MATH 423 Linear Algebra II Lecture 8: Subspaces and linear transformations. Basis and coordinates. Matrix of a linear transformation.

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1 MATH 423 Linear Algebra II Lecture 8: Subspaces and linear transformations. Basis and coordinates. Matrix of a linear transformation.

2 Linear transformation Definition. Given vector spaces V 1 and V 2 over a field F, a mapping L : V 1 V 2 is linear if L(x + y) = L(x) + L(y), L(rx) = rl(x) for any x,y V 1 and r F. Basic properties of linear mappings: L(r 1 v r k v k ) = r 1 L(v 1 ) + + r k L(v k ) for all k 1, v 1,...,v k V 1, and r 1,...,r k F. L(0 1 ) = 0 2, where 0 1 and 0 2 are zero vectors in V 1 and V 2, respectively. L( v) = L(v) for any v V 1.

3 Subspaces and linear maps Let V 1, V 2 be vector spaces and L : V 1 V 2 be a linear map. Given a set U V 1, its image under the map L, denoted L(U), is the set of all vectors in V 2 that can be represented as L(x) for some x U. Theorem If U is a subspace of V 1 then L(U) is a subspace of V 2. Proof: U is nonempty = L(U) is nonempty. Let u,v L(U). This means u = L(x) and v = L(y) for some x,y U. By linearity, u + v = L(x) + L(y) = L(x + y). Since U is a subspace of V 1, we have x + y U so that u + v L(U). Similarly, if u = L(x) for some x U then for any r F we have ru = rl(x) = L(rx) L(U).

4 Subspaces and linear maps Let V 1, V 2 be vector spaces and L : V 1 V 2 be a linear map. Given a set W V 2, its preimage (or inverse image) under the map L, denoted L 1 (W), is the set of vectors x V 1 such that L(x) W. Theorem If W is a subspace of V 2 then its preimage L 1 (W) is a subspace of V 1. Proof: Let 0 1 be the zero vector in V 1 and 0 2 be the zero vector in V 2. By linearity, L(0 1 ) = 0 2. Since W is a subspace of V 2, it contains 0 2. Hence 0 1 L 1 (W). Let x,y L 1 (W). This means that L(x), L(y) W. Then L(x + y) = L(x) + L(y) is in W since W is closed under addition. Therefore x + y L 1 (W). Similarly, if L(x) W for some x V 1 then for any r F we have L(rx) = rl(x) W so that rx L 1 (W).

5 Range and null-space Let V,W be vector spaces and L : V W be a linear mapping. Definition. The range (or image) of L is the set of all vectors w W such that w = L(v) for some v V. The range of L is denoted R(L). The null-space (or kernel) of L, denoted N(L), is the set of all vectors v V such that L(v) = 0. Theorem (i) The range R(L) is a subspace of W. (ii) The null-space N(L) is a subspace of V. Proof: R(L) = L(V), N(L) = L 1 ({0}).

6 Dimension Theorem Theorem Let L : V W be a linear mapping of a finite-dimensional vector space V to a vector space W. Then dim R(L) + dim N(L) = dim V. The null-space N(L) is a subspace of V. It is finite-dimensional since the vector space V is. Take a basis v 1,v 2,...,v k for the subspace N(L), then extend it to a basis v 1,v 2,...,v k,u 1,u 2,...,u m for the entire space V. Claim Vectors L(u 1 ), L(u 2 ),...,L(u m ) form a basis for the range of L. Assuming the claim is proved, we obtain dim R(L) = m, dim N(L) = k, dim V = k + m.

7 Claim Vectors L(u 1 ), L(u 2 ),...,L(u m ) form a basis for the range of L. Proof (spanning): Any vector w R(L) is represented as w = L(v), where v V. Then v = α 1 v 1 + α 2 v α k v k + β 1 u 1 + β 2 u β m u m for some α i,β j F. By linearity of L, w = L(v) = α 1 L(v 1 )+ +α k L(v k )+β 1 L(u 1 )+ +β m L(u m ) = β 1 L(u 1 ) + + β m L(u m ). Note that L(v i ) = 0 since v i N(L). Thus R(L) is spanned by the vectors L(u 1 ),...,L(u m ).

8 Claim Vectors L(u 1 ), L(u 2 ),...,L(u m ) form a basis for the range of L. Proof (linear independence): Assume that t 1 L(u 1 ) + t 2 L(u 2 ) + + t m L(u m ) = 0 for some t i F. Let u = t 1 u 1 + t 2 u t m u m. Since L(u) = t 1 L(u 1 ) + t 2 L(u 2 ) + + t m L(u m ) = 0, the vector u belongs to the null-space of L. Therefore u = s 1 v 1 + s 2 v s k v k for some s j F. It follows that t 1 u 1 +t 2 u 2 + +t m u m s 1 v 1 s 2 v 2 s k v k = u u = 0. Linear independence of vectors v 1,...,v k,u 1,...,u m implies that t 1 = = t m = 0 (as well as s 1 = = s k = 0). Thus the vectors L(u 1 ), L(u 2 ),...,L(u m ) are linearly independent.

9 Let V 1, V 2 be vector spaces and L : V 1 V 2 be a linear map. Definition. The map L is one-to-one if it maps different vectors from V 1 to different vectors in V 2. That is, for any x,y V 1 we have that x y implies L(x) L(y). The map L is onto if any element y V 2 is represented as L(x) for some x V 1. If the map L is both one-to-one and onto, then the inverse map L 1 : V 2 V 1 is well defined. Theorem A linear map L is one-to-one if and only if the nullspace N(L) is trivial. Proof: Let 0 1 be the zero vector in V 1 and 0 2 be the zero vector in V 2. If a vector x 0 1 belongs to N(L), then L(x) = 0 2 = L(0 1 ) so that L is not one-to-one. Conversely, assume that N(L) is trivial. By linearity, L(x y) = L(x) L(y) for all x,y V 1. Therefore L(x) = L(y) = x y N(L) = x = y. Thus L is one-to-one.

10 Basis and coordinates If {v 1,v 2,...,v n } is a basis for a vector space V, then any vector v V has a unique representation v = x 1 v 1 + x 2 v x n v n, where x i F. The coefficients x 1, x 2,...,x n are called the coordinates of v with respect to the ordered basis v 1,v 2,...,v n. The coordinate mapping vector v its coordinates (x 1, x 2,...,x n ) establishes a one-to-one correspondence between V and F n. This correspondence is linear.

11 Let v 1,v 2,...,v n be elements of a vector space V. Define a map f : F n V by f (x 1, x 2,...,x n ) = x 1 v 1 + x 2 v x n v n. Theorem (i) The map f linear. (ii) If vectors v 1,v 2,...,v n are linearly independent then f is one-to-one. (iii) If vectors v 1,v 2,...,v n span V then f is onto. (iv) If vectors v 1,v 2,...,v n form a basis for V then f is one-to-one and onto. Proof: The map f is linear since (x 1 +y 1 )v 1 + (x 2 +y 2 )v (x n +y n )v n = (x 1 v 1 +x 2 v 2 + +x n v n ) + (y 1 v 1 +y 2 v 2 + +y n v n ), (rx 1 )v 1 +(rx 2 )v 2 + +(rx n )v n = r(x 1 v 1 +x 2 v 2 + +x n v n ) for all x i, y i, r F. Further, linear independence of v 1,...,v n means that the null-space of f is trivial, which is equivalent to f being one-to-one. Finally, statement (iii) is obvious while statement (iv) follows from (ii) and (iii).

12 Examples. Coordinates of a vector v = (x 1, x 2,...,x n ) F n relative to the standard basis e 1 = (1, 0,...,0, 0), e 2 = (0, 1,...,0, 0),..., e n = (0, 0,...,0, 1) are (x 1, x 2,...,x n ). Coordinates of a matrix relative to the basis ( ) are (a, c, b, d). ( ), ( ) a b c d ( M 2,2 (F) ), ( ) 0 1, 0 0 Coordinates of a polynomial p(x) = a 0 + a 1 x + + a n x n P n relative to the basis 1, x, x 2,...,x n are (a 0, a 1,...,a n ).

13 Matrix of a linear transformation Let V,W be vector spaces and L : V W be a linear map. Let α = [v 1,v 2,...,v n ] be an ordered basis for V and β = [w 1,w 2,...,w m ] be an ordered basis for W. Definition. The matrix of L relative to the bases α and β is an m n matrix whose consecutive columns are coordinates of vectors L(v 1 ), L(v 2 ),...,L(v n ) relative to the basis β. Notation. [w] β denotes coordinates of w relative to the ordered basis β, regarded as a column vector. [L] β α denotes the matrix of L relative to α and β. Then [L] β α = ( [L(v 1 )] β, [L(v 2 )] β,...,[l(v n )] β ).

14 Examples. D : P 2 P 1, (Dp)(x) = p (x). Let α = [1, x, x 2 ], β = [1, x]. Columns of the matrix [D] β α are coordinates of polynomials D1, Dx, Dx 2 w.r.t. the basis 1, x. ( ) D1 = 0, Dx = 1, Dx 2 = 2x = [D] β α = L : P 2 P 2, (Lp)(x) = p(x + 1). Let us find the matrix [L] α α: L1 = 1, Lx = 1 + x, Lx 2 = (x + 1) 2 = 1 + 2x + x = [L] α α =