Lec7.pdf

Lecture 7: Linear Transformationa and their matrices Examples of linear transformations. (1) Rotation by π/2 in R2 followed by reflection about the x-axis. Remember that rotation by π/2 is given by     x −y R = y x and reflection about the x-axis is accomplished by the linear transformation     x x S = y −y One followed by the other is then the linear transformation            x x x −y −y T = (S ◦ R) =S R =S = y y y x −x (2) Rotation in R2 by an angle α counterclockwise     x x cos α − y sin α T = y y cos α + x sin α (3) Integration. V = Pn , W = Pn+1 . Define T : V −→ W Z T (p) = p(x) dx.

This is a linear transformation because of Z Z Z (p(x) + q(x))dx = p(x) dx + q(x) dx Z Z αp(x) dx = α p(x) dx if α ∈ R. (4) Shifting the graph of a function to the right. Let V = W = Pn and define T : V −→ W by (T p)(x) = p(x − 1) 2

For example if p(x) = x +3 then T p is the polynomial (T p)(x) = (x−1)2 +3. This is a linear transformation: (T (p + q))(x)

= =

(p + q)(x − 1) p(x − 1) + q(x − 1)

=

(T p)(x) + (T q)(x)

and (T (αp))(x) = (αp)(x − 1) = αp(x − 1) = α(T p)(x). Convince yourself that moving a graph up or down (T p)(x) = p(x) + 1 is not a linear transformation. 1

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(5) Taking the transpose of a matrix. Let V = W = M2×2 and define a transformation T : M2×2 −→ M2×2 by T (A) = AT , i.e.     a b a c T = c d b d This is a linear transformation because of    ′    a b a b′ a + a′ b + b ′ T + = T c d c′ d′ c + c′ d + d′   a + a′ c + c′ = b + b′ d + d′    ′  a c a c′ = + b d b′ d′    ′  a b a b′ = T +T c d c′ d′ and   a T α c

b d



= = = =



 αb T αd   αa αc αb αd   a c α b d   a b αT c d αa αc

Finding the matrix of a Linear Transformation between vector spaces. We begin the discussion by considering a linear transformation T : Rn → Rm ,   x1   where Rn consists as usual of all column vectors x =  ...  with real entries. In

xn the textbook they sometimes write (x1 , . . . , xn )T (to save space). We assume that ej are the standard basis vectors (the only nonzero entry being ’1’ in row #j) so that x = x1 e1 + · · · xn en . We write 

  T (e1 ) = a1 =  

a11 a21 .. . am1





     , T (e2 ) = a2 =   

and we organize all the numbers aij  a11 a12  a21 a22  A= . ..  .. . am1

am2

a12 a22 .. . am2





     , . . . , T (en ) = an =   

into a matrix  · · · a1n · · · a2n   ..  = [a1 , . . . , an ]. ··· .  · · · amn

a1n a2n .. . amn



  , 

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Then, since T is linear, we calculate T (x) = T (x1 e1 + · · · xn en ) = x1 T (e1 ) + · · · xn T (en ) = x1 a1 + · · · xn an = A·x recalling the definition of matrix-vector multiplication. Summarizing, In order to find the matrix A of a linear transformation T : Rn → Rm with respect to the standard bases on Rn and Rm we compute the vectors       a11 a12 a1n  a21   a22   a2n        T (e1 ) =  .  , T (e2 ) =  .  , . . . , T (en ) =  .  ,  ..   ..   ..  am1

am2

and collect them together in a matrix  a11  a21  A= .  .. am1

a12 a22 .. . am2

amn

··· ··· ··· ···

a1n a2n .. . amn



  . 

We will illustrate this on some of the previous examples.

Example (1): Rotation by π/2 in R2 followed by reflection about the x-axis. We get         1 0 0 −1 T = ,T = 0 −1 1 0 and the matrix of the transformation is given by   0 −1 −1 0 Example (2): Rotation in R2 by an angle α counterclockwise. We get         1 cos α 0 − sin α T = ,T = 0 sin α 1 cos α and the matrix of the transformation is given by   cos α − sin α sin α cos α For further examples we need to discuss how to find the matrix of a linear transformation between general vector spaces. Assume that V, W are vector spaces with bases e1 , . . . , en and f1 , . . . , fm respectively. Moreover, let T : V −→ W be a linear transformation. The bottomline is that after choosing these bases, the linear transformation T can be described by an (m × n)-matrix.

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Each vector T (ek ), 1 ≤ k ≤ n, can be written as a linear combination of the basis vectors f1 , . . . , fm , i.e. there are scalars ajk with 1 ≤ j ≤ m, 1 ≤ k ≤ n such that m X ajk fj (1) T (ek ) = j=1

(the double indices ajk are just a convenient way to keep track of the coefficients). Let now v ∈ V be a vector in V . We can represent it in terms of the basis vectors as follows: v = v1 e1 + · · · + vn en . We then compute ! n X T (v) = T vk ek k=1

= = = =

n X

T (vk ek )

k=1 n X

vk T (ek )

k=1 n X m X

ajk vk fj

k=1 j=1 m X

(aj1 v1 + · · · + ajn vn )fj

j=1

Using the bases of V and W we will write vectors in V and W as column vectors as follows: We will associate a column vector   v1  ..  n  . ∈R vn

with a vector v = v1 e1 + · · · + vn en in V and similarly we will associate a column vector   w1  ..  m  . ∈R wm

with a vector w = w1 f1 + · · · + wm fm ∈ W . We can then write        v1 a11 v1 + · · · + a1n vn v1 a11 · · · a1n      .. .. ..   ..  T  ...  =  = . . .  .  vn am1 v1 + · · · + amn vn vn am1 · · · amn

Summarizing, we need to compute T (ek ), relate them to the basis vectors fk , and formula (1) yields the matrix of T .

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Here are some examples: Reflection about the x-axis in R2 We choose V = W = R2 but we choose the following basis instead of the standard basis     1 1 h1 = , h2 = 0 1 In the previous notation we have n = m = 2 and ej = fj = hj . Let T : R2 → R2 be the linear transformation which is reflection about the x-axis, i.e.        1 0 x1 x1 x1 = = T x2 0 −1 −x2 x2

The above matrix is the matrix of T with respect to the standard basis. We want to compute the matrix of T with respect to the basis h1 , h2 . We calculate     1 1 T (h1 ) = T = = h1 0 0 and



       1 1 −1 2 = = + = −h2 + 2h1 1 −1 −1 0 Comparing with formula (1) the matrix of T with respect to the basis h1 , h2 is given by   1 2 . 0 −1 Integration of polynomials Let V = P2 , W = P3 . Define T : V Z−→ W T (h2 ) = T

T (p) =

p(x) dx.

We pick the following bases for P2 and P3 :

{e1 , e2 , e3 } = {1, x, x2 } ⊂ P2 and {f1 , f2 , f3 , f4 } = {1, x, x2 , x3 } ⊂ P3 . We get by integrating 1 1 1 1 2 x = f3 , T (e3 ) = x3 = f4 2 2 3 3 Identifying all vectors with column vectors this amounts to            0 0 0 1 0 0  1   0   0    1 =  0 = T 0 =  0 ,T  1/2  , T  0 0 0 1 0 0 1/3 T (e1 ) = x = f2 , T (e2 ) =

   

and the matrix of T with respect to the bases {e1 , e2 , e3 } and {f1 , f2 , f3 , f4 } is given by   0 0 0  1 0 0     0 1/2 0  . 0 0 1/3 Note that different choices of bases lead to a different matrix.