Problems about Vector spaces

• READ THIS FIRST
  
• Problems about Vector spaces
  
  • Level 1 problems
    
  • Level 2 problems
    

READ THIS FIRST

Some problems can't be solved without the knowledge about matrices and systems of linear equations.

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Vector spaces

Level 1 problems

• 
  

  Show that the set V = {(x,y,z) | x,y,z in R and x+y = 11} is not a subspace of R3 .

  
  

  Since (0,0,0) is not in V, V is not a subspace of R³.

• 
  

  Show that the set V = {(x,y,z) | x,y,z in R and x.x = z.z } is not a subspace of R3.

  
  

  (1,1,1) and (1,1,-1) are in V but the sum (1,2,0) is not!

• 
  

  Show that the set V = {(x,y,z) | x,y,z in R and x + 2y + z = 0} is a subspace of R3.

  
  

  For each r,s in R, and for each (x,y,z) and (x',y',z') in V, we have
  

   
  x + 2y + z = 0 ; x' + 2y' + z' = 0  and
   
  r(x,y,z)+s(x',y',z') = (rx+sx',ry+sy',rz+sz')
   
  Now, it is sufficient to prove that (rx+sx',ry+sy',rz+sz') is in V.
  
          (rx+sx',ry+sy',rz+sz') is in V
  
  <=>     rx + sx'+2ry + 2 rsy' + rz + sz' = 0
  
  <=>     r(x + 2y + z) + s(x' + 2y' + z') = 0
  
  and this is true since x + 2y + z = 0 and x' + 2y' + z' = 0
  

• 
  

  Examine whether or not M = {(r,r+2,0) | r in R} is a subspace of R3.

  
  

  Since (0,0,0) is not in M, M is not a subspace of R³.

• 
  

  S = {(2,5,3)} and T = {(2,0,5)} The intersection of span(S) and span(T) is a vector space. Find this space.

  
  

   
  span(S) = {r.(2,5,3) | r in R}
  span(T) = {r'.(2,0,5) | r'in R}
  The only common vector of span(S) and span(T) is (0,0,0).
  The intersection of span(S) and span(T) is a vector space {(0,0,0)}.
  This space is called the zero space or the null space.
  

• 
  

  Show that {(1,2,3) , (2,3,4) , (3,4,5) } is not a basis of R3.

  
  

   
          The three vectors are linear dependent
  
  <=>     There is a  k,l,m not all zero such that
           l(1,2,3)+m(2,3,4)+n(3,4,5)=(0,0,0)
  
  <=>     The following system has a non trivial solution
                  1.l + 2.m + 3.n = 0
                  2.l + 3.m + 4.n = 0
                  3.l + 4.m + 5.n = 0
  
  <=>             | 1     2       3|
                  | 2     3       4| = 0
                  | 3     4       5|
  And this is true!
  

• 
  

  Find a basis of R3 containing the vectors (1,2,5) and (0,1,2).

  
  

  Since the dimension is 3, we have to find 1 vector (a,b,c) such that (1,2,5) ;(0,1,2) and (a,b,c) are linear independent.

   
          The three vectors are linear dependent
  
  <=>     There is a  k,l,m not all zero such that
           l(1,2,5)+m(0,1,2)+n(a,b,c)=(0,0,0)
  
  <=>     The following system has a non trivial solution
                  1.l + 2.m + 5.n = 0
                  0.l + 1.m + 2.n = 0
                  a.l + b.m + c.n = 0
  
  <=>             | 1     2       5|
                  | 0     1       2| = 0
                  | a     b       c|
  

  If the last determinant is 0, the three vectors are independent. We must find an independent vector(a,b,c). So, we choose a,b and c such than the determinant is not zero. Example (0,0,1). (1,2,5) ;(0,1,2) and (0,0,1) form a basis of R³.
• 
  

  Assume that v and w are linear independent vectors. Prove that v , w and (v + w) are linear dependent vectors.

  
  

  Since v + w = 1.v + 1.w , the vector (v + w) is a linear combination of v and w.

• 
  

  Find the coordinates of the vector (3,2,1) with respect to the basis ((1,0,2),(2,1,0),(0,3,5)) in R3.

  
  

   
  Say (x,y,z) are those coordinates, then
  (3,2,1) = x(1,0,2)+y(2,1,0)+z(0,3,5)
  <=>
          / x+2y+0z = 3
          | 0x+y+3z = 2
          \ 2x+0y+5z= 1
  <=>
          x = 1/17 ; y = 25/17 ; z = 3/17
  

• 
  

  Find the solution space of the linear system 3x + 2y + 6z = 0 -x - y + 2z = 0 2x + y + 8z = 0

  
  

   
  The third equation is a linear combination of the previous equations.
  
  So, the system is equivalent with
  
          3x + 2y + 6z = 0
          -x - y  + 2z = 0
  
  For each value of z, we have just one solution of the system.
  
          x = -10z ; y = 12z ; z = z
  
  The solutions are {(-10z,12z,z) | with z in R}
  
  These solutions form a one dimensional vector space with basis (-10,12,1)
  

• 
  

  All polynomials p(x) with degree not greater than 2 constitute a vector space V. Replace in (1, 1 + x2 , b(x) ) the polynomial b(x) such that it becomes an ordered basis for that vector space. Calculate the coordinates of (2x2 - 7x) with respect to the chosen basis.

  
  

  The vectors 1 and (1 + x²) are linear independent. It is easy to see that x can't be written as a linear combination of the given vectors. So, (1, 1 + x² , x ) is an ordered basis for that vector space.

  Now we write (2x² - 7x) as a linear combination of the vectors of the ordered basis.

  We find -2.(1) + 2.(1 + x²) - 7.(x) = (2x² - 7x)

  The coordinates of (2x² - 7x) are (-2,2,-7).

Level 2 problems

• 
  

  S = {(2,5,3),(1,0,2)} and T = {(2,0,5),(3,5,5)} The intersection of span(S) and span(T) is a vector space. What is the dimension of that space.

  
  

   
  span(S) = { r(2,5,3) + s(1,0,2) | r,s in R}
  span(T) = { r'(2,0,5) + s'(3,5,5) | r',s' in R}
  (x,y,z) is an element of the intersection if and only if
  there is an r, s , r', s' such that
          (x,y,z) = r(2,5,3) + s(1,0,2) = r'(2,0,5) + s'(3,5,5)
  
  <=> there is an r, s , r', s' such that
           2r + s =  2r' + 3s'
           5r     =        5s'
           3r +2s =  5r' + 5s'
  
  <=> there is an r, s , r', s' such that
           2r + s -  2r' - 3s' = 0
           5r     -        5s' = 0
           3r +2s -  5r' - 5s' = 0
  This is a homogeneous system of the second kind.
  The rank of the coefficient matrix is 3.
  We choose a main matrix
          [2  1  -2]
          [5  0   0]
          [3  2  -5]
  s' is the side unknown and s' is an arbitrary number.
  Hence, for each choice of s' there is a vector of the intersection.
  The dimension of the the intersection space is 1.
  

• 
  

  Assume that v and w are linear independent vectors. Prove that v and (v + w) are linear independent vectors.

  
  

   
          rv + s(v + w) = 0
  =>      (r+s)v + sw = 0
                  since v en w are linear independent
  =>      r+s = 0 and s = 0
  
  =>      r = s = 0
  
  =>      v and (v + w) are linear independent vectors.
  
  

• 
  

  Find the condition for r and s such that the vectors (r,2,s) , (r+1,2,1) and (3,s,1) are linear dependent.

  
  

   
       (r,2,s) , (r+1,2,1) and (3,s,1) are linear dependent.
  <=>
          There is a (k,l,m) not (0,0,0) such that
          k(r,2,s) + l(r+1,2,1) + m(3,s,1) = 0
  <=>
          The system
          / r.k  +  (r+1).l  + 3.m = 0
          | 2.k  +     2. l  + s.m = 0
          \ s.k  +     1. l  + 1.m = 0
          has a  solution different from (0,0,0).
  <=>
          | r     r+1     3|
          | 2      2      s| = 0
          | s      1      1|
  <=>
          rs2  +s2  - rs -6s + 4 = 0
  

• 
  

  Find ,for each m, the dimension of the row space of the matrix [2 m m-1] [3 m 5 ] [1 0 m+1]

  
  

• 
  

  Find ,for each m, the solution space of the linear system 3x + 2y + mz = 0 mx - y + 4z = 0 2x + y + 3z = 0

  
  

   
  
          |3      2       m|
          |m      -1      4| = ...=(m-5)(m+1)
          |2      1       3|
  Case 1 :   m is different from 5 and -1
          The only solution is x = y = z = 0.
          The solution space is {(0,0,0)}
  Case 2 :   m = -1 ; The system becomes
  
          3x + 2y -  z = 0
          -x - y  + 4z = 0
          2x + y  + 3z = 0
  
          The third equation is a linear combination of the previous ones.
          The system is equivalent to
  
          3x + 2y -  z = 0
          -x - y  + 4z = 0
  
          and this system is equivalent to
  
          x = -7z
          y = 11z
  
          The solution space is {(-7,11,1).z | with z in R}
  
  Case 3 :  m = 5 ;       The system becomes
  
          3x + 2y + 5z = 0
          5x - y  + 4z = 0
          2x + y  + 3z = 0
  
          The third equation is a linear combination of the previous ones.
          The system is equivalent to
  
          3x + 2y + 5z = 0
          5x - y  + 4z = 0
  
          and this system is equivalent to
  
          x = -z
          y = -z
  
          The solution space is {(-1,-1,1).z  | with z in R}
  

• 
  

  In the vector space V = R3, we take a set S = {(4,5,6) , (r,5,1) , (4,3,2)} Find the values of r such that the vector space spanned by S is not V.

  
  

  If the three vectors of S are linear independent, the vector space spanned by S is V. If this is not the case, then (r,5,1) has to be a linear combination of (4,5,6) and (4,3,2).

   
  This is equivalent to:
          |r      5       1|
          |4      5       6| =0
          |4      3       2|
  
  <=>     r = 9
  

• 
  

  In R3 we have basis B = ((1,0,1) , (0,2,0) , (1,2,3)) and a basis C = ((1,0,0) , (2,0,1) , (0,0,3)) The coordinates of a vector v with respect to B are (x,y,z). The coordinates of a vector v with respect to C are (x',y',z'). Write the relation between these coordinates in matrix notation.

  
  
• 
  

  M = span { (1+m, 4, 2); (5,6,-1-m) } N = span { (5+2m, 10,0) } Show that (1+m, 4, 2) and (5,6,-1-m) are linear independent for all real m values. Calculate m such that M+N is a direct sum.

  
  
  (Ans: m = 1 or m = 15 )