TA: Geoff Williams
Tutorials: Thursday 1:30-2:20 in TSH-120
E-mail:williagg@math.mcmaster.ca
Office hours: Wednesday 1:30-3:30 at the Math Help Centre
Tutorial # 10
Given vectors X1 , X2 , . . . , Xk in Rn , a vector of the form
t1 X1 + t2 X2 + . . . + tk Xk
(where the ti are scalars) is called a linear combination of the Xi .
The set of all such linear combinations is called the span of the Xi and
is denoted
span{X1 , X2 , . . . , Xk } = {t1 X1 + t2 X2 + . . . + tk Xk | ti in R}
5.1 #2 In each case determine if X lies in U = span{Y, Z}
 
 


1
1
0
 0 
 1 
 2 
 
 

(b). X = 
 15  , Y =  0  , Z =  0 
11
1
1
Solution:
Because U = span{Y, Z} every vector in U can be written as a linear combination of Y and Z. So in this case can we find coefficients a and b such that
X = aY + bZ?

  
 
1
0
1

  
 0 
 + b 1  =  2 
a
 0 
 0   15 
11
1
1


1 0 1
 0 1 2 


 0 0 15 
1 1 11
Notice that in the third row we have a · 0 + b · 0 = 15 which is not possible.
So the system doesn’t have a solution. Hence X is not in U .
1





−1
2
2




 5 
 , Y =  −1  , Z =  2 
(d). X = 
 2 
 0 
 8 
−3
5
3
Solution:
Again, can we find coefficients a and b such that X = aY





1 −2 −5
1 −2 −5
2 −1 2



 −1
3 12
2 5 
 →  2 −1 2  →  0





 0
0
1 4
0
1 4
2 8
0
7 28
5 −3 3
5 −3 3

+ bZ?.


1
 0

→

 0
0
0
1
0
0

3
4 

0 
0
Hence a = 3, b = 4 and X = 3Y + 4Z. We conclude that X lies in U .
For A an m × n matrix we have the following two subspaces
null(A) = {X in Rn | AX = 0}
im(A) = {AX | X in Rn }
• null(A) is a subspace of Rn
• im(A) is a subspace of Rm
To answer the next question we need to remember the following fact: To
show A = B where A and B are two sets we have to prove that
1. A ⊆ B
2. B ⊆ A
Then we can conclude that A = B.
5.1 #14
Prove that if A is an m × n matrix and V is an n × n invertible matrix then
im(A) = im(AV ).
Proof:
2
1. We want to show that im(A) ⊆ im(AV ).
By definition
im(A) = {AX : X in Rn }
im(AV ) = {AV Y : Y in Rn }
Let’s take any arbitrary vector Z in im(A) and show that it also lies
in im(AV ). If Z is in im(A) then
Z = AX
for some vector X in Rn . Since X is just some vector in Rn and since
V is invertible we can easily find another vector Y in Rn such that
X = V Y and Y = V −1 X. Hence
Z = AX = AV Y
for some Y in Rn which proves that Z lies in im(AV ).
2. We want to show that im(AV ) ⊆ im(A).
Let Z 0 be a vector in im(AV ). We want to show that it also lies in
im(A). Since Z 0 lies in im(AV ) we have that
Z 0 = AV Y
for some Y in Rn . But V Y will give us some vector X in Rn (V Y = X).
Hence
Z 0 = AV Y = AX
which shows that Z 0 lies in im(A).
Hence im(A) = im(AV ).
3
A set of vectors {X1 , X2 , . . . , Xk } is linearly independent if it satisfies
the conditions:
If t1 X1 + t2 X2 + . . . + tk Xk = 0 then t1 = t2 = . . . = tk = 0
5.2 #1(a)
Are the vectors

 
 
3
3
1
 −1  ,  2  ,  5 
−2
−1
0

linearly independent?
Solution
Is the only solution to





  
1
3
3
0
a  −1  + b  2  + c  5  =  0 
0
−1
−2
0
a = b = c = 0?
From row 3 we have
−b − 2c = 0
⇒ b = −2c
Row 2 gives us the equation
−a + 2b + 5c
Subbing in b = −2c we get
a = 2(−2c) + 5c
⇒a=c
Finally row 1 gives us
a + 3b + 3c = 0
Subbing in b = −2c and a = c
c + (−3c) + 3c = 0
⇒ −2c = 0
⇒c=0 ⇒a=0 ⇒b=0
So the vectors are linearly independent.
4
5.2 #1(a)
Are the vectors
 
2
1
 −1   0
 

 1 , 1
0
−1


0
  −2 

,
  1 
0
 
linearly independent?
Solution
No.

 
1
2
 −1   0
 
−2 
 1 + 1
−1
0
 
0
0
  −2   0
+
 
  1 = 0
0
0


This can be found either by inspection

1 2 0
 −1 0 −2

 1 1 1
−1 0 0
5




or row reducing the system

0
0 

0 
0
If U is a subspace of Rn , a set of vectors in U is called a basis of U if it
satisfies the following two conditions:
1. {X1 , X2 , . . . , Xm } is linearly independent.
2. U = span{X1 , X2 , . . . , Xm }.
Further, the dimension of U is m. (dimU = m)
5.2 #3(a).
Find a basis and calculate the dimension of the following subspace of R4 .

   

1
2
1 




  3   9 
−1
   

span 
 2  ,  0  ,  −6 





0
3
6
Solution
First check, are our 3 vectors linearly independent?


1 2 1 0
 −1 3 9 0 


 2 0 −6 0 
0 3 6 0

1 2
 0 5
→
 0 −4
0 3

1 2
 0 1
→
 0 1
0 1

1 2
 0 1
→
 0 0
0 0
1
10
−8
6
1
2
2
2
1
2
0
0

0
0 

0 
0

0
0 

0 
0

0
0 

0 
0
z = t, y = −2t, x = 3t
6

3
= t  −2 
1

So
 
  

1
2
1
 
  
 −1 
 − 2 3  +  9  = 
3
 0   −6  
 2 
6
3
0



  
1
2 
1




 9 
  3 
−1






is in span 
Therefore 
,
−6 
2   0 





6
3
0
Are these two vectors linearly independent?


0
0 

0 
0
Definitely.

 
1
2
 −1   3
 
So our basis is 
 2 , 0
0
3


 and the dimension is 2.

7