Math 2350

Week 04 Worksheet: Vector spaces

To receive the grade, solve the selected practice problems and hand in. Complete all unfinished
problems at home. The solution of this worksheet will be made available on Canvas later.
1. (Review) Vector spaces, linear independence, span, basis, dimension, null space.
2. (Practice) Determine if the following sets are subspaces of Pn for an appropriate value
of n, assuming a is in R. Justify your answers.
(a) All polynomials of the form p(x) = ax2 .
(b) All polynomials of the form p(x) = a + x2 .
(c) All polynomials of degree at most 3, with integers as coefficients.
Solution:
Let’s recall the definition of subspace.
Def: A subspace of a vector space V is a subset H of V that has two properties:
(i) For each u and v in H, u + v is in H. (H is closed under vector addition.)
(ii) For each u in H and each scalar c, cu is in H. (H is closed under multiplication by
scalars.)
(1) Yes.
Suppose p1 (x) = a1 x2 , p2 (x) = a2 x2
(i) p1 (x) + p2 (x) = (a1 + a2 )x2 = ax2 is in the subspace where a = a1 + a2 is real.
(ii) cp1 (x) = ca1 x2 is in the subspace where ca1 is real.
(2) No.
Suppose p1 (x) = a1 + x2 , p2 (x) = a2 + x2 , p1 (x) + p2 (x) = (a1 + a2 ) + 2x2 , hence it is not
closed under addition. Or you can check it is not closed under scalar multiplication.
(3) No.
Obviously it doesn’t meet the property (ii). If we use a real number to scale the “integer
vector”, we cannot always get an “integer vector”.
3. (Practice) Let W be the set of all vectors of the form ⟨s + 3t, s − t, 2s − t, 4t⟩. Is W a
subspace of R4 ? Find a basis for W .
Solution:  

 s + 3t 

s − t
  
4
(1) W =   : s, t are real ⊆ R .
 

 2s − t 
 4t 
   
s1 + 3t1 s2 + 3t2
 s1 − t1   s2 − t2 
Suppose there are two vectors u1 =  2s1 − t1  and u2 = 2s2 − t2  in W .
  
4t1 4t2
 
(s1 + s2 ) + 3(t1 + t2 )
 (s1 + s2 ) − (t1 + t2 ) 
(i) u1 + u2 =  2(s1 + s2 ) − (t1 + t2 ) is in W .

4(t1 + t2 )
  
cs1 + 3ct1
 cs1 − ct1 
(ii) For each scalar c, cu1 = 
2cs1 − ct1  is in W .

4ct1
Therefore, W is a subspace of R4 .
     
s + 3t 1 3
s−t 1 −1
2s − t = s 2 + t −1,
(2) Since      
4t 0 4
   
1 3
1 −1
W = span{v1 , v2 } where v1 = 
2 , v2 = −1 .
  
0 4
v1 and v2 are linearly independent, so {v1 , v2 } is a basis for W .
4. (Practice) Find any basis for the set of vectors in R3 in the plane x + 2y + z = 0. Can
you find an orthogonal basis?
Solution:  
x
(1) If the vector y  is in the plane, the components x, y, z satisfy the equation x + 2y +

z
z = 0.            
x −2y − z −2 −1 −2 −1
Since y =
   y  = y 1 +z 0
    where  1  and  0  are linearly
z  z   0 1 0 1
 −2 −1 
independent,  1 , 0  form a basis for the set of the vectors in the plane.
 
0 1
 

(2) In general, you can use Gram-Schmidt Process to find an orthogonal basis for a given
subspace W . Here let’s use a more intuitive way to find one. The equation indicates that
the normal vector of this plane is < 1, 2, 1 > (Since < 1, 2, 1 > · < x, y, z >= 0). We
already know a vector in the plane, e.g. < −1, 0, 1 >. To find the other vector, we just
need to calculate the cross product of the normal vector and the first vector < 1, 2, 1 >
× < −1, 0, 1 >=< 2, −2, 2 >, or you can use < 1, −1, 1 > which  is also in 
the plane
 and
 1 −1 
perpendicular to the first vector. So an orthogonal basis can be −1 ,  0  .
1 1
 

5. (Practice) Find a basis for the set of vectors in R2 on the line y = 5x.
Solution:  
x
If the vector is in the line, the components x, y satisfy the equation y = 5x.
y
       
x x 1 1
Since = =x , forms a basis for the set of the vectors in the line.
y 5x 5 5
6. (Practice) The first four Hermite polynomials, which form a basis for P3 , are 1, 2x,
−2 + 4x2 , and −12x + 8x3 . Let p(x) = 7 − 12x − 8x2 + 12x3 . Find the coordinate vector
of p(x) relative to the first four Hermite polynomials.
Solution:
 Let p(x) = 7 − 12x − 8x2 + 12x3 = a · 1 + b · 2x + c(−2 + 4x2 ) + d(−12x + 8x3 ).


 a − 2c = 7

2b − 12d = −12
Compare the coefficients, we have .

 4c = −8

8d = 12



 a=3

 b=3
Solving this linear system gives .

 c = −2

d = 3/2


So the coordinate vector of p(x) is (3, 3, −2, 3/2).

7. (Practice) Determine a basis for the null space of
 
1 1 1 0
A = 1 1 0 1 .
1 0 1 1

Solution:
Use row operations to construct rref(A):
     
1 1 1 0 1 1 1 0 1 1 1 0
1 1 0 1 → 0 0 −1 1 → 0 0 −1 1
1 0 1 1 1 0 1 1 0 −1 0 1
   
1 1 1 0 1 1 0 1
→ 0 −1 0 1 → 0 −1 0 1
0 0 −1 1 0 0 −1 1
   
1 0 0 2 1 0 0 2
→ 0 −1 0 1 → 0 1 0 −1
0 0 −1 1 0 0 1 −1
 
x1
x2 
Then, if the vector 
x3  is in the null space of A, the components x1 , x2 , x3 , x4 satisfy

x4
the equations 
x1 = −2x4

x2 = x4 .

x = x
3 4
.        
x1 −2x4 −2   −2  
x2   x4   1   1 
Since   = 
    = x4  ,   forms a basis for the null space of A.
   
x3 x4  1 
 1 
x4 x4 1  1  