Rotation by 90 degree counterclockwise

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1 0.1. Injective an suejective linear map. Given a linear map T : V W of linear spaces. It is natrual to ask: Problem 0.1. Is that possible to fin an inverse linear map T 1 : W V? We briefly review those examples Suppose V is a -imensional linear space. T is a linear transformation by rotating V 90 egree counterclockwise. Then the inverse of T is rotating V 90 egree clockwise Rotation by 90 egree clockwise Rotation by 90 egree counterclockwise If V is a -imensional linear space with selecte x,y axis. T is a linear transformation by projecting V own to a horizontal linex-axis. Then we can not recover T since for any w on x-axissee re point in picture, there are infinitely many v such that T v = wnot unique. For any w not on x-axis see blue point in picture, we can t fin any v such that T v = w. y Noboy projects to me! Sa. No existence every green people projects to me! Cheers! x No uniqueness Therefore, for the linear map T : V W to have inverse, then for any w W, there must exists unique v V, such that T v = w. Any failure of uniqueness or existence will make T non-invertible. 0.. Surjective linear map, Existence of v, the Image. 1

2 Definition 0..1 A linear map T : V W is calle as surjective if for any w W, there always exists a v V such that T v = w. If V is -imensional linear space in above example. T : V V is a projection own to x-axis. Then T is not a surjective map because the blue point missing a vector to map onto it. Let W V be the x-axis. Then W is a subspace of V. We may restrict the target of V to W. Then the linear map T : V W, is a projection of V onto W. This map is surjective because any vectors in W always have some v V such that T v = W. The case for that blue point can not happen because W oes not contain the blue point. Consier V = PX 3 = {X3 a + X b + Xc +, where a, b, c, R}, W = PX = {X a + Xb + c, where a, b, c R}. Let : V W be the linear map by taking erivative of elements in V. Then : V W is a surjective because any quaratic polynomial can be a erivative of a cubic polynomial. But : V V is not a surjective because no cubic polynomial can have cubic polynomial as erivative. From above example. We iscovere that to be a surjective map, we will nee the target to be small enough to only contain those elements from the omain. That s clear not all linear map can be surjective because the target is not small enough, we use the following concept to measure the failure of being a surjective map. Definition 0.. For a linear map T : W V, the image ImT is efine to be the subset of elements of the form T v for v V In other wors, the image of a linear map is the minimal subspace for T to be surjective. Proposition 0.1 Suppose e 1 e n is a basis of omain. We have the following statements 1 ImT = span T e 1 T e T e n w ImT if an only if there exists v V such that T v = w We have the following equivalent statements. Proposition 0. Let T : V W be a linear map, let ɛ1 ɛ ɛ m be a baiss in W. An e1 e e n be a basis in V, an. Then the following statements are equivalent. T e 1 e e n = ɛ1 ɛ ɛ m P

3 1 T is a surjective. ImT = W 3 T e 1 T e T e n spans the whole space. 4 The reuce row echelon form of P has leaing one occupie all its rows. 5 rankp = m Furthermore, by thinking the leaing one, we have the following property: Proposition 0.3 If T : V W is a surjective linear map, then imv imw Surjective map preserves the property of span the whole space. Proposition 0.4 If T : V W is a surjective linear map, v 1 v n spans the whole space of V. Then T v1 T v n spans the whole space W We en up this section by an example of fining the image. Problem 0.. Let e 1 e e 3 be a basis for linear space V. w1 w w 3 is a basis for linear space W. Suppose the matrix representation of T is T e 1 e e 3 = w1 w w 3 6 Fin a basis of ImT Since ImT = span T e 1 T e T e n, to fin a basus of this subspace, we only nee to get ri of reunant vectors. By oing row operations. r r 1 r 3 8 r r 3 6 r r 3 r

4 r 1 3 r r 3 +4 r From the last column of reuce row echelon form we iscovere that T e 3 is a reunant vector. T e1 T e is linearly inepenent. Then T e1 = w 1 1+w +w 3 8 an T e = w 1 3+w +w 3 consists a basis of ImT. : Problem 0.3. Consier V = PX 3 = {X3 a + X b + Xc +, where a, b, c, R},. Let V V be the linear map by taking erivative of elements in V. Fin a basis of Im 1 X X X 3 is a basis for V. Applying the linear map. The image is the span of the vectors T 1 X X X 3 = 0 1 X 3X in the target space. Since among vectors of 0 1 X 3X. 0 is the reunant vector an 1 X 3X is linearly inepenent. So 1 X 3X is a basis of Im 0.3. Injective linear map, Uniqueness of v, the Kernel. Definition A linear map T : V W is calle as injective if for any w W, there is either no v V, or exists a unique v V such that T v = w. Let W V be the x-axis. Then W is a subspace of V. We may restrict the target of V to W. Then the linear map T : V W, is a projection of V onto W. This map is not an injective map because any vectors in W always have more than one v V such that T v = w. To measure the failure from being an injective. Consier the case if there are two vectors maps to one. Suppose T v 1 = T v, by linearity, T v 1 v = 0. So the non-uniqueness actually comes from the nonuniqueness of vector v such that T v = 0 Definition For a linear map T : W V, the kernel ker T is efine to be the subsets of W that maps to 0. Namely, ker T = {w W T w = 0} Proposition 0.5 Suppose e 1 e n is a basis of omain, then we have the following statements: 1 v ker T if an only if T v = 0 4

5 If v ker T an v = e 1 e n T e1 T e n a 1. a n. Then a 1. a n is a linear relation of Proposition 0.6 Let T : V W be a linear map, let ɛ1 ɛ ɛ m be a baiss in W. An e1 e e n be a basis in V, an T e 1 e e n = ɛ1 ɛ ɛ m P. Then the following statements are equivalent. 1 T is an injective. ker T = {0} 3 T e 1 T e T e n is linearly inepenent. 4 The reuce row echelon form of P has leaing one occupie all its columns. 5 rankp = n Furthermore, by thinking the leaing one, we have the following property: Proposition 0.7 If T : V W is an injective linear map, then imv imw Injective map preserves the property of linearly inepenent. Proposition 0.8 If T : V W is an injective linear map, v 1 v n is linearly inepenent. Then T v1 T v n is also linearly inepenent. We en up this section by an example of fining the kernel. Problem 0.4. Let e 1 e e 3 be a basis for linear space V. w1 w w 3 is a basis for linear space W. Suppose the matrix representation of T is T e 1 e e 3 = w1 w w 3 6 Fin a basis of ker T Since the coorinate of ker T is a linear relation on T e 1 T e T e n, So is a linear relation on the coorinate matrix 6. Since row operation oes not change linear relation on columns, so we o 5

6 r r 1 r 3 8 r r 3 6 r r 3 r r 1 3 r r 3 +4 r From this we iscovere all linear relation is all linear combination of is a basis of ker T 1 1. So e 1 +e e 3 : Problem 0.5. Consier V = PX 3 = {X3 a + X b + Xc +, where a, b, c, R},. Let V V be the linear map by taking erivative of elements in V. Fin a basis of ker 1 X X X 3 is a basis for V. Applying the linear map. The image is the span of the vectors T 1 X X X 3 = 0 1 X 3X Clear all linear relation on those vectors has the form t. So a basis of kernel can be choose as 1 There is an important ientity of the imension of kernel an image. Theorem 0.6 Suppose T : V W is a linear map. Then imker T + imimt = im V 0.4. Isomorphism. 6

7 Definition A linear map T : V W is calle as isomorphism if T is both surjective an injective. Proposition 0.9 If T : V W is an isomorphism, then T is invertible. There exists an unique inverse T 1 : W V such that T 1 T = i V ; T T 1 = i W. 7