3
where the final equation is of the form 0 = 0. This means you have a consistent set of equations with several solutions. One solution is a 3 = 1, a 2 = 0, a 1 = 3. That is,
3
Notice that three equations in three unknowns give several solutions. This happens because one of the equations is redundant in the presence of the other two: it provided no new information, which is why it disappeared on gaussian elimination. There were essentially two equations in three unknowns, and hence several solutions. This, in turn, happens because one of the vectors vi is redundant. That is, any one of the three vectors {vi}^3 i=1 lies in the span of the other two, and hence
a third vector does not provide an additional, independent direction. For example, of the three vectors in equation (1), 1 2
That is, one can add some amounts of the vectors to cancel one another out, and get the zero vector. For example, 1 2
We can add any amounts of the zero vector (equation (3)) to equation (2) to get several solutions to our problem. Suppose we add the zero vector n times, we get equation (2) + n× equation (3):
3
(^) + n(^12
or
(3 + n 2 )
(^) − n 2
(^) + (1 − n)
Hence, in general, a 1 = 3 + n 2 , a 2 = − n 2 , a 3 = 1 − n for all integers n. The above issue does not arise when the vectors v 1 , v 2 , v 3 are linearly independent. Loosely speaking, a set of vectors {vi}ni=1 is linearly independent if the only way of getting the zero vector by combining them is to take a zero amount of each vector.
1 Linear Independence
Definition A set of vectors {vi}ni=1 ⊂ V is linearly independent if ∑ni=1 aivi = 0 ⇔ ai = 0 f or i = 1, 2 , ...n Definition A set of vectors {vi}ni=1 ⊂ V that is not linearly independent is linearly dependent. Example 5: The vectors:
v 1 =
(^) , v 2 =
(^) , v 3 =
are linearly dependent because ∑ v 3 = v 1 + v 2. That is, because v 1 + v 2 − v 3 = 0. Hence, (^3) i=1 aivi = 0 does not imply that ai = 0 f or i = 1, 2 , 3.
- Solve the set of linear equations Aa = 0 using gaussian elimination. How many solutions does it have? (a) We know that there is one solution for these equations, it is a = 0. (b) If there is exactly one solution, and no more, the vectors {vi}ni=1 are linearly indepen- dent. (c) If there are several solutions, the vectors {vi}ni=1 are not linearly independent
Example 8: Is the set of vectors:
v 1 =
,^ v^2 =
,^ v^3 =
,^ v^4 =
linearly independent? In other words, the question is, are there values a 1 , a 2 , a 3 and a 4 , not all zero, such that a 1 v 1 + a 2 v 2 + a 3 v 3 + a 4 v 4 = 0 or (^)
a 1 a 2 a 3 a 4
=^0
Now, the above are four equations is four unknowns, for which we already know one solution: a 1 = a 2 = a 3 = a 4 = 0. So the set of vectors is linearly independent only if there are no more solutions. Gaussian elimination, as always.
a 1 a 2 a 3 a 4
=^0
Swap for a non-zero pivot. Remember to swap rhs as well, though, in this case, because the rhs is zero, it doesn’t matter. (^)
a 1 a 2 a 3 a 4
=^0
a 1 a 2 a 3 a 4
=^0
a 1 a 2 a 3 a 4
=^0
Because the final equation contains a single unknown, there is only one solution to this set of equations. We know that a 1 = a 2 = a 3 = a 4 = 0 is a solution, now we know it is the only solution. (You can check that the gaussian elimination gives you that solution: − 4 a 4 = 0, or a 4 = 0. Working backwards, you get that all four unknowns are zero.) We could have determined the same by checking if the matrix corresponding to the equations was invertible. If it was, then there would be exactly one solution. We didn’t invert the matrix because this works only when there are n equations in n unknowns, which is not always the case.
3 Spans and Linear Equations
Consider a set of n equations in m unknowns. We have observed examples of the following facts:
- The set of equations Ax = b has a solution (that is, at least one solution) if b ∈ Span{Ai}mi= where Ai represents the ith^ column of A. That is, the equations have a solution if b can be expressed as a linear combination of {Ai}mi=1.
- The converse is also true: the set if equations has no solution if b is not in Span{Ai}mi=1.
- If b ∈ Span{Ai}mi=1 (the set of equations has at least one solution), then: (a) The set of equations has more than one solution if {Ai}mi=1 are not linearly independent. (b) The converse is also true. The set of equation has only one solution if {Ai}mi=1 are linearly independent.
4 Bases of a Vector Space
Consider the vector space R^2. Consider any vector in it. This vector can be written as a linear combination of v 1 =
[ 1
]
, v 2 =
[ 0
]
[ (^) x y
]
= x
[ 1
]
[ 0
]
= xv 1 + yv 2
That is R^2 ⊆ Span({v 1 , v 2 })
Example 9:
v 1 =
,^ v^2 =
,^ v^3 =
,^ v^4 =
is a basis for R^4. We have already shown the above set is linearly independent (see Example 8). To show that it is a basis, we also need to show that R^4 is its span. For this, we need to consider any vector in R^4 :
w x y z
and show that it is a linear combination of the vectors v 1 , v 2 , v 3 , and v 4. That is, we need to show that there exist a 1 , a 2 , a 3 , a 4 such that
a 1 v 1 + a 2 v 2 + a 3 v 3 + a 4 v 4 =
w x y z
or
a 1
+^ a^2
+^ a^3
+^ a^4
w x y z
Again, the equations may be written as:
a 1 a 2 a 3 a 4
w x y z
which may be reduced:
a 1 a 2 a 3 a 4
w x − w y − w z − w
a 1 a 2 a 3 a 4
w y − w x − w z − w
a 1 a 2 a 3 a 4
w y − w x − w z − w − (y − w) = z − y
a 1 a 2 a 3 a 4
w y − w x − w z − y + x − w
a 4 = w−x+ 4 y−z, a 3 = w−x− 4 y+z, a 2 = w+x− 4 y−z, a 1 = w+x+ 4 y+z Thus, given any vector in R^4 , we have obtained its coefficients wrt {vi}^4 i=1, hence R^4 ⊆ Span({vi}^4 i=1). Further, because each of the vi is a four dimensional column, linear combinations of the vi cannot lie outside the four-dimensional space over the real numbers. Hence, Span({vi}^4 i=1) ⊆ R^4. Hence Span({vi}^4 i=1) = R^4. Example 8 shows that {vi}^4 i=1 are linearly independent, hence {vi}^4 i=1 forms a basis for R^4.
6 The Dimension of a Vector Space
Definition The number of vectors in a finite basis of a vector space is its dimension. Fact: The number of vectors in a basis for Rn^ is n. This gives us one additional fact connecting linear equations to vector spaces. Consider a set of n equations in n unknowns, represented by the matrix equation Ax = b. Notice that A is square. We now add to the observations of section 3 for the special case of the square matrix.
- The equations have exactly one solution if A is invertible, or, equivalently, if all its columns are independent, or, equivalently, if the columns form a basis for Rn.
7 Exercises
- Which of the following sets of vectors is linearly independent? a.
v 1 =
(^) , v 2 =
(^) , v 3 =
b.
v 1 =
(^) , v 2 =
(^) , v 3 =
Solutions
- Which of the following sets of vectors is linearly independent? Answer: Try to find non-zero coefficients that combine the vectors to get 0. If you can find the non-zero coefficients, the vectors are not linearly independent. Otherwise, they are. a.
v 1 =
(^) , v 2 =
(^) , v 3 =
Answer: Solve Aa = 0
or (^)
a 1 a 2 a 3
a 1 a 2 a 3
a 1 a 2 a 3
Final equation of form 0 = 0 hence there is more than one solution to the equations, hence the vectors are not linearly independent.
b.
v 1 =
(^) , v 2 =
(^) , v 3 =
Answer: Solve Aa = 0 or (^)
a 1 a 2 a 3
a 1 a 2 a 3
a 1 a 2 a 3
Final equation of form 0 = 0 hence there is more than one solution to the equations, hence the vectors are not linearly independent.
c.
v 1 =
(^) , v 2 =
(^) , v 3 =
Answer: Solve Aa = 0 or (^)
a 1 a 2 a 3
a 1 a 2 a 3
a 1 a 2 a 3
The system has exactly one solution, a 1 = a 2 = a 3 = a 4 = 0. Hence the vectors are linearly independent.
- Show that
w 1 =
(^) , w 2 =
(^) , w 3 =
is a basis for R^3.
Answer: We need to show that w 1 , w 2 and w 3 are linearly independent, and that Span({wi}^3 i=1) = R^3. First we show linear independence. Find a 1 , a 2 , a 3 such that:
a 1 a 2 a 3
a 1 a 2 a 3
a 1 a 2 a 3
a 1 a 2 a 3
Has exactly one solution, a 1 = a 2 = a 3 = 0. Hence the vectors are linearly independent. Next we show that Span({vi}^3 i=1) = R^3. First, we observe that the vectors are three dimensional, hence all combinations are also three dimensional, and Span({vi}^3 i=1) ⊆ R^3. Next we wish to show that R^3 ⊆ Span({vi}^3 i=1). We consider any vector in R^3 , (x, y, z). We see if it can be expressed as a linear combination of {vi}^3 i=1.
a 1 a 2 a 3
x y z
a 1 a 2 a 3
y x z
a 1 a 2 a 3
y x z − y
a 1 a 2 a 3
y x z − y − x
a 3 = x+ 2 y− z, a 2 = x−y 2 +z, a 1 = −x+ 2 y +z. Hence R^3 ⊆ Span({vi}^3 i=1). Hence we have shown that the given vectors form a basis of R^3.
- Express the vectors in (a)-(d) as linear combinations of the set of vectors:
w 1 =
(^) , w 2 =
(^) , w 3 =
a. (^)
Answer: Solve the equation:
a 1 a 2 a 3
a 1 a 2 a 3
2
a 1 a 2 a 3
(^52) 2
a 3 = 2, a 2 = 5, a 1 = − 2
b. (^)
Answer: Solve the equation:
a 1 a 2 a 3
a 1 a 2 a 3
a 1 a 2 a 3
a 3 = 1, a 2 = 4, a 1 = − 2
c. (^)
Answer: Solve the equation:
a 1 a 2 a 3
