Problems about Linear transformations

• READ THIS FIRST
  
• Problems about Linear transformations
  
  • Level 1 problems
    
  • Level 2 problems
    
  • Level 3 problems
    

READ THIS FIRST

The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the problem before you read the given solution.

Problems about Linear transformations

Level 1 problems

• 
  

  Is the following transformation linear t : R x R --> R x R : (x,y) --> (2x - y, 0)

  
  

   
  For all vectors (x,y) , (x',y')  and all real numbers r, s we have
          t(r(x,y) + s(x',y')) = t(rx+sx',ry+sy') =(2rx + 2sx' -ry -sy',0)
  Otherwise
          r.t(x,y) + s.t(x',y') = r.(2x - y, 0) + s.(2x' - y', 0)
                  = (2rx + 2sx' -ry -sy',0)
  Hence t(r(x,y) + s(x',y')) = r.t(x,y) + s.t(x',y') and
  t is a linear transformation.
  

• 
  

  Find the matrix of the following linear transformations with respect to a natural basis. t1: R x R --> R x R : (x,y) --> (2x - y, 0) t2: R x R --> R x R : (x,y) --> (2x - y, x) t3: R x R x R --> R x R x R: (x,y,z) --> (2x - y, 0, y +z) t4: R x R x R --> R x R x R: (x,y,z) --> (0, 0,y)

  
  

   
  
  The basis of R x R  is ( (1,0) , (0,1) )
  t1(1,0) = (2,0) and has coordinates (2,0)
  t1(0,1) = (-1,0) and has coordinates (-1,0)
  The matrix of t1 is then
          [2   -1]
          [0    0]
  
  t2(1,0) = (2,1) and has coordinates (2,1)
  t2(0,1) = (-1,0) and has coordinates (-1,0)
  The matrix of t2 is then
          [2   -1]
          [1    0]
  
  The basis of R x R x R  is ( (1,0,0), (0,1,0), (0,0,1) )
  t3(1,0,0) = (2,0,0) and has coordinates (2,0,0)
  t3(0,1,0) = (-1,0,1) and has coordinates (-1,0,1)
  t3(0,0,1) = (0,0,1) and has coordinates (0,0,1)
  The matrix of t3 is then
          [2   -1   0]
          [0    0   0]
          [0    1   1]
  
  t4(1,0,0) = (0,0,0) and has coordinates (0,0,0)
  t4(0,1,0) = (0,0,1) and has coordinates (0,0,1)
  t4(0,0,1) = (0,0,0) and has coordinates (0,0,0)
  The matrix of t4 is then
          [0    0   0]
          [0    0   0]
          [0    1   0]
  
  

• 
  

  Let t be a linear transformation of V. M is the set of all fixed points of t. Show that M is a subspace of V.

  
  

  Take two random fixed points v and w of t.
  

   
     t(v) = v and t(w) = w.
  
  Since t is a linear transformation, we have for all real numbers r and s
  
      t( rv + sw )
  
  =   r t(v) + s t(w)
  
  =   r v + s w
  

  So for all real numbers r and s we have that r v + s w is a fixed point of t. From this it follows immediately that M is a subspace of V.

• 
  

  Let V be the vector space with the complex numbers as vectors. We choose the vectors 1 and i as a basis. t is a linear transformation of V with matrix [ 3 1 ] [ 4 3 ] Find the fixed points of t.

  
  

   
      r + s i with coordinates (r,s) is a fixed point of t
  <=>
    [ 3   1 ] [r] = [r]
    [ 4   3 ] [s]   [s]
  <=>
    / 3 r + 1 s = r
    \ 4 r + 3 s = s
  <=>
    / 2 r +   s = 0
    \ 4 r + 2 s = 0
  <=>
      s = - 2 r
  

  The fixed points are the complex numbers r - 2ri = r(1 - 2i).
  These fixed points form a line in the Gauss-plane.

• 
  

  Find the image of the vector (-2,4) with respect to the following linear transformations. Do this first without the matrix, and next with the matrix of t. t1 : R x R --> R x R : (x,y) --> (2x - y, 0) t2 : R x R --> R x R : (x,y) --> (2x - y, x)

  
  

   
  
  t1(-2,4)=(-8,0)
  With the matrix of t1
  
          [2   -1][-2]            [-8]
          [0    0][4]    =        [ 0]
  
  t2(-2,4)=(-8,-2)
  With the matrix of t2
  
          [2   -1][-2]            [-8]
          [1    0][4]    =        [-2]
  
  
  

• 
  

  Take an origin O in the plane and orthonormal basis vectors e1 and e2. A homothetic transformation h with center O and real factor k is a transformation of the plane such that h(v) = k v for all vectors v of that plane. Show that a homothetic transformation is a linear transformation and determine the matrix with respect to the basis (e1, e2). Find the eigenvectors of h?

  
  

  For all vectors a and b and all real numbers r and s we have

   
    t( r a + s b ) = k (r a + s b ) = k r a + k s b    (*)
  
    r t(a) + s t(b) = r k a +  s k b       (**)
  

  Since (*) = (**) h is a linear transformation.

  The image of e₁ is k e₁ and has coordinates (k,0).
  The image of e₂ is k e₂ and has coordinates (0,k).

  The matrix of h is

   
     [ k  0 ]
     [ 0  k ]
  

  Since each vector is transformed in a multiple of itself, all vectors, different from zero, are eigenvectors and the eigenvalue is k.

• 
  

  V is the vector space R x R. We define a transformation t of V such that the vector (x,y) is transformed in the vector (2x+3y,x-2y). Show that t is a linear transformation We choose the natural basis ((1,0), (0,1)). Find the matrix A of t. We choose a new basis ((2,-1), (3,0)) in V. Take the vector v=(7,3). Find the coordinates of v with respect to the new basis. Find the matrix B of t with respect to the new basis. Find the coordinates of t(v) with respect to the new basis and write the connection between t(v) and these coordinates.

  
  

  1. For all vectors (x,y) and (x',y') and all real numbers r and s we have

      
         t(r(x,y) + s(x',y'))
     
      =  t( rx+sx', ry+sy')
     
      =  ( 2(rx+sx') + 3(ry+sy') , rx+sx' - 2(ry+sy') )
     
      =  ( 2rx + 2sx' + 3ry + 3sy' , rx + sx' - 2ry - 2sy' )      (*)
     

     On the other hand we have

      
         r.t(x,y) + s.t(x',y')
     
      =  r (2x+3y,x-2y) + s (2x'+3y',x'-2y')
     
      =  ( 2rx + 3ry , rx - 2ry) + ( 2sx' + 3sy', sx'- 2sy')
     
      =  ( 2rx + 2sx' + 3ry + 3sy' , rx + sx' - 2ry - 2sy' )     (**)
     

     Since (*) = (**) , t is a linear transformation

  2. We calculate the coordinates of the images of the basis vectors.

      
       t(1,0) = (2,1) and has coordinates (2,1)
      
       t(0,1) = (3,-2) and has coordinates (3,-2)
      
     
     The matrix  of t with respect to the natural basis is A =
     
     
       [2  3]
       [1 -2]
     

  3. • direct method

        
          v = (7,3) has coordinates (a,b) with respect to the new basis
       <=>
           7 = a (2,-1) + b (3,0)
       <=>
           7 = 2a + 3b
           3 = -a
       <=>
           a = -3
           b = 13/3
       
          So,  v = (7,3) has coordinates (-3, 13/3).
       

     • Second method : calculation of the new coordinates from the old coordinates

       The basis is now ((2,-1), (3,0)). All vectors have new coordinates. The new coordinates are connected with the old coordinates by a transformation matrix C. The columns of the transformation matrix C are the coordinates of the new basis vectors with respect to the original (old) basis vectors.

        
          C = [ 2   3]
              [-1   0]
       
          v = (7,3) and the old coordinates are (7,3).
       
          (old coordinates of v) = C . (new coordinates of v)
       <=>
          C-1 . (old coordinates of v) = (new coordinates of v)
       
           [ 2   3]-1 [7]  =  [ -3 ]
           [-1   0]   [3]     [13/3]
       
          The new coordinates of v are (-3, 13/3).
       

  4. • First method: deduce the matrix B from the old matrix the old matrix A.
       The connection between B and A is B = C^-1 A C.

        
          B = [ -4, -3 ]
              [ 3,  4  ]
       

     • Second method: direct calculation of B.

       The new basisvectors are (2,-1) and (3,0). The images of these vectors, by t, are the vectors (1,4) and (6,3).

       After calculation we find that the vector (1,4) has coordinates (-4,3).
       After calculation we find that the vector (6,3) has coordinates (-3,4).
       

       The columns of B are the coordinates of the images of the new basis vectors.

        
          B = [ -4, -3 ]
              [ 3,  4  ]
       

  5. The coordinates of t(v) with respect to the new basis are

      
     
        B . [ -3 ]  = [ -4, -3 ][ -3 ] = [  -1 ]
            [13/3]    [ 3,  4  ][13/3]   [ 25/3]
     
     
        t(v) = (23,1) and has new  coordinates (-1, 25/3).
     
        So,    (23,1) = -1 (2,-1) + 25/3 (3,0)
     

• 
  

  Find the eigenvalues and the characteristic vectors of t1 : R x R --> R x R : (x,y) --> (2x - y, 0) t2 : R x R --> R x R : (x,y) --> (2x - y, x) t3 : R x R x R --> R x R x R: (x,y,z) --> (2x - y, 0, y +z) t4 : R x R x R --> R x R x R: (x,y,z) --> (0, 0,y)

  
  

   
  
  
  First, we calculate the eigenvalues r of t1
  The characteristic equation  is
          |2-r   -1|
          |        | = 0  <=> -r(2-r) = 0
          |0    0-r|
  The eigenvalues are r = 0 and r = 2
  For r=0 the characteristic vectors are the non trivial solutions of the
  system
          2x - y = 0
          0x +0y = 0
  The characteristic vectors are all the multiples of (1,2).
  For r=2 the characteristic vectors are the non trivial solutions of the
  system
          2x - y = 2x
          0x +0y = 2y
  This system is equivalent with 0x + y = 0
  The characteristic vectors are all the multiples of (1,0).
  
  You find the eigenvalues and characteristic vectors of t2 in the same way.
  
  Next, we calculate the eigenvalues r of t3
  The characteristic equation  is
          |2-r  -1   0 |
          |0    -r   0 | = 0  <=> (r-1).r.(2-r) = 0
          |0     1  1-r|
  The eigenvalues are r = 0 , r = 1 and r = 2
  For r=0 the characteristic vectors are the non trivial solutions of the
  system
          2x - y + 0z = 0
          0x +0y + 0z = 0
          0x + y +  z = 0
  This system is equivalent with
          2x - y = 0
          y + z = 0
  The characteristic vectors are all the multiples of (1,2,-2).
  For r=1 the characteristic vectors are the non trivial solutions of the
  system
          2x - y + 0z = x
          0x +0y + 0z = y
          0x + y +  z = z
  This system is equivalent with
          x - y = 0
            y = 0
  The characteristic vectors are all the multiples of (0,0,1).
  For r=2 the characteristic vectors are the non trivial solutions of the
  system
          2x - y + 0z = 2x
          0x +0y + 0z = 2y
          0x + y +  z = 2z
  This system is equivalent with
          y=0
          z=0
  The characteristic vectors are all the multiples of (1,0,0).
  You find the eigenvalues and characteristic vectors of t4 in the same way.
  
  

• 
  

  Take an origin O in the plane and orthonormal basis vectors e1 and e2. The linear transformation t1 is the orthogonal reflection in the line y = x. The linear transformation t2 is the orthogonal projection on the x-axis. Let t = t2 o t1 Find the matrix and the eigenvectors of the transformation t.

  
  

   
  t1 has matrix
  
    [0  1]
    [1  0]
  
  t2 has matrix
  
   [1  0]
   [0  0]
  
  t = t2 o t1 has matrix
  
    [1  0]  [0  1]  =   [0  1]
    [0  0]  [1  0]      [0  0]
  
  The eigenvalues of t are the solutions of
  
   | -r  1 |  = 0  <=>  r2 = 0 <=> r = 0
   | 0  -r |
  
  The eigenvectors v(x,y) are obtained by
  
   [0  1] [x]  = [0]    <=>  y = 0
   [0  0] [y]    [0]
  
  The eigenvectors are the vectors with image point on the x-axis.
  

• 
  

  A linear transformation t has a matrix [ 1-m 2 4 ] [ 3 -1 0 ] [ m m -2 ] Find the m-values such that the null-space of t is different from {0}. Find the null-space corresponding with each of these m-values.

  
  

  The kernel of t is the set of all vectors v(x,y,z) such that

   
    [ 1-m    2   4 ][x]   [0]
    [  3    -1   0 ][y] = [0]
    [  m     m  -2 ][z]   [0]
  
    This system has solutions different from the zero solution
  
       | 1-m    2   4 |
  <=>  |  3    -1   0 | = 0
       |  m     m  -2 |
  
  <=> m = -1
  
  We take m = -1 and we calculate the null-space.
  This null-space of t is the set of all vectors v(x,y,z) such that
  
    [  2    2   4 ][x]   [0]
    [  3   -1   0 ][y] = [0]
    [ -1   -1  -2 ][z]   [0]
  

  The third equation is a multiple of the first one. The system is equivalent with

   
     x   + y   = - 2 z
     3 x - y   =   0
  <=>
     y = 3 x
     z = -2 x
  
  The kernel is the set of vectors with coordinates
  
    { (r, 3r , -2r) | r in R }
  
  It is a space with dimension 1 and generated by the vector w(1,3,-2).
  

• 
  

  In V = R2 we take a natural basis B. The set S = { (3 r, 7 r) | r in R and r not zero } is a set of characteristic vectors of a linear transformation t. In V we take a new basis B. The basis-vectors are (3,-1) and (2,-1). Find the set of all the coordinates of the characteristic vectors in S with respect to the new basis B.

  
  

  We start with the characteristic vector (3,7).

   
         (x,y) are the coordinates of  (3,7) with respect to B
  <=>
           (3,7) = x (3,-1) + y (2, -1)
  <=>
              3 = 3 x + 2 y
              7 = - x  -  y
  <=>
             x = 17 ; y = -24
  
  The  set of all the coordinates of the characteristic vectors in S
  with respect to the new basis B is
  
        { (17 r, -24 r) | r in R and  r not 0}
  

• 
  

  We start with two vectors v and w of a vector space V. t is a linear transformation of V. Show that: t(v) = t(w) <=> v - w is in ker(t)

  
  

   
       t(v) = t(w)
  <=>
       t(v) - t(w) = 0
  <=>
       t(v - w)  = 0
  <=>
      v - w  is in ker(t)
  

• 
  

  t is a linear transformation of a vector space V. u(-1,1) is a fixed point of t. v(2,-1) is in ker(t). Find the matrix of t.

  
  

   
  The required matrix has the form
  
       [a  b]
       [c  d]
  
  Now, we must have
  
  
     [-1] = [a  b] [-1]  en [0] = [a  b][ 2]
     [ 1]   [c  d] [ 1]     [0]   [c  d][-1]
  
  <=>
     ...
  <=>
     a = -1 ; b = -2 ; c = 1 ; d = 2
  
  The  matrix is
  
     [-1 -2]
     [ 1  2]
  

• 
  

  Take an origin O in the plane and orthonormal basis vectors e1 and e2. The linear transformation t has matrix [[3,-1],[2,-1]]. t transforms all points of the line r with equation y = 3x in another line r'. Find the equation of r'.

  
  

  t transforms (0,0) in (0,0).
  t transforms (1,3) in (0,-1).
  The required line is the line y = -x.

Level 2 problems

• 
  

  V is a vector space with basis e1 en e2. t is a linear transformation of V. t( u(1,3) ) = u'(5,8) t( v(2,-1) ) = v'(3,-5) Find the matrix of the linear transformation t with respect to the given basis.

  
  

   
  The required matrix has the form
       [a  b]
       [c  d]
  Now, we must have
  
     [5] = [a  b] [1] and [ 3] = [a  b][ 2]
     [8]   [c  d] [3]     [-5]   [c  d][-1]
  
  <=>
      a + 3 b = 5
     2a -   b = 3
      c + 3 d = 8
     2c -  d  = -5
  
  <=>
     a = 2 ; b = 1 ; c = -1 ; d = 3
  
  The required matrix is
  
     [ 2  1]
     [-1  3]
  

• 
  

  In V = R2 we take a natural basis B. The set S = { (3 r, 7 r) | r in R and r not zero } is a set of characteristic vectors of a linear transformation t. The linear transformation t' has matrix [[2,1],[4,-1]] with respect to the basis B. Find the set S' of all vectors v(x,y) such that t'(v) belongs to S.

  
  

   
          v(x,y) is in S'
  <=>
           t'(v) is in S
  <=>
        There is an r (not 0) such that
         [ 2  1 ] [x] = r [3]
         [ 4 -1 ] [y]     [7]
  <=>
        There is an r (not 0) such that
         [x] = r  [ 2  1 ]-1 [3]
         [y]      [ 4 -1 ]   [7]
  <=>
        There is an r (not 0) such that
         [x] = r [ 1/6, 1/6  ] [3]
         [y]     [ 2/3, -1/3 ] [7]
  <=>
        There is an r (not 0) such that
         [x] = r [ 5:3  ]
         [y]     [ -1:3 ]
  <=>
        There is an r (not 0) such that
         [x] = r [  5 ]
         [y]     [ -1 ]
  

  The set S' consists of the real, non-zero, multiples of the vector (5,-1).

• 
  

  The rotation, with angle u radians (u not 0), about a fixed point o is a linear transformation of the vector space of all vectors in a plane. Find the matrix of a rotation with respect to an orthonormal basis (e1,e2) in the plane Write this matrix for u = pi, and calculate the eigenvalues and the characteristic vectors.

  
  

   
  The coordinates of the image of e1 are (cos(u),sin(u))
  The coordinates of the image of e2 are (cos(u+pi/2),sin(u+pi/2))
                                  or ( -sin(u) , cos(u) )
  The matrix of the rotation is
  
          [cos(u)     -sin(u)]
          [sin(u)      cos(u)]
  
  The characteristic equation is
  
          [cos(u)-r           -sin(u)]
          [                          ] = 0
          [sin(u)            cos(u)-r]
  
  <=>     (cos(u) - r)2  + sin2 u = 0
  
  <=>     r2 -2.cos(u).r + 1 = 0
  
  The discriminant is 4.cos2 (u) -4 is negative for all u not 0.
  So, there are no real eigenvalues and no real characteristic vectors.
  

• 
  

  Let A = matrix of a linear transformation t. Prove that t has an eigenvalue 0 if and only if A is singular.

  
  

  If t has an eigenvalue 0, there is a characteristic vector v such that
  t(v) = 0.v <=> t(v) = 0 .

  Let (x,y,z) be the coordinates of v. (x,y,z) is not (0,0,0).

   
            [x]  [0]
          A.[y] =[0]   has a solution different from (0,0,0)
            [z]  [0]
  
  From the theory about homogeneous systems, we know that this is equivalent
  with det(A) = 0 <=> A is singular.
  

• 
  

  The linear transformation t has, with respect to an orthonormal basis (e,u) , the matrix [m m] [1 2] a) Calculate m such that (1,1) are the coordinates of a characteristic vector v. b) Calculate the coordinates of a characteristic vector w, linear independent of v and such that w is a unit vector. c) What is the matrix of t if we take v and w as a new basis in the vector space. d) Calculate the coordinates of e and u with respect to this new basis.

  
  

  a)(1,1) are the coordinates of a characteristic vector v

   
  <=>     [m    m][1]  = r. [1]  <=> 2 m = r
          [1    2][1]       [1]       3  = r
  
  <=>  m = 3/2 and r = 3
  

  b)The eigenvalues are the solutions of

   
          |3/2 -r    3/2|  =  0  <=> ... <=> r = 3 or r = 1/2
          |   1     2- r|
  

  Each characteristic vector corresponding with r = 1/2 is a solution of the system

   
          / 3/2 x + 3/2 y = 1/2 x
          |                         <=> ... <=> 2x+3y = 0
          \     x +  2  y = 1/2 y
  

  The characteristic vectors have coordinates (3t,-2t).
  The magnitude has to be = 1.

   
    _____________
   |    2      2                   ____
  \| 9 t  + 4 t   = 1  <=> t = 1 /V 13
                                       ____        ____
  A unit vector w has coordinates (3/ V 13  , -2/ V 13 )
  

  c)

   
          [ 3,  0  ]
          [ 0, 1/2 ]
  

  d) We have

   
  v = e + u   and
          ____       ___
  w = 3/ V 13 e -2/ V 13 u
  
  Solving this for w end u, we find
               ____
  e = 2/5 v + V 13 /5 w
               ____
  u = 3/5 v - V 13 /5 w
  
  The coordinates of e and u with respect to this
  new basis are
             ____
  e =(2/5 , V 13 /5 )
              ____
  u =(3/5 ,- V 13 /5)
  

• 
  

  The vector v = (1,-1) is a characteristic vector of the linear transformation with matrix A = [4 3] [7 8] a) What is the corresponding eigenvalue? b) Calculate 1998 A v

  
  

   
          [4    3][ 1]  =  [ 1]
          [7    8][-1]     [-1]
  
  So the eigenvalue is 1 and A transforms v in v
  
                   1998
                  A     v = v
  

• 
  

  Calculate all eigenvalues of the linear transformation t with matrix [1 u v] [0 1 u] [0 0 1] Here, u and v are constant, non zero real numbers. Give for each eigenvalue the dimension of the associated vector space.

  
  
  
  The characteristic equation is

   
          (1-r)3 = 0 ; so the only eigenvalue r = 1
  The coordinates of the  characteristic vectors are the
  solutions of the system
  
          0x + uy + vz = 0
          0x + 0y + uz = 0
          0x + 0y + 0z = 0
  
  The solutions are all real multiples of (1,0,0) .
  The dimension of the associated vector space is 1.
  

• 
  

  Take an origin O in the plane and orthonormal basis vectors e1 and e2. The linear transformation p is a non-orthogonal projection on a line through O. The image of u(4,5) by p is u'(2,-1). Find the matrix of p.

  
  

   
  Let
       [a  b]
       [c  d]
  be the required matrix.
  
  Then we must have
  
    [ 2] = [a  b] [4]  en [ 2] = [a  b][ 2]
    [-1]   [c  d] [5]     [-1]   [c  d][-1]
  
  <=>
       4 a + 5 b = 2
       2 a -   b = 2
       4 c + 5 d = -1
       2 c -   d = -1
  <=>
      a = 6/7 ; b = -2/7 ; c = -3/7 ; d = 1/7
  
  
  The matrix is
  
     [6/7    -2/7]
     [-3/7   1/7 ]
  

Level 3 problems

• 
  

  Let A = matrix of a linear transformation t and C is a regular matrix. Prove that A and B = C-1 .A.C have the same eigenvalues.

  
  

   
  The eigenvalues of B are the roots of the characteristic equation
  
          det(B - rI) = 0     (I is the identity matrix)
  
  <=>     det(C-1 .A.C- rI) = 0
  
  <=>     det(C-1.A.C- r.C-1.I.C) = 0
  
  <=>     det(C-1.(A - rI).C) = 0
  
  <=>     det(C-1).det(A - rI).det(C) = 0
  
                  det(C) and det(C-1) are not 0 since C is a regular matrix.
  
  <=>     det(A - rI) = 0
  
  And the roots of this equation are the eigenvalues of A.
  
  

• 
  

  A of a linear transformation t is a 2x2 matrix with two different eigenvalues. Show that the characteristic vectors corresponding with different eigenvalues are linear independent. Prove that, if you choose two linear independent characteristic vectors, as a new basis, the matrix of t is a diagonal matrix.

  
  

  The product of all eigenvalues of a matrix A is not 0. Show that A is a regular matrix.

  
  

   
  Suppose that A is a singular matrix, then det(A) = 0 and
  the characteristic equation det(A - r I)=0 has the solution r=0.
   
  Thus, there is at least one eigenvalue = 0 and then
  the product of all eigenvalues of a matrix A is 0.
  

• 
  

  Take an origin O in the plane and orthonormal basis vectors e1 and e2. The linear transformation t has as matrix [[2,-2],[1,1]]. t transforms all points of the circle with equation x2+y2 = 2 in points of another curve. Find the equation of this other curve.

  
  

  By t, the vector v(x,y) is transformed in v'(x',y'). So, the image point P(x,y) of vector v is transformed in the image point P'(x',y') of vector v'.

   
     [ 2  -2] [x] = [x']
     [ 1   1] [y]   [y']
  
  <=>
     [x] = (1/4) [ 1  2] [x']
     [y]         [-1  2] [y']
  <=>
      x = (1/4) (x' + 2 y')
      y = (1/4) (-x'+ 2 y')
  
  Now, we have :
  
     x2 + y2 = 2
  <=>
     (1/16) (x' + 2 y')2 + (1/16)(-x'+ 2 y')2 = 2
  <=>
     x'2 + 4 y'2 = 16
  
  Point P' describes an ellipse with equation
  
      x'2      y'2
    ------ + ------- = 1
      16       4
  

• 
  

  V is the real vector space of all polynomials in x with real coefficients and with a degree lower than three. In V we take the basis B = (x2 , x , 1). We take a linear transformation t such that t(x2) = x + m t(x) = (m - 1)x t(1) = x2 + m Find : The matrix of t with respect to the given basis. Ker(t) for all values of the parameter m. The image of t for all values of the parameter m.

  
  

  1. The columns of the matrix are the coordinates of the images of the basis vectors. The matrix T of t is

      
          [0   0   1]
     T =  [1  m-1  0]
          [m   0   m]
     

  2.  
         Vector v(d,e,f) belongs to ker(t)
     <=>
           [0   0   1] [d]   [0]
           [1  m-1  0] [e] = [0]         (*)
           [m   0   m] [f]   [0]
     

     • First case: the determinant of T is not zero

        
             |0   0   1|
             |1  m-1  0| = m (1-m)  is not 0
             |m   0   m|
       
         <=>  (m not 0) and (m not 1)
       
         In this case the system (*) has only the solution (0,0,0).
         ker(t) = {0}.
       

     • Second case: m = 0

        
            Vector v(d,e,f) belongs to ker(t)
       <=>
             [0   0   1] [d]   [0]
             [1  -1   0] [e] = [0]
             [0   0   0] [f]   [0]
       <=>
            d = e  ; f = 0
       <=>
            ker(t) = { d x2 + d x | d in R }
       

     • Third case : m = 1

        
           Vector v(d,e,f) belongs to ker(t)
       <=>
             [0   0   1] [d]   [0]
             [1   0   0] [e] = [0]
             [1   0   1] [f]   [0]
       <=>
             d = 0 ; e = random  ; f = 0
       <=>
            ker(t) = { e x | e in R }
       

  3.  
        Vector v(d,e,f) belongs to the image of t
     <=>
        There is a vector  w(a,b,c) such that t(w) = v
     <=>
          (a,b,c) exists such that
     
           [0   0   1] [a]   [d]
           [1  m-1  0] [b] = [e]
           [m   0   m] [c]   [f]
     

     • First case: determinant of T is not zero

        
             |0   0   1|
             |1  m-1  0| = m (1-m)  is not 0
             |m   0   m|
       
         <=>  (m not 0) and (m not 1)
       
         In this case there is for each v(d,e,f) a suitable  w(a,b,c).
         The image of t is V.
       

     • Second case: m = 0

        
          Vector v(d,e,f)  belongs to the image of t
       <=>
          There is a vector  w(a,b,c) such that t(w) = v
       <=>
            (a,b,c) exists such that
       
            [0   0   1] [a]   [d]
            [1  -1   0] [b] = [e]
            [0   0   0] [c]   [f]
       <=>
           f = 0
       <=>
           The image of tconsists of all vectors  v(d,e,0).
       
       <=>
           The image of t is { dx2 + e x | d and e in R}
       

     • Third case: m = 1

        
          Vector v(d,e,f)  belongs to the image of t
       <=>
          There is a vector  w(a,b,c) such that t(w) = v
       <=>
            (a,b,c) exists such that
       
             [0   0   1] [a]   [d]
             [1   0   0] [b] = [e]
             [1   0   1] [c]   [f]
       <=>
            (a,b,c) exists such that
       
            c = d ; a = e ; a+c = f
       <=>
            d + e = f
       <=>
           The image of t consists of all vectors v(d,e,d+e)
       <=>
            The image of t is { d x2 + e x + d+e  | d and  e in R}