sohum/ma330/sp12/lectures/lec4.pdf · ImportantObservations. 1 LetA beamatrixwithrownum(A) = m andcolnum(A) = n. LetL = T A. Thenrank(A) ≤min(m,n). 2 Ifn > m,thenrank(A)

Sections 1.8, 1.9

Ma 322 Spring 2011

Ma 322

Jan. 31, Feb.1-4

Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 1 / 14

Summary

Linear Transformations from <n to <m.Matrix of a given transformation.Subspaces associated to a Linear Transformation.Injective (one-to-one) and Surjective (onto) transformations.Transformation with desired properties.


Summary



Summary



Summary



Summary



Summary



Summary



A Linear Transformation.

A Linear Transformation extends the idea of a function sothat the domain is <n rather than just the field of realnumbers.The word “Linear” also means that it has the simplestpossible formula consistent of ordinary linear functions.Def.1: A Linear Transformation is a map L : <n → <m

satisfying two properties:1 L(v + w) = L(v) + L(w) for all v, w ∈ <n and2 L(cv) = cL(v) for all v ∈ <n and c ∈ <.

The map defined by L((

xy

)) =

x + yx − y

2x + 3y

defines a

Linear Transformation from <2 to <3.






xy

)) =

x + yx − y

2x + 3y

defines a







xy

)) =

x + yx − y

2x + 3y

defines a







xy

)) =

x + yx − y

2x + 3y

defines a







xy

)) =

x + yx − y

2x + 3y

defines a







xy

)) =

x + yx − y

2x + 3y

defines a







xy

)) =

x + yx − y

2x + 3y

defines a



Verification of the definition.This is checked from definition thus:Let v =

(v1v2

)and w =

(w1w2

). Then L(v + w) equals:

L((

v1 + w1v2 + w2

)) =

v1 + w1 + v2 + w2v1 + w1 − v2 − w2

2v1 + 2w1 + 3v2 + 3w2

which simplifies: v1 + v2

v1 − v22v1 + 3v2

+

w1 + w2w1 − w2

2w1 + 3w2

= L(v) + L(w).

L(cv) = L((

cv1cv2

)) =

cv1 + cv2cv1 − cv2

2cv1 + 3cv2

= cL(v).



(v1v2

)and w =

(w1w2


L((

v1 + w1v2 + w2

)) =

v1 + w1 + v2 + w2v1 + w1 − v2 − w2

2v1 + 2w1 + 3v2 + 3w2


v1 − v22v1 + 3v2

+

w1 + w2w1 − w2

2w1 + 3w2

= L(v) + L(w).

L(cv) = L((

cv1cv2

)) =


2cv1 + 3cv2

= cL(v).



(v1v2

)and w =

(w1w2


L((

v1 + w1v2 + w2

)) =

v1 + w1 + v2 + w2v1 + w1 − v2 − w2

2v1 + 2w1 + 3v2 + 3w2


v1 − v22v1 + 3v2

+

w1 + w2w1 − w2

2w1 + 3w2

= L(v) + L(w).

L(cv) = L((

cv1cv2

)) =


2cv1 + 3cv2

= cL(v).



(v1v2

)and w =

(w1w2


L((

v1 + w1v2 + w2

)) =

v1 + w1 + v2 + w2v1 + w1 − v2 − w2

2v1 + 2w1 + 3v2 + 3w2


v1 − v22v1 + 3v2

+

w1 + w2w1 − w2

2w1 + 3w2

= L(v) + L(w).

L(cv) = L((

cv1cv2

)) =


2cv1 + 3cv2

= cL(v).



(v1v2

)and w =

(w1w2


L((

v1 + w1v2 + w2

)) =

v1 + w1 + v2 + w2v1 + w1 − v2 − w2

2v1 + 2w1 + 3v2 + 3w2


v1 − v22v1 + 3v2

+

w1 + w2w1 − w2

2w1 + 3w2

= L(v) + L(w).

L(cv) = L((

cv1cv2

)) =


2cv1 + 3cv2

= cL(v).


When does the definition fail?

The reason that the calculations work is that the formulasare homogeneous linear expressions in the coordinates of thevectors in <n.Thus

S((

xy

)) =

x + y + 1x − y

2x + 3y

and T ((

xy

)) =

x + yxy

2x + 3y

both fail the definition.This should be checked.




S((

xy

)) =

x + y + 1x − y

2x + 3y

and T ((

xy

)) =

x + yxy

2x + 3y





S((

xy

)) =

x + y + 1x − y

2x + 3y

and T ((

xy

)) =

x + yxy

2x + 3y





S((

xy

)) =

x + y + 1x − y

2x + 3y

and T ((

xy

)) =

x + yxy

2x + 3y



Matrix of a Linear Transformation.

Here is an example of a map guaranteed to give a LinearTransformation. Let A be a matrix with real entries havingm rows and n columns.Def.2: The transformation TA.Define the map TA : <n → <m by the formula TA(X) = AX .Then the following calculation shows that TA is a LinearTransformation.

TA(v + w) = A(v + w) = Av + Aw = TA(v) + TA(w)

andTA(cv) = A(cv) = cAv = cTA(v).

This can be called as the Linear Transformation defined by A.




















How to find the Matrix of a LinearTransformation?

We first need some notation.Notation: Define a vector en

i to be a column with n entrieswhich are all zero except the i-th entry is 1.Thus for n = 3 we have:

e31 =

100

, e32 =

010

, e33 =

001

.

While working with <n for a fixed n, we often drop thesuperscript n to simplify our display.


How to find the Matrix of a LinearTransformation?

We first need some notation.Notation: Define a vector en

i to be a column with n entrieswhich are all zero except the i-th entry is 1.Thus for n = 3 we have:

e31 =

100

, e32 =

010

, e33 =

001

.

While working with <n for a fixed n, we often drop thesuperscript n to simplify our display.


The matrix calculated.

Given a map L : <n → <m, we calculate the n columns

v1 = L(en1 ), v2 = L(en

2 ), · · · , vn = L(enn).

Let A be the matrix with n columns v1, v2, · · · , vn in order.Then the theorem is: L is a Linear Transformation iffL(X) = AX for all X ∈ <n.In other words, L = TA.




v1 = L(en1 ), v2 = L(en

2 ), · · · , vn = L(enn).





v1 = L(en1 ), v2 = L(en

2 ), · · · , vn = L(enn).





v1 = L(en1 ), v2 = L(en

2 ), · · · , vn = L(enn).





v1 = L(en1 ), v2 = L(en

2 ), · · · , vn = L(enn).



Spaces associated with a Linear Transformation..

Given any matrix A with m rows and n columns, we havetwo natural sets associated with it.Def.3: Column and Null spaces of a Matrix.

1 Col(A) = {AX | X ∈ <n}.2 Nul(A) = {X | AX = 0 and X ∈ <n}.

We now consider a Linear Transformation L : <n → <m.Def.4: Kernel and Image of a Linear Transformation.

1 Image(L) = {L(X) | X ∈ <n}.2 Ker(L) = {X | L(X) = 0 and X ∈ <n}.

We shall later define subspaces and show that Col A is asubspace of <m and Nul A is a subspace of <n.


























































Relationship with properties of a LinearTransformation.

Def.5: Injective or 0ne-to-one transformation. Afunction is said to be one-to-one, if it maps different elementsto different elements. For a linear transformation L, it isenough to check L(v) 6= 0 if v 6= 0.In other words if Ker(L) contains only the zero vector. In thiscase, we call the transformation to be injective or one-to-one.Def.6: Surjective or onto transformation. A function issaid to be onto if every element of the target space is animage of some element.For a linear transformation L : <n → <m, the target space is<m and thus the condition reduces to Image(L) = <m. Inthis case, we call the transformation to be surjective or onto.











The criteria for Injectivity and Surjectivity.

When L = TA, we know that Ker(L) = Nul(A) and so wehave L = TA is injective iff Nul(A) = 0, orrank(A) = colnum(A) = n.When L = TA, we also know that Image(L) = Col(A) and sowe have L = TA is surjective iff Col(A) = <m, orrank(A) = rownum(A) = m.A Fundamental Fact: The rank of a matrix with m rows andn columns: It is obvious that rank(A) is less than or equal tomin(m, n).This gives some easy but important conclusions which arewell worth memorizing!

















Important Observations.

1 Let A be a matrix with rownum(A) = m and colnum(A) = n.Let L = TA. Then rank(A) ≤ min(m, n).

2 If n > m, then rank(A) < n = colnum(A) and hence Nul(A)is non zero. Consequently, Ker(L) 6= 0 and L cannot beinjective.

3 If m > n, then rank(A) < m = rownum(A) and hence Col(A)is smaller than <m. Consequently, Image(L) 6= <m and henceL cannot be surjective.

4 If n = m then rank(A) may be equal to this common value ormay be smaller.We discuss this next.
































Observations continued.

1 If rank(A) equals this common value n = m, then the map Lis both surjective and injective.Def.7: Isomorphism. The Linear transformation L is saidto be an isomorphism if it is both surjective and injective. Inthis case, it is a one-t-one onto map from <n to <n.

2 If rank(A) < n = colnum(A), then L = TA is not injective.3 If rank(A) < m = rownum(A), then L = TA is not surjective.


















Creating a Suitable Linear Transformation.We now show how to use the matrix of a transformation tocreate a Linear Transformation with a desired property.Suppose we want L : <3 → <3 to rotate all vectors about thez-axis.Recall that the vectors e3

1, e31, e3

1 are along the x , y, z axesrespectively.It is clear that we want

L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.

Thus L = TA where A =

0 −1 01 0 00 0 1

.Now, we can find the rotated image of any desired vector, say

L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.



1, e31, e3


L(e31) = e3

2, L(e32) = −e3

1, L(e33) = e3

3.


0 −1 01 0 00 0 1


L

1

11

=

0 −1 01 0 00 0 1

1

11

=

−111

.


Documents

sohum/ma330/sp12/lectures/lec4.pdf · ImportantObservations. 1 LetA beamatrixwithrownum(A) = m andcolnum(A) = n. LetL = T A. Thenrank(A) ≤min(m,n). 2 Ifn > m,thenrank(A)