sohum/ma330/sp12/lectures/lec4.pdf · ImportantObservations. 1 LetA beamatrixwithrownum(A) = m...
Transcript of sohum/ma330/sp12/lectures/lec4.pdf · ImportantObservations. 1 LetA beamatrixwithrownum(A) = m...
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 2 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 3 / 14
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).
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 4 / 14
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).
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 4 / 14
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).
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 4 / 14
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).
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 4 / 14
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).
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 4 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 5 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 5 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 5 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 5 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 6 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 6 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 6 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 6 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 7 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 7 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 8 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 8 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 8 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 8 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 8 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 9 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 10 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 10 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 10 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 10 / 14
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!
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 11 / 14
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!
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 11 / 14
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!
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 11 / 14
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!
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 11 / 14
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!
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 11 / 14
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!
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 11 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 12 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 12 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 12 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 12 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 12 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 12 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 13 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 13 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 13 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 13 / 14
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.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 13 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14
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
.
Avinash Sathaye (Ma 322) Summary Week 4 Jan. 31, Feb.1-4 14 / 14