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Solutions to Tutorial 5
Simon Rose
April 24, 2013
These are problems from section 5 of the notes.
Recall that a set of vectors {v1 , . . . , vn } ⊂ V is linearly dependent if there
are real numbers x1 , . . . , xn , not all of which are zero, such that
0 = x1 · v 1 + · · · + xn · v n
Such a set is linearly independent if it is not linearly dependent; that is, if
whenever we write
0 = x1 · v 1 + · · · + xn · v n
for any real numbers x1 , . . . , xn , then we must in fact have
x1 = · · · = xn = 0
Consider the following vectors in the real vector space R2 :
v1 = (1, 0) v2 = (3, 0) v3 = (0, 0) v4 = (1, 1) v5 = (2, 1)
For each of the following sets, specify whether they are linearly dependent or
linearly independent.
Problem 1 (g): {v3 , v4 }
In this case, we note that because v3 = 0, we must always have that
α · v3 = 0. Thus we can choose any non-zero real number α and we have
that
0 = α · v3 + 0 · v4
which yields that this set is linearly dependent (since we chose α 6= 0).
Problem 1 (h): {v4 , v5 }
These two vectors are linearly independent. Let us see why. Suppose that
x4 , x5 are real numbers so that
0 = x4 · v 4 + x5 · v 5
This is equivalent to the pair of linear equations
0 = x4 + 2x5
0 = x4 + x5
1
We can then back-substitute: The second equation reads that x4 = −x5 ,
which in the first yields that
0 = (−x5 ) + 2x5 = x5
and so it follows that x5 = 0, and hence also that x4 = 0. Thus the only
solution to
0 = x4 · v 4 + x5 · v 5
is the trivial (x4 = x5 = 0 one, i.e. they are linearly independent.
Problem 1 (l): {v1 , v2 , v4 , v5 }
These vectors are linearly dependent. As we will see later on, there are
simply too many vectors in this list to not be linearly dependent; as we
have not discussed the relation between the size of a spanning set and
the dimension of a vector space, however, we must solve this in the usual
manner.
Suppose that we have the relation
0 = x1 · v 1 + x2 · v 2 + x4 · v 4 + x5 · v 5
for some real numbers x1 , x2 , x4 , x5 . This is equivalent to the linear equations
0 = x1 + 3x2 + x4 + 2x5
0 = x4 + x5
Similar to before, we must have that x4 = −x5 . In the first equation this
becomes
0 = x1 + 3x2 + (−x5 ) + 2x5 = x1 + 3x2 + x5
Morally, we are done because we have one equation and three unknowns,
which has infinitely many solutions. That said, we should actually provide
a solution to ensure that we are correct.
Let us assume that x5 = 0. We could pick it to be any number, but this
will make things simpler. This equation now reads
0 = x1 + 3x2
So if we let x2 be any non-zero real number and let x1 = −3x2 , then we see
that this equation is satisfied. In particular, if we choose (x1 , x2 ) = (−3, 1)
we find an explicit example. We then see that the values
x1 = −3 x2 = 1 x4 = 0 x5 = 0
satisfy the equation
0 = x1 · v1 + x2 · v2 + x4 · v4 + x5 · v5
Since not all of the coefficients are zero, it follows that this set is linearly
dependent.
2
Consider the following vectors in the real vector space Rˆ4 :
 
 
 
 
 
2
1
0
0
1
1
1
0
1
0









v1 =   v2 =   v3 =   v4 =   v5 =  
0
1
1
0
0
0
0
1
1
0
For each of the following sets, specify whether they are linearly dependent or
linearly independent.
Problem 2 (c): {v2 }
Suppose that we have the equation
0 = x2 v2
This is equivalent to the linear equations
0 = 2x2
0 = x2
0=0
0=0
with some obvious redundancies. The second equation reads that x2 = 0,
and so it follows that the only solution is the trivial one; that is, this set
is linearly independent.
Problem 2 (j): {v1 , v4 , v5 }
Suppose that we have the equation
0 = x1 v1 + x4 v4 + x5 v5
This is equivalent to the linear equations
0 = x1
0 = x5
0 = x4
0 = x4 + x5
This is easy then: The first three equations read exactly that our three
coefficients must be zero; thus there is again, only the trivial solution. Our
set is then linearly independent.
Problem 2 (l): {v1 , v2 , v4 , v5 }
Suppose that we have the equation
0 = x1 v1 + x2 v2 + x4 v4 + x5 v5
3
This is equivalent to the linear equations
0 = x1 + 2x2
(1)
0 = x2 + x5
(2)
0 = x4
(3)
0 = x4 + x5
(4)
Similar to before, we see first (equation 3) that x4 must be zero. But this
yields (equation 4) that x5 is also zero. Combining this with equation 2,
we see that x2 is zero. But finally this shows (equation 1) that x1 is also
zero. So as with the other cases, we have only the trivial solution, and so
this set is linearly independent as well.
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