LECTURE 7: LINEAR TRANSFORMATION (CHAPTER 4 IN THE BOOK)

LECTURE 7: LINEAR TRANSFORMATION (CHAPTER 4 IN THE BOOK) Everything marked by is not required by the course syllabus In this lecture, F is a fixed field and all vector spcaes are over F. One can assume F = R or C. 1.1. Linear maps. 1. Definitions Definition 1.1. A mapping L from a vector space V into a vector space W is said to be a linear transformation (or linear map or linear homomorphism) if for all v 1, v 2 V and for all scalars α and β. Remark: L(αv 1 + βv 2 ) = αl(v 1 ) + βl(v 2 ) 1. A map L from V to W could be denoted by L V W. 2. Now from the definiotion, L is a linear map is equivalent to say (a) L(v 1 + v 2 ) = L(v 1 ) + L(v 2 ) and (b) L(αv) = α(v). 3. If L V W is a linear map, the we have (a) L(0 V ) = 0 W. (b) L(α 1 v 1 +α 2 v 2 + +α k v k ) = α 1 L(v 1 )+α 2 L(v 2 )+ +α k L(v k ) (the image of a linear combination of some vectors is a linear combination of the image of these vectors) (c) L( v) = L(v). 4. A linear map L V V (from V to V itself) is called a linear operator on V or a linear endomorphism of V Example 1.2. (1) The idenity map L V V defined by L(v) = v. (2) The map L V V defined by L(v) = 2v is a linear map/ linear operator on V (this a scaler multiplication). (3) The map L V V defined by L(v) = v + a is not a linear operation on V unless a is the zero vector 0. (4) Rotation let R θ R 2 R 2 be the map that send a vector in 2d plan to its rotation by degree θ counterclockwise. Then R θ ((x, y) T ) = (x cos θ y sin θ, x sin θ + y cos θ) = ( cos θ sin θ sin θ cos θ ) (x y ) 1

2 LECTURE 7: LINEAR TRANSFORMATION (5) In general, let A be an m n matrix. The map L A defined by L A F n F m x Ax is a linear map from F n to F m. The claim follows form the arthematics of matrices: L A (αx 1 + βx 2 ) =A(αx 1 + βx 2 ) = A(αx 1 ) + A(βx 2 ) =αax 1 + βax 2 = αl A (x 1 ) + βl A (x 2 ). (6) Recall that F [x] is the space of polynomials with coefficient in F. Let p = a n x n + a n 1 x n 1 + +a 1 x+a 0 be a polynomial in F [x]. The dirivative map D F [x] F [x] defined by D(p) = d dx p = na nx n 1 + (n 1)a n 1 x n 2 + + a 1 is a linear map. Similarly, we have indefinite integration is a linear map I F [x] F [x] defined by 1 I(p) = p dx = n + 1 a nx n+1 + 1 n a n 1x n + + 1 2 a 1x 2 + a 0 x. The following examples are about function spaces, you can skip it as your wish. (7) Recall that C k ([a, b]) is the space of conitnouns differencaible functions of order k on the intervial [a, b]. Then D C n ([a, b]) C n 1 ([a, b]) defined by D(f) = d dxf is a linear map. (8) Now consider the integrations: L C[a, b] R defiend by L(f) = a b f(x)dx. is a linear transformation. (9) Fix a point x 0 [a, b] the evalutation map eval x0 C([a, b]) R given by eval x0 (f) = f(x 0 ) is a linear map. (10) Definite integration gives linear maps: L C([a, b]) R defined by L(f) = b a f(x)dx. I C([a, b]) C 1 ([a, b]) defined by (I(f))(t) = t a f(x) d dx. 1.2. Kernel and image. Definition 1.3. Let L V W be a linear map. The kernel of L, denoted ker(l), is defined by ker(l) = { v V L(v) = 0 W }. Definition 1.4. Let L V W be a linear map and let S be a subspace of V. The image of S, denoted L(S), is defined by L(S) = { w W w = L(v) for some v S } The image of the entire vector space, image(l) = L(V ), is called the range or image of L. Lemma 1.5. If L V W is a linear transformation and S is a subspace of V, then (i) ker(l) is a subspace of V. (ii) L(S) is a subspace of W.

LECTURE 7: LINEAR TRANSFORMATION 3 Proof. (1) It is obvious that ker(l) is nonempty since 0 V ker(l). Now we show that ker(l) is closed under scalar multiplication and addition. Let v, v 1, v 2 ker(l) and let α F. Closure under scalar multiplication: Therefore, αv ker(l). For closure under addition: L(αv) = αl(v) = α0 W = 0 W. L(v 1 + v 2 ) = L(v 1 ) + L(v 2 ) = 0 W + 0 W = 0 W Therefore, v 1 + v 2 ker(l) and hence ker(l) is a subspace of V. (2) L(S) is nonempty, since 0 W = L(0 V ) L(S). For closure under scalar multiplication If w L(S), then w = L(v) for some v S. For any scalar α, αw = αl(v) = L(αv) Since αv S, it follows that αw L(S), and hence L(S) is closed under scalar multiplication. For closur under addition If w 1, w 2 L(S), then there exist v 1, v 2 S such that L(v 1 ) = w 1 and L(v 2 ) = w 2. Thus, w 1 + w 2 = L(v 1 ) + L(v 2 ) = L(v 1 + v 2 ). and hence L(S) is closed under addition. It follows that L(S) is a subspace of W. Remark: One can visualize the relationship between these spaces by following diagram which is called a short exact sequance. (Here 0 reprensents the zero verctor space.) 0 ker(l) V L(V ) 0. Example 1.6. (1) Let A be an m n and consider the map L A F n F m defined by L A (x) = Ax. Then kernel ker(l A ) of L A is the solution space of the equation Ax = 0. The image L A (V ) of L is the column space of A. (2) Recall that P n = { p F [x] p has degree less than or equal to n }. Recall dirivative operator D P n P n. We have ker(d) = P 0, i.e. the space of constant polynomials and D(P n ) = P n 1. 1.3. Quick Review about maps. Recall that a map between two sets, say f A B is called an injection or injective if for each element b B there is at most one a A such that f(a) = b. f is called a surjection or surjective if f(a) = B. If f is both injective and surjective, it is called a bijection or bijective. For a bijection f A B, each element in a A is corresponding to a unique element b B. Therefore, we can define the inverse map f 1 B A by f(a) a. The inverse map f 1 is characterised by the property that it is the unique map, say h, such that h f = id A where id A A A is the identity map given by a a. Two maps f A B and g B C could be compose to form a map g f A C by a g(f(a)). 1.4. Linear isomorphism. Lemma 1.7. Let L V W be a linear map. If ker(l) = 0 if and only if L is a injection. Proof. ( ) Suppose ker(l) = 0. Suppose there is two elements v 1 v 2 V such that L(v 1 ) = L(v 2 ) = w for some w W. Hence L(v 1 v 2 ) = 0 W, i.e. 0 V v 1 v 2 ker(l). This is contradict to ker(l) = 0. ( ) Suppose that L is an injection, we claim that ker(l) is the zero space. We prove by contradiction, suppose ker(l) is not zero, then there is a nonzero vector v ker(l).

4 LECTURE 7: LINEAR TRANSFORMATION We have L(v) = 0 W v 0 V. = L(0 V ). This contradict to that L is an injection, since Definition 1.8. A linear map L V W is called a linear isomorphism if ker(l) = 0 and L(V ) = W. We could denote V L W or V W. Suppose L V W is a linear isomorphism then it is a bijection. Its inverse L 1 W V is well defined and also a linear isomorphism. Exercise (Spaces which are linear isomorphic to each other have the same dimnsion). Suppose L V W is a linear isomorphism, then dim V = dim W. (Hint: check that L sends any basis of V into linearly independent set of W, hence dim V dim W ; apply the same argument to L 1 W W, get dim W dim V ; down) Example 1.9 (an important example). Now let V be a finite dimensional vector space. Suppose V has dimension n and S = { v 1,, v n } is a basis of V. Then the map send each vector v in V to its coordinate [v] S with respect to the basis S is give a linear isomorphism form V to F n. Similarly, the inverse of the above map is also an isomorphism which is given by (x 1, x 2,, x n ) x 1 v 1 + x 2 v 2 + + x n v n. In summary, the example says, V F n if V is has dimension n. Roughly speaking, if two vector spaces V and W are isomorphic, one could think them are the same in many cases. However isomorphisms are not unique. (For example, the isomorphism V F n are all different for different basis.) Notation. Let V be an n-dimensional vector space and S = { v 1, v 2,, v n }. Let ι S F n V denote the isomorphism from F n to V given by the basis S. More precisely, and ι S ((x 1,, x n )) = x 1 v 1 + +x 2 v 2 + + x n v n ι 1 S (v) = [v] S for v V. for (x 1, x 2,, x n ) F n Example 1.10. Consider the basis S = { 1, x, x 2,, x n } of P n. Then ι S ((a 0, a 1,, a n )) =a 0 + a 1 x + + a n x n P n ι 1 S (a 0 + a 1 x + + a n x n ) =(a 0, a 1,, a n ) F n+1 gives the isomorphism between F n+1 and P n. Example 1.11 (Change of basis: from standard basis). Consider the case V = F n and a basis S = { v 1, v 2,, v n } as column vectors. Define matrix (by abuse of notation) S = (v 1 v 2 v n ). Then, for x = (x 1,, x n ) F n and v V = F n, we have ι S (x) = Sx and [v] S = ι 1 S (v) = S 1 v. (Note that S is an invertible matrix since { v 1,, v n } is a basis!)

1.5. Composition of linear maps. LECTURE 7: LINEAR TRANSFORMATION 5 Definition 1.12. Suppose L V W and T W U are two linear maps. Define the composition T L V U of L and T by T L(v) = L(T (v)). Exercise. Check that T L is a linear map. Notation (Commutative diagram ). Let Q V W be a linear map. Suppose Q = T L then we say the following diagram is commutative. V L U Q Example 1.13. Let L A F n F m and L B F m F t be two maps represented by matrices A and B respectivly. Then L B L A = L BA. T W In fact, L B L A (x) = L B (L A (x)) = L B (Ax) = B(Ax) = (BA)x = L BA (x). 2. Matrix Representations of Linear Transformations The following theorem says, all linear maps from F n to F m is given by a matrix. Theorem 2.1. If L F n F m is a linear map from F n to F m, then there is an m n matrix A such that L(x) = Ax for each x F n. In fact, A = (a 1 a 2 a n ) where the j-th column a j is given by (e j s from the standard basis) a j = L(e j ). Proof. This is clear in fact. Write each vector x = (x 1,, x n ) T in F n interms of standard baisis: x = x 1 e 1 + x 2 e 2 + + x n e n Now L(x) =x 1 L(e 1 ) + x 2 L(e 2 ) + + x n L(e n ) =x 1 a 1 + x 2 a 2 + + x n a j x 1 x 2 = (a 1 a 2 a n ) =Ax Notation. We will identify an n m matrix with the corresponding linear map L A F n F m by abuse of notation. Theorem 2.2 (Matrix Representation Theorem). If S = { v 1, v 2,, v n } and T = { w 1, w 2, w m } are ordered bases for vector spaces V and W, respectively, then corresponding to each linear transformation L V W, there is an m n matrix A such that x n [L(v)] T = A[v] S for each v V.

6 LECTURE 7: LINEAR TRANSFORMATION Proof. Consider the diagram: V ι S L W ι T F n L A F m Similarly, let ι T F n W corresponding to T. Then ι 1 T by certain matrix A. Now [L(v)] T = ι 1 T (L(v)) = (ι 1 T L ι S )(ι 1 S (v)) = A[v] S. L ι S F n F m is represented Definition 2.3. Under the notation of above theorem. We will let [L] S,T matrix A representing L under the basises S and T. denote the 2.1. Changing of basis down right. Let V be a vector spaces of dimension n. S = { v 1,, v n } and T = { u 1,, u 2 }. The change of basis matrix is viewed via the below diagram: V ι S F n id [id] S,T V F n ι T Here [id] S,T is represented by the change of basis matrix. Clearly we have [id] T,S = ([id] S,T ) 1 When V = F n and v i, u j s as column vectors and matrices (abuse notation) S = (v 1 v 2 v n ) and T = (u 1 u 2 u n ) Then the change of basis matrix is For any vector v V, we have [id] S,T = ι 1 T ι S = T 1 S. [v] T = ι 1 T (v) =ι 1 T ι S (ι 1 S (v)) = ι 1 T ι S ([v] S ) = T 1 S [v] S Notation. Let E denote the standard basis { e 1, e 2,, e n }. Clearly, (e 1 e 2 e n ) = I n. By discussions in Example 1.11, [id] S,E = S and [id] E,S = S 1. 2.2. Matrix prsentation and changing of basis. Theorem 2.4. Let S and S be two basis of V. Let T and T be two basis of W. Let L V W be a linear map. Let [L] S,T be the matrix representing L under the basis S and T. Then [L] S,T = [id W ] T,T [L] S,T [id V ] S,S Proof. Clear from the following diagram V id V V L W id W W ι S ι S ι T ι T F n [id V ] S,S F n [L] S,T F m [id W ] T,T F m

LECTURE 7: LINEAR TRANSFORMATION 7 Lemma 2.5. Let S = { v 1,, v n } and T = { u 1,, u m } be ordered bases for F n and F m, respectively. If L F n F m is a linear transformation, then In particular, the j-the column of [L] S,T is where T = (u 1,, u m ). [L] S,T = T 1 (L(v 1 ) L(v 2 ) L(v n )). T 1 L(v j ). Proof. Let S = (v 1 v 2 v n ). Let A be the matrix representing L with respect to the standard basis. Now [L] S,T =[id W ] E,T A [id V ] S,E = T 1 AS =T 1 A (v 1 v 2 v n ) = (T 1 Av 1 T 1 Av 2 T 1 Av n ) = (T 1 L(v 1 ) T 1 L(v 2 ) T 1 L(v n )) The colloray below give a way to calculate [L] S,T by row operations. Corollary 2.6. Under the assumption of Lemma 2.5, the reduced row echelon form of (u 1 u m L(v 1 ) L(v n )) is (I m [L] S,T ). Proof. Follows from the fact that is row equivalent to (T L(v 1 ) L(v n )) T 1 (T L(v 1 ) L(v n )) = (I m T 1 L(v 1 ) T 1 L(v n )) = (I m [L] S,T ). 2.3. Similarity. Theorem 2.7. Let T = { v 1,, v n } and T = { u 1,, u n } be two ordered bases for a vector space V, and let L be a linear operator on V. Let S be the transition matrix representing the change from T to T. If A is the matrix representing L with respect to S, and B is the matrix representing L with respect to T, then B = S 1 AS. Proof. By definition, S = [id] T,T, S 1 = [id] T,T, A = [L] T,T and B = [L] T,T. Now Theorem 2.4 implies B = [L] T,T = [id] T,T [L] T,T [id] T,T = S 1 AS. Definition 2.8. Let A and B be n n matrices. B is said to be similar to A if there exists a nonsingular matrix S such that B = S 1 AS.

8 LECTURE 7: LINEAR TRANSFORMATION Upshot, similar matrices could be think as the matrix representations of a linear map under different basises. A very important problem is to find a nonsignular matrix S such that S 1 AS is as simple as possible. Example 2.9. Following are some very important facts (we will not give a proof right now) (1) A 2 2 matrix over R is similar to one of the following matrices (with a 1 a 2 R, λ R, θ [0, 2π) ) ( a 1 0 ), ( λ 1 θ sin θ ), (cos 0 a 2 0 λ sin θ cos θ ) (2) Any matrix over C is similar to an upper triangler matrix. (In fact, one can say more). (3) Any symmatric matrix over R is similar to a diagonal matrix. 2.4. Nullity-Rank theorem (revisited). Let L V W be a linear map between finite dimensional spaces V and W. Definition 2.10. (i) The nullity of L is defined as the dimension of ker(l). It is denoted by null(l) (ii) The rank of L is defined as the dimension of L(V ). It is denoted by rank(l). Theorem 2.11. Let L V W be a linear map. We have null(l) + rank(l) = dim V. Proof. Suppose that dim V = n and dim W = m. Fix any basis S of V and T of W. Then [L] S,T is an n m matrix. Recall that N([L] S,T ) is the null space of the matrix [L] S,T. Let Col([L] S,T ) be the column space of [L] S,T. Now ι S (N([L] S,T )) = ker(l) and ι T (Col([L] S,T ) = L(V ). Therefore dim ker(l) = dim N([L] S,T ) and dim L(V ) = dim Col([L] S,T = dim Row([L] S,T ). Hence the theorem follows from the nullity-rank theorem for matrices.