Themes > Science > Mathematics > Algebra > Foci of a conic section > Topics and Problems > Linear transformations

..Linear transformations
..A linear transformation
..Image of the vector 0.
..Criterion for the linearity of a transformation of V
..Building linear transformations
..Matrices and linear transformations
..Matrix of a linear transformation with respect to a basis in V.
..Null-space of a linear transformation
..The null-space is a subspace of V
..Nullity
..Matrix and the change of a basis
..Example
..Similar matrices
..Property of similar matrices.
..Sum of linear transformations
..Scalar multiplication of a linear transformation with a real number
..Product of two linear transformations
..Eigenvectors or characteristic vectors and eigenvalues
..Definition
..Eigenvalues and eigenvectors in a space with dimension 3.
..Vector space associated with an eigenvalue.
..Linear independent characteristic vectors.
..Characteristic vectors as a basis of a vector space
..Diagonal matrices and characteristic vectors.
..Corollary
..Example.
..Power of a diagonal matrix
..Power of a matrix
..Fibonacci


Linear transformations

A linear transformation

In a vector space V we define
 
        A transformation t of V is linear

                <=>

        For all vectors u , v and all real numbers r
        t(u+v) = t(u)+t(v) and t(r.u) = r.t(u)
The set of all linear transformations of V is L(V).
Examples : 
 
t : R x R  -> R x R  : (x,y)  -> (x+y,x)
t : R x R  -> R x R  : (x,y)  -> (0,y)
t : R -> R     : x -> 6x

Image of the vector 0.

Let t be a linear transformation of V, then 
 
        t(0) = t(0v) = 0.t(v) = 0
Hence, the image of the vector 0 is 0.

Criterion for the linearity of a transformation of V

Theorem :
Take a transformation t of V.
 
                t is in L(V)
                   <=>
        For all vectors u, v and all real numbers r, s
        t(r.u + s.v) = r.t(u) + s.t(v)
Proof :
Part 1 : If t is in L(V) then
 
        t(r.u + s.v) = t(r.u) + t(s.v) =  r.t(u) + s.t(v)
Part 2 : If t(r.u + s.v) = r.t(u) + s.t(v) for all r, s then 
 
        take r = s = 1   t(u+v) = t(u)+t(v)
        take s = 0     t(r.u) = r.t(u)
Q.E.D.

Building linear transformations

We show this for dimension(V) = 3, but all can easily be generalized.
Theorem :
If (e1, e2, e3) is an ordered basis of V, and if (u1, u2, u3) is an ordered random set of three vectors from V.
Then, there is just one linear transformation t of V such that
t(e1) = u1 t(e2) = u2 t(e3) = u3
Prove :
A random vector v in V can be written as v = k.e1+l.e2+m.e3.
A random vector w in V can be written as w = k'.e1+l'.e2+m'.e3.
Then u + v = (k+k')e1 + (l+l')e2 + (m+m')e3.
We start from a transformation t of V defined by 
 
        t(v) = k.u1 + l.u2 + m.u3
t is linear because : 
 
        t(v + w)  = t( (k + k')e1  +  (l + l')e2  +  (m + m')e3 )
                = (k + k')u1  +  (l + l')u2  +  (m + m')u3
                =  k.u1 + l.u2 + m.u3  +  k'.u1 + l'.u2 + m'.u3
                =  t(v)  +  t(w)

        t(r.v)   = t(rk.e1 + rl.e2 + rm.e3)
                = rk.u1 + rl.u2 + rm.u3
                = r(k.u1 + l.u2 + m.u3)
                = rt(v)
Furthermore 
 
        t(e1) = t(1.e1 + 0.e2 + 0.e3) = 1.u1 + 0.u2 + 0.u3 = u1
        t(e2) = t(0.e1 + 1.e2 + 0.e3) = 0.u1 + 1.u2 + 0.u3 = u2
        t(e3) = t(0.e1 + 0.e2 + 1.e3) = 0.u1 + 1.u2 + 1.u3 = u3
Finely, t is unique because
Suppose t and t' two linear transformations such that
t(e1) = u1 and t'(e1) = u1
t(e2) = u2 and t'(e2) = u2
t(e3) = u3 and t'(e3) = u3
Then, for each v in V 
 
        t(v) = t(k.e1+l.e2+m.e3)
             = k.u1 + l.u2 + m.u3
and
        t'(v)= t'(k.e1 + l.e2 + m.e3)
             = t'(k.e1) + t'(l.e2) + t'(m.e3)
             = k.t'(e1) + l.t'(e2) + m.t'(e3)
             = k.u1 + l.u2 + m.u3
So t = t'
Example:
There is just one linear transformation of R x R such that
t(1,0) = (3,2)
t(0,1) = (5,4)
We calculate the image of (-1,5) .
t(-1,5) = t( -1(1,0) + 5(0,1) ) = -1(3,2) + 5(5,4) = (22,18)

Matrices and linear transformations

Matrix of a linear transformation with respect to a basis in V.

We show this for dimension(V) = 3, but all can easily be generalized.
If (e1, e2, e3) is an ordered basis of V, and if (u1, u2, u3) is an ordered random set of three vectors from V.
Then, we know there is just one linear transformation t of V such that
t(e1) = u1
t(e2) = u2
t(e3) = u3
And these vectors u1, u2, u3 can be expressed in e1, e2, e3 . 
 
        u1 = a.e1 + b.e2 + c.e3
        u2 = d.e1 + e.e2 + f.e3
        u3 = g.e1 + h.e2 + i.e3
A random vector v = k.e1+l.e2+m.e3 is transformed by t in t(v) = k.u1+l.u2+m.u3
So, 
 
        t(v) = k.u1 + l.u2 + m.u3
<=>
        t(v) =  k.(a.e1 + b.e2 + c.e3) +
                l.(d.e1 + e.e2 + f.e3) +
                m.(g.e1 + h.e2 + i.e3)
<=>
        t(v) =  (k.a + l.d + m.g).e1 +
                (k.b + l.e + m.h).e2 +
                (k.c + l.f + m.i).e3
The coordinates of v with respect to the basis (e1, e2, e3) are (k,l,m).
We call (k',l'm') the coordinates of t(v) with respect to the basis (e1, e2, e3).
Then, 
 
        k' = k.a + l.d + m.g
        l' = k.b + l.e + m.h
        m' = k.c + l.f + m.i

<=>

        [k']    [a  d  g]   [k]
        [l'] =  [b  e  h] . [l]
        [m']    [c  f  i]   [m]
The last matrix formula is the transformation formula associated with t with respect to the basis (e1, e2, e3). The matrix 
 
         [a  d  g]
         [b  e  h]
         [c  f  i]
is called the matrix of the linear transformation with respect to the basis (e1, e2, e3). The columns of this matrix are the coordinates of (u1, u2, u3).

Example :
There is just one linear transformation of R x R such that
t(1,0) = (3,2)
t(0,1) = (5,4)
The matrix of the linear transformation with respect to the basis is 

 
        [3  5]
        [2  4]

Null-space of a linear transformation

The null-space of a linear transformation t is the set of all vectors v such that t(v) = 0

The null-space is a subspace of V

Since t(0) = 0, the null-space is not empty. If u an v are in the null-space, then 
 
        t(r.u + s.v) = r.t(u) + s.t(v)
                     = r.0 + s.0
                     = 0
        So, r.u + s.v is in the null-space.
Hence the null-space is a subspace of V.

Nullity

The dimension of the null-space is called the nullity of the linear transformation t.

Matrix and the change of a basis

We show this for dimension(V) = 3, but all can easily be generalized.
If (e1, e2, e3) is an ordered basis of V an t is a linear transformation of V with matrix Ao. 
 
             [a  d  g]
        Ao = [b  e  h]
             [c  f  i]
The linear transformation t transforms a random vector
v = k.e1+l.e2+m.e3 in t(v) = k'.e1+l'.e2+m'.e3 and then
 
        [k']   [a  d  g] [k]
        [l'] = [b  e  h].[l]
        [m']   [c  f  i] [m]

          [k']            [k]
Let Ko' = [l']   and Ko = [l]
          [m']            [m]

Then,    Ko' = Ao.Ko            (*)
Now we take a new basis in V. Then, all the vectors of V have new coordinates. From the theory of vector spaces, we know that these new coordinates are linked to the old ones with a transformation matrix C.
Denote the new coordinates in matrix form as Kn.
Then Ko = C.Kn and Ko' = C.Kn'
We write (*) with the new coordinates.
 
        C.Kn' = Ao.C.Kn

<=>       Kn' = C-1  .Ao.C.Kn      (**)
The last formula gives the connection between the coordinates of v and t(v) with respect to the new basis.
Denote An as the matrix of t with respect to the new basis. Then we have
 
        Kn' = An.Kn      (***)
From (**) and (***) we see that 
 
        An = C-1  .Ao.C
The last formula gives us the possibility to calculate the new matrix An of t from the old matrix Ao.

Example

In a 2-dimensional space with basis (e1, e2), a linear transformation t has matrix 
 
        [3  1]
        [-1 1]
Now we take a new basis 
 
        e1' = e1 + e2
        e2' = e1 - e2

Then the transformation matrix C is

        [1  1]
        [1 -1]

and from this C-1   is

        [1/2   1/2]
        [1/2  -1/2]

The matrix of the linear transformation t with respect to the new
basis is

        [1/2   1/2] [3  1] [1  1]
        [1/2  -1/2] [-1 1] [1 -1]


 =      [2   0]
        [2   2]

Similar matrices

Two n x n matrices A an B are similar if and only if there is a nonsingular n x n matrix C such that 
 
        B  = C-1. A .C
As a corollary from previous formula we see that the matrices of a linear transformation, with respect to a different basis, are similar.

Property of similar matrices.

Say A and B are similar matrices. Then 
 
        B  = C-1. A .C

=>     |B| =|C-1|.|A|.|C|

=>     |B| =|C-1|.|C|.|A|


=>     |B| = |A|
So, similar matrices have the same determinant.

Sum of linear transformations

The sum of two linear transformations t and t' is defined by 
 
        t+t' : V -> V : v -> t(v) + t'(v)
It can easily be proved that t+t' is a linear transformation and that the matrix of t+t' is equal to the sum of the matrices of t and of t'.

Scalar multiplication of a linear transformation with a real number

The scalar multiplication of a linear transformation t with a real number r is defined by 
 
        r.t : V -> V : v -> r.t(v)
It can easily be proved that r.t is a linear transformation and that the matrix of r.t is equal to r.(matrix) of t.

Product of two linear transformations

The product t.t' of two linear transformations t and t' is defined by 
 
        t.t' : V -> V : v -> t(t(v))
It can easily be proved that t.t' is a linear transformation and that the matrix of t.t' is equal to (matrix of t).(matrix of t') .

Eigenvectors or characteristic vectors and eigenvalues

Definition

Say t is a linear transformation of a vector space V. 
 
   u is called an eigenvector or characteristic vector with respect to t

                      if and only if

                     u is not 0
       and there is a real number r such that t(u) = r.u
The real number r is called the eigenvalue of u.

Eigenvalues and eigenvectors in a space with dimension 3.

Say t is a linear transformation of a vector space V with dimension 3.
We fix a basis in V. With respect to that basis, t has a unique matrix . 
 
          [a  b  c]
          [d  e  f]
          [g  h  i]

We denote the co(u) = (x,y,z).

Now,     u(x,y,z) is a characteristic vector of t with eigenvalue r

                          <=>

                 t(u) = r.u  and  u not 0

                          <=>

                [a  b  c] [x]     [x]       [x]     [0]
                [d  e  f].[y] = r.[y]  with [y] not [0]
                [g  h  i] [z]     [z]       [z]     [0]


                          <=>

               [a  b  c] [x]     [1  0  0] [x]   [0]         [x]     [0]
               [d  e  f].[y] - r.[0  1  0].[y] = [0]    with [y] not [0]
               [g  h  i] [z]     [0  0  1] [z]   [0]         [z]     [0]


                         <=>

               [a  b  c] [x]     [r  0  0] [x]   [0]         [x]     [0]
               [d  e  f].[y]  -  [0  r  0].[y] = [0]    with [y] not [0]
               [g  h  i] [z]     [0  0  r] [z]   [0]         [z]     [0]


                        <=>
              The homogeneous system in x,y,z

               [a-r  b   c ] [x]    [0]
               [d   e-r  f ].[y]  = [0]
               [g    h  i-r] [z]    [0]

              has a solution different from (0,0,0).

                        <=>

                   |a-r  b   c |
                   |d   e-r  f | = 0
                   |g    h  i-r|
The last equation is called the characteristic equation of t with respect to the fixed basis in V. If r is a solution of this equation, then the system 
 
                [a  b  c] [x]     [x]
                [d  e  f].[y] = r.[y]
                [g  h  i] [z]     [z]
has a solution (x,y,z) different from (0,0,0). With this solution corresponds a characteristic vector u(x,y,z) of t.

This way of thinking can be extended to vector spaces with dimension n.

Vector space associated with an eigenvalue.

Theorem:
The set of all characteristic vectors associated with an eigenvalue k, form a vector space, together with 0.
Proof:
Say u and v are characteristic vectors with eigenvalue k, then t(u) = k.u and t(v) = k.v.
Hence, for all real r and s we have 
 
t(r.u + s.v) = r.t(u) + s.t(v) = r.k.u + s.k.v = k.(r.u + s.v)
So, for all real r and s, (r.u + s.v) is a characteristic vector with eigenvalue k.

Linear independent characteristic vectors.

Theorem:
Take a vector space with dimension 2 and a linear transformation t.
If two characteristic vectors correspond with different eigenvalues, then these characteristic vectors are linear independent.

Proof:
Let v = characteristic vector with eigenvalue k.
Let w = characteristic vector with eigenvalue l.
Suppose v and w are linear dependent, then there is a scalar r such that

 
        w = r v
=>      t(w) = t(r v)
=>      l w = r.t(v)
=>      l w = r. k v
=>      l r v = r. k v
=>        k = l
This gives a contradiction with the fact that the two characteristic vectors correspond with different eigenvalues.

This theorem can be extended for a vector space with dimension n. Take a vector space with dimension n and a linear transformation t.
If the n chacteristic vectors correspond with all different eigenvalues, then these characteristic vectors are linear independent.

Characteristic vectors as a basis of a vector space

From previous theorem we know:
Take a vector space V with dimension n and a linear transformation t.
If n characteristic vectors correspond with all different eigenvalues, then these characteristic vectors are linear independent. The characteristic vectors can be used as a basis of V.

Diagonal matrices and characteristic vectors.

Take a vector space V with dimension 2 and a linear transformation t.
If two characteristic vectors v and w correspond with different eigenvalues k and l, then these characteristic vectors are linear independent and they constitute a basis for V. The images of v and w are 
 
        t(v) = k.v = k.v + 0.w
        t(w) = l.w = 0.v + l.w

The matrix of t with respect to the basis ( v , w ) is

        [ k   0 ]
        [ 0   l ]
We say that the matrix of t is diagonal.

This can be extended for a vector space with dimension n.
Take a vector space V with dimension n and a linear transformation t.
If n characteristic vectors correspond with all different eigenvalues, then these characteristic vectors are linear independent. The characteristic vectors can be used as a basis of V. The matrix of t with respect to that basis is diagonal and the eigenvalues are the diagonal elements of the matrix.

Corollary

Say a linear transformation t has a matrix A with respect to a basis and t has n characteristic vectors corresponding with all different eigenvalues.
If we take these characteristic vectors as a new basis, the new matrix of t is a diagonal matrix D. We know that there is a formule that connect A and D. 
 
        D = C-1 . A . C
Here, C is the transformation matrix. The columns of C are the coordinates of the new basis-vectors (characteristic vectors) with respect to the original basis in V.

Example.

In a vector space with dimention 2 and with basis (e1, e2) a linear transformation t has matrix A = 
 
        [4 -1]
        [2  1]
Calculating the eigenvalues we find 3 and 2.
As corresponding characteristic vectors we choose v1(1,1) and v2(1,2).
Take these characteristic vectors as a new basis.
The transformation matrix is 
 
        [1  1]
        [1  2]
Then we have 
 
                        -1
        [3  0]  = [1  1]  [4  -1] [1  1]
        [0  2]    [1  2]  [2   1] [1  2]

Power of a diagonal matrix

Using mathematical induction, it is easy to prove that

diag(a, b, ... l) = diag(an, bn, ... ln)

Power of a matrix

Suppose that it is possible to transform a matrix A to a diagonal matrix D. Then there is a matrix C such that 
 
    D = C-1.A.C  <=> A = C.D.C-1
Then

    An = C.D.C-1 . C.D.C-1 . C.D.C-1 . ... . C.D.C-1

    An = C.Dn.C-1
Since it is easy to calculate Dn, An can be calculated.
As an illustration the power of this result we'll find the formula for the n-th term of the fibonacci sequence starting from the recursive formula.

Fibonacci

The Fibonacci sequence is 1, 1, 2, 3, 5, 8, ...

The recursive formula is un = un-1 + un-2

Starting from this, we'll find the formula for the n-th term of the fibonacci sequence.

First we write un-1 + un-2 = un in matrix notation. 

 
    [ 1   1 ]   [ un-1]
    [ 1   0 ]   [ un-2]    =

    [  un  ]
    [ un-1 ]


Let M =
        [ 1   1 ]
        [ 1   0 ]
and
Let F =
        [1]
        [1]



Then
     [u3]
     [  ]     = M. F
     [u2]


Then [u4]       [u3]
     [  ]   = M.[  ]     =   M2. F
     [u3]       [u2]

...

Then [un  ]
     [    ]      =     Mn-2. F                (1)
     [un-1]
Now, we'll calculate Mn-2. To use the method from above, we need the eigenvalues and characteristic vectors connected with the matrix M.

The characteristic equation is r2 - r - 1 = 0 .

The eigenvalues are (1 + sqrt(5))/2 and (1 - sqrt(5))/2.
We call these eigenvalues respectively k and l.

Note that k - l = sqrt(5) and k.l = 1.

You'll find that (k,1) is a characteristic vector corresponding with k
and (l,1) is a characteristic vector corresponding with l.

If we choose these characteristic vectors as a new basis, then we have the connection between M and the diagonal matrix. 

 
    [k  0]
    [0  l]     =

          -1
    [k  l]               [k  l]
    [1  1]      . M .    [1  1]


This is equivalent with

    M =

    [k  l]  [k  0]   [k  l] -1
    [1  1]  [0  l]   [1  1]
From this we can calculate Mn-2 
 
    Mn-2 =


    [k  l]  [kn-2  0]   [k  l] -1
    [1  1]  [0  ln-2]   [1  1]


Now
    [k  l] -1
    [1  1]          =

    [1/(k-l)    -l/(k-l)]
    [-1/(k-l)    k/(k-l)]    =

    [1/sqrt(5)  -l/sqrt(5)]
    [-1/sqrt(5)  k/sqrt(5)]

Then, we have

   Mn-2 =

    [k  l]  [kn-2  0]  [1/sqrt(5)  -l/sqrt(5)]
    [1  1]  [0  ln-2]  [-1/sqrt(5)  k/sqrt(5)]

Writing only the first row of this product we have

    = (1/sqrt(5)) . [kn-1 - ln-1     -l.kn-1+ln-1.k ]

Now from (1) above we can write

  un =  (1/sqrt(5)) .( kn-1 - ln-1 -l.kn-1+ln-1.k )

<=>

  un =   (1/sqrt(5)) .(kn-1.(1-l) - ln-1.(1-k))

<=>

  un =   (1/sqrt(5)) .(kn - ln)


 with k = (1 + sqrt(5))/2  and l = (1 - sqrt(5))/2.
This is the formula for the n-th term of the fibonacci sequence.


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