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CHAPTER ONE
Linear Mapping (Linear Transformation)
1.1 Defintion: Let U, V be vector spaces over the same field K. A mapping
F : V −→ U
is called a linear mapping or linear transformation if it satisfies two conditions
(1) For any vector v, w ∈ V, F (v + w) = F (v) + F (w)
(2) For any scalar k and vector v ∈, F (kv) = kF (v)
Namely F:V−→ U is linear if it preserves the basic operations of a vector space. Note: Substituting k = 0 into condition (2) we obtain F(0) = 0. Therefore every linear mapping takes the zero vector into the zero vector. Alternative Definition: For any scalars a, b ∈K and any vectors v, w ∈V, if we have
F (av + bw) = aF (v) + bF (w)
It is worthy of note that the term linear transformation rather than the linear mapping is frequently used for linear mapping of the form
F : Rn^ −→ Rm
EXERCISE: Show that the two definitions of linear mappings above are equivalent.
1.2 Examples of Linear Mapping
(1) F: R^3 −→ R^3 by F(x, y, z) = (x, y, 0) (A projection into the xy-plane) (2) G: R^2 −→ R^2 by G(x, y) = (x + 1, y + 2) ( A translation) i
(3) Derivative Let V be the vector space of polynomials P(t) over the real field R, D: V −→ V be the derivative mapping d dt(u^ +^ v) =^
du dt and^
d dt(ku) =^ k
du dt for any u, v ∈P(t). (4) Integral Let Ψ: V −→ R be the integral mapping
Ψ(f (t)) =
∫ (^1) 0 f^ (t)dt (5) Zero Mapping: Let F:V−→U be the mapping that assigns the zero vector 0 ∈U to every vector v ∈V. Then for any vector v, w ∈ V and any scalar k ∈ K , we have
F (v + w) = F (u) + F (w)
(6) Identity Mapping: The mapping I: V−→V which maps each vector v ∈ V into itself.
1.3 Theorem: Let V and U be vector spaces over a field K. Let {v 1 , v 2 , ..., vn} be the basis of V and let {u 1 , u 2 , ..., un} be any vectors in U. Then there exists a unique linear mapping F : V −→ U such that F (v 1 ) = u 1 , F (v 2 ) = u 2 , ..., F (vn) = un
1.3.1 Remark: The theorem states that a linear mapping is completely determined by its values on the elements of a basis. 1.3.2 Illustration: Let F : R^2 −→ R^2 be the linear mapping for which F(1, 2) = (2, 3) and F(0, 1) = (1,
ii
which maps each vector v ∈ V into its coordinate vector [v]s is an isomorphism between V and Kn.
1.6 Kernel and Image of a Linear Mapping: Let F: V −→ U be a linear mapping. The Kernel of F written ker F is the set of elements in V that map into the zero vector 0 in U; that is
KerF = {v ∈ V : F (v) = 0 }
ImF = {u ∈ U : ∃ v ∈ V for which F (v) = u}
i.e the set of image points in U. 1.6.1 Theorem Let F: V −→ U be a linear mapping, then the Kernel of F is a subspace of V and Im F is a subspace of U. 1.6.2 Theorem Suppose v 1 , v 2 , ..., vn span a vector space V and suppose F: V −→ U is linear. Then
F (v 1 ), F (v 2 ), ..., F (vn) span Im F
1.7 Kernel and Image of Matrix Mapping Consider say a 3 x 4 matrix A and the usual basis {e 1 , e 2 , e 3 , e 4 } of K^4 (written as columns). That is
A =
a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 c 1 c 2 c 3 c 4
^ e^1 =
e 2 =
e 3 =
e 4 =
Recall that A may be viewed as linear mapping A : K^4 −→ K^3 , where the vectors in K^4 and K^3 are viewed as column vector. Now the usual basis vectors span K^4. Therefore, by the above theorem, their images Ae 1 , Ae 2 , Ae 3 , Ae 4 span the Image of A. But the vectors Ae 1 , Ae 2 , Ae 3 , Ae 4 are precisely the columns of A. That is
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Ae 1 = (a 1 b 1 c 1 )T^ , Ae 2 = (a 2 b 2 c 2 )T^ , Ae 3 = (a 3 b 3 c 3 )T^ , Ae 4 = (a 4 b 4 c 4 )T^. Thus the Image of A is precisely the Column Space of A. On the other hand, the Kernel of A consists of all vectors for which Av = 0. This means that the Kernel of A is the solution space of the homogeneous system
AX = 0 called the Null Space of A
1.8 Rank and Nullity of a linear Mapping Let F: V −→ U be a linear mapping. The rank of F is defined to be the dimension of its Image and the nullity of F is defined to be the dimensionof its kernel. Namely rank(F) = dim(Im F) nullity(F) = dim(Ker F) 1.8.1 Theorem Let V be of finite dimension and let F: V −→ U be linear. Then
dim V = dim(Ker F) + dim(Im F) = nullity(F) + rank(F)
Illustration: Let F: R^4 −→ R^3 be the linear mapping defined by F(x, y, z, t) = (x - y + z + t, 2x - 2y
- 3z + 4t, 3x - 3y +4z + 5t).
(a) Find a basis and the dimension of the Image of F; (b) Find a basis and dimension of the kernel of F (c) Verify the last theorem for F.
Solution To be given in class.
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1.10 Singular and Nonsingular Linear Mappings: ISOMORPHISMS Let F: V −→ U be a linear mapping. Recall that
F (0) = 0
A mapping F is said to be Singular if the image of some nonzero vector is 0. That is, if there ∃ v 6 = 0 such that F (v) = 0.
Thus F :V−→ U is nonsingular if the zero vector is the only vector whose image under F = 0. KerF = { 0 }
Theorem Let F :V−→ U be nonsingular linear mapping. Then the image of any linearly independent set is linearly independent.
Isomorphism proposition: A linear mapping F :V−→U is one-to-one iff F is nonsingular.
Theorem Suppose V has finite dimension and dim V = dim U. Suppose F:V−→U is linear. Then F is an isomorphism iff F is nonsingular.
Theorem Let V and U be vector spaces over field K. Then the collections of linear mappings from V into U with the following operations of addition and scalar multiplication form a vector space over K.
(1) (F + G)(v) = F(v) + G(v) (2) (kF)(v) = kF(v)
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The vector space of linear mappings in the above theorem is denoted by Hom(V, U)
Theorem Suppose dim V = m and dim(U) = n. Then
dim(Hom(V, U)) = mn
1.11 Composition of Linear Mappings Now suppose V, U, W are vector spaces over the same field K and suppose F: V−→U and G: U−→W are linear mappings The Composition Function G o F is the mapping from V into W defined by
(G o F)(v) = G(F(v)).
Note G o F is linear when F and G are linear.
1.12 Algebra A(V) of Linear Operators Let V be the vector space over a field K. Linear operator or linear transformation on V is a linear mapping from the vector space V into itself. That is linear mapping of the form F: V−→V. We write A(V) instead of Hom(V, V) for the space of all such mappings. 1.12.1 Square Matrices as Linear Operators Let M = Mn,n be the vector spaces of all square matrices over K. Then any matrix in M defines a linear mapping
FA : Kn^ −→ Kn^ by F (u) = Au
where the vectors in Kn^ are written as columns. Since the mapping is from Kn^ into itself, the square matrix A is a linear operator, not simply a linear mapping.
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CHAPTER ONE
2.1 Coordinate vector: Let S = {u 1 , u 2 , ..., un} be a basis of a vector space V over a field K. For any vector v ∈ V, suppose v = {c 1 u 1 + c 2 u 2 + ... + cnun}.
Then the coordinate vector of v relative S is represented and defined by a column vector
[v]S = [c 1 , c 1 , ..., cn]T
2.2 Matrix Representation of a Linear Operator: Let F be a linear operator from a vector space V into itself, and suppose S = {u 1 , u 2 , ..., un is a basis of V. Then F (u 1 ), F (u 2 ), ..., F (un) are vectors in V, and so each is a linear combination of the vectors in the basis S: that is
F (u 1 ) =c 11 u 1 + c 12 u 2 + ... + c 1 nun F (u 2 ) =c 21 u 1 + c 22 u 2 + ... + c 2 nun ........................................................ F (un) =cn 1 u 1 + cn 2 u 2 + ... + cnnun
Using matrix notation, we have
F
u 1 u 2 .... un
c 11 c 12 ... c 1 n c 21 c 22 ... c 2 n ... ... ... ... cn 1 cn 2 ... cnn
u 1 u 2 .... un
The matrix representation of T relative to the basis S, or simply the matrix of T in the basis S is the transpose of the matrix of coefficients above. Usually it is denoted by mS (T ) or [T ]S. That is
mS (F ) = [F ]S =
c 11 c 12 ... c 1 n c 21 c 22 ... c 2 n ... ... ... ... cn 1 cn 2 ... cnn
T
(6)
x
Note that the columns of mS (F ) are the coordinate vectors of F (u 1 ), F (u 2 ), ..., F (un) rspectively. 2.3 Algorithm For Finding Matrix Representations For each given basis vector ui in S carry out the following
(a) Find F (ui) (b) Write F (ui) as a linear combination of the basis vectors u 1 , u 2 , ..., un. (c) Form the matrix [F ]S whose columns are the coordinate vectors in step b above.
Illustration 1 Let F be the linear operator on R^3 defined by F(x, y, z) = (2y + z, x - 4y, 3x). Find the matrix representation of F relative to the basis
S = {u 1 , u 2 , u 3 } = {(1, 1 , 1), (1, 1 , 0), (1, 0 , 0)}
Solution: To be given in the class. Note: The matrix representation of any n ×n square matrix A over a field K relative to the usual basis E = {e 1 , e 2 , ..., en} of Kn^ is the matrix A itself; that is
[A]E = A Illustration 2 Find the matrix representation of F above relative to the usual basis {e 1 , e 2 , e 3 } = {(1, 0 , 0), (1, 0 , 0), (1, 0 , 0)} Solution: To be given in the class. Exercises
(a) Let D denotes the differential operator: that is D(f(t)) = df/dt. Find the matrix representing D for the basis { 1 , t, sin 3 t, cos 3 t} of the vector space V of functions. (b) Let V be the vector space of 2 × 2 matrices. Consider the following matrix M and usual basis E of V: M =
^ a^ b c d
and E = {
^1 0 0
,
^0 0 0
,
^0 0 1
,
^0 1 0
} (7)
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linear combination of the vectors in S; say
v 1 =c 11 u 1 + c 12 u 2 + ... + c 1 nun v 2 =c 21 u 1 + c 22 u 2 + ... + c 2 nun ........................................................ un =cn 1 u 1 + cn 2 u 2 + ... + cnnun
Let P be the transpose of the above matrix of coefficient; that is, P = [pij ], where pij = cij. Then P is called the Change-of-basis matrix from basis S to the basis S′. Remarks: Analogously there is a change of matrix Q from the basis S′ to the basis S.
Proposition: Let P and Q be the above change of basis marices. Then
Q = P −^1
Now suppose S = {u 1 , u 2 , ..., un} is a basis of a vector V, suppose [pij ]is any NONSIN- GULAR matrix. Then the n vectors
vi = p 1 iv 1 + p 2 iv 2 + ... + pnivn i = 1, 2 , ...
corresponding to the columns of P are linearly independent. They form another basis S′. Moreover, P will be the change-of- basis matrix from S to the basis S′. Illustration: Consider the following two bases of R^2 S = {u 1 , u 2 } = {(1, 2), (3, 5)} S′ = {v 1 , v 2 } = {(1, −1), (1, −2)}
(a) Find the change of basis matrix P from S to the S′ (b) Find the change of basis matrix Q from the S′ back to the basis S. (c) Show that Q = P −^1 xiii
Solution To be given in the class. exercise E = {e 1 , e 2 , e 3 } = {(1, 0 , 0), (0, 1 , 0), (0, 0 , 1)} S = {u 1 , u 2 , u 3 } = {(1, 0 , 1), (2, 1 , 2), (1, 2 , 2)}
Proposition: Let the change of basis matrix from the usual basis E of Kn^ for any basis S of Kn^ is the matrix P whose columns are respectively the basis vectors of S.
Applications of Change-of-Basis Matrix
(1) Effects on Coordinates of vector in a space V
Theorem Let P be the change-of-basis matrix froma basis S to a basis S′ in a vector space V. Then for any vector v ∈,V we have P [v]S′^ = [v]S and hence P −^1 [v]S = [v]S′ Namely, if we multiply the coordinates of V in the original basis S by P −^1 , we get the coordinates of V in the basis S′. (2) Effects on the matrix representation of a linear operator.
Theorem D Let P be the Change-of-basis matrix from a basis S to a basis S′ in a vector space V. Then for any linear operator T on V. [T ]S′^ = P −^1 [T ]S P Thus, if A and B are the matrix representation of T relative, respectively to S and S′ then B = P −^1 AP xiv
Diagonalizable Operator A linear operator T is said to be diagonalizable if there exists a basis of V such that T is represented by a diagonal matrix: the basis is then said to diagonalize T.
Theorem Let A be the matrix representation of a linear operator T. Then T is diagonalizable if and only if there exists an invertible natrix P such that P −^1 AP is a diagonal matrix.
Matrices and Linear Mapping Suppose V and U are vector spaces over the same field K and say dim V = m and dim U = n. Further more, suppose
S = {v 1 , v 2 , ..., vm} and S′ = {u 1 , u 2 , ..., un}
are arbitrary but fixed bases repectively of V and U. Suppose F :V−→ U is a linear mapping. Then the vectors
F (v 1 ), F (v 2 ), ..., F (vm)
belong to U and so each is a linear combination of the basis vectors in S′ : say
F (v 1 ) =c 11 u 1 + c 12 u 2 + ... + c 1 nun F (v 2 ) =c 21 u 1 + c 22 u 2 + ... + c 2 nun ........................................................ F (vm) =cn 1 u 1 + cn 2 u 2 + ... + cmnun
The transpose of the above matrix of coefficients denoted by mS,S′^ (F ) or FS,S′^ is called the matrix representation of F relative to the bases S and S′. Theorem For any vector v ∈ V [F ]S,S′^ [v]S = [F (v)]S′
That is, multiplying the coordinate of V in the basis S of V by [F ]S,S′ , we obtain the coordinate of F(v) in the basis S′ of U.
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Theorem Let P be the Change-of-basis matrix from a basis e to a basis e′ in V, and let Q be the Change-of-basis of basis matrix from a basis f to basis f ′ in U. Then for any linear mapping F : V−→ U [F ]e′^ ,f ′^ = Q−^1 [F ]e,f P
The last theorem shows that any linear mapping from one vector space V into another vector space U can be represented by a simple matrix.
Theorem Let F :V−→ U be linear and, say rank (F) = r. Then there exist bases of V and U such that the matrix representation of F has the form
A =
^ Ir^0 0 0
(12)
where Ir is the r-square identity matrix.
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(3) If A has two identical rows (columns), then
|A| = 0
(4) If A is triangular i.e A has zeros above or below the diagonal, then |A| = product of diagonal elements. Then in particular |I| = 1 , where I is the identity matrix.
(5) If two rows (columns) of A were interchanged to obtain B, then
|B| = −|A|
(6) If a row (columns) of A were multiplied by a scalar k to obtain to obtain B, then
|B| = k|A|
(6) If a multiple of a row (column) of A were added to another row (columns) of A to obtain B, then |B| = |A|
(7) The determinant of a product of two matrices A and B is the product of their determinants: that is det(AB) = det(A)det(B)
(8) Let A be a square matrix. Then the following are equivalent
(i) A is invertible, that is A has an inverse A−^1 (ii) AX = 0 has only the zero solution (iii) The determinant of A is not zero that is
det(A) 6 = 0
(9) Suppose A and B are similar matrice, then
|A| = |B|
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Minors And Cofactors Consider an n-square A = [aij ]. Let Mij denote the (n - 1) square submatrix of A obtained by deleting its ith^ and jth^ column. The determinant |Mij | is called the minor of the element aij of A and we define the cofactor of aij denoted by Aij , to be the signed minor. Aij = (−1)i+j^ |Mij |
Note that signs (−1)i+j
Illustration Let
A =
^ (17)
Find the following minors and cofactors
(a) |M 23 | and A 23 (b) |M 31 | and A 31
Laplacian Equation The determinant of a square matrix A = [aij ] is equal to the sum of the products obtained by multiplying the elements of any row (column) by their respective cofactors.
|A| = ∑^ n j=
= aij Aij
|A| = ∑^ n i=1^ =^ aij^ Aij xx
