Span & Linear Independence (Pop Quiz)

Transcription

1 Span & Linear Independence (Pop Quiz). Consider the following vectors: v = 2, v 2 = 4 5, v 3 = 3 2, v 4 = Is the set of vectors S = {v, v 2, v 3, v 4 } linearly independent? Solution: Notice that the number of vectors in the set S (which is four) is greater than the number of components of each vector (which is three). This means that at least one column of the matrix A = [ v v 2 v 3 v 4 ] will not have a pivot. Therefore, the system Ax = has at least one free variable which means that the system has infinitely many nontrivial solutions. Equivalently, there exist constants x, x 2, x 3, and x 4 not all zero such that x v + x 2 v 2 + x 3 v 3 + x 4 v 4 =. Thus, by definition, S is linearly dependent.. 2. Consider the following vectors: v =, v 2 = 2, v 3 = Is the set of vectors S = {v, v 2, v 3 } linearly independent? Solution: In this problem, the number of vectors in the set S is equal to the number components of each vector. Therefore, neither linear independence nor dependence can be established using this fact. However, each column of the matrix A = v v 2 v 3 has a pivot. Therefore, S is linearly independent. 3. Suppose that in Problem 2, we change the vector v 3 to v 3 =. Is S linearly independent in this case? 5. 5 Solution: In this case, S is now linearly dependent. We see this just be using the fact that the matrix A = [ v v 2 v 3 ] has no pivot in the last column. We can also use the fact that v 3 is in Span{v 2 } since v 3 is a constant multiple of v 2. Since one of the vectors in the set depends on at least one of the other vectors (e.g. v 3 depends on v 2 ), S must be linearly dependent.

2 Note that Problems 2 and 3 illustrate the fact that it is possible for a set of vectors to be either linearly independent or linearly dependent if the number of vectors in the set is equal to the number of components of each vector. This actually also holds true if the number of vectors in a set is less than the number of components of each vector. 4. Consider the following vectors: v = 2 3, v 2 = Is the set of vectors S = {v, v 2 } linearly independent? Moreover, does S span R 3? Solution: Since v 2 = 2v, v 2 lies in Span{v }. Therefore, S is linearly dependent. To answer the next question, we need to determine if every vector in R 3 is a linear combination of the vectors in S. To access every vector in R 3, it makes sense that we would need at least three vectors to work with. Since S only contains two vectors, there isn t enough vectors to generate all of the vectors in R 3. Geometrically, Span{v, v 2 } is a line in R 3 through the origin which is not all of R 3. Thus, S does not span R 3. We can certainly answer both of the above questions by using pivots. Note that the reduced echelon matrix of A = v v 2 is 2. The boxed entry is the only pivot. Clearly, at least one row and at least one column of the reduced echelon matrix has no pivot. Therefore, S is a linearly dependent set that does not span R

3 Sections.8 &.9: Linear Transformations In your past precalculus courses, you studied the concept of a function. Recall that a function, f, is a rule that assigns to each input (variable x) a unique output (variable y = f(x)). For example, let f(x) = 2x. If the input is x =, then the output is f() = 2. This is an elementary idea, but let s stop and think about what the function actually does to each input. If we plug in any input value, the function will always double it since that is exactly what the rule is. The general idea here is that each inputted object (i.e. a number) is transformed into a different object (it could be a similar object such as another number) according to what the function (the rule) is. Now how does this idea extend to the world of matrices and vectors? Let s look at an example. Example: Let s define Ax = b where x and b are vectors in R 2 and 2 A =. We know that for each x in R 2 that we plug in, we obtain a new vector b in R 2. Therefore, we can think of T (x) = Ax as being the [ function, and x and b = T (x) as the input and output, respectively. Let s plug in x = and see what we get. It follows that 3] [ 2 3] }{{}}{{} A x Notice [ the effect that the matrix A has on the vector [ x geometrically. The matrix stretches 5 x = and rotates it to produce a new vector b =. (See Figure.) In general, all that 3] 2] a matrix A does to a vector x (when A is multiplied to x) is that A stretches (or compresses) x and then relocates it (by rotation or reflection). = [ 5 2] }{{} b Figure : Geometric description of matrix-vector multiplication (Ax = b).

4 The matrix A in the previous example is an example of a transformation. Definition: A transformation (or function or mapping) T from R n to R m is a rule that assigns to each vector x in R n a vector T (x) in R m. The set R n is called the domain of T, and R m is called the codomain of T. Notation: The notation T : R n R m simply means that T is a function that transforms vectors in R n (domain) to vectors in R m (codomain). Definition: For x in R n, the vector T (x) in R m is called the image of x under T. The set of all images T (x) is called the range of T. Matrix Transformations When T is defined to be T (x) = Ax, where A is an m n matrix and x is in R n, then T is called a matrix transformation. The transformation itself can be denoted by x Ax, which simply means that x is being mapped to the vector Ax. The matrix A is called the standard matrix for the linear transformation T. Note: If the size of A is m n and the size of x is n, then the size of b = Ax is m. Example: Define T : R 3 R 2 to be a transformation that projects every vector in R 3 onto the xy-plane. What is the image of x = 2 under T? 3 Solution: Since we are projecting x onto the xy-plane, we are expecting that the third component (the z-coordinate) of the image is zero. Note that the codomain is R 2, which means that we will need to express the output as a two-component vector. Thus, T [ 2 = 2] 3 The standard matrix that does this particular transformation is A =

5 Let s verify this. Ax = 2 = 3 [ () + (2) + (3) = () + (2) + (3) 2] Example: Let A = [ ] and u = (a) Define T : R 3 R 2 by T (x) = Ax. Determine T (u). Solution: Applying the same procedure as in the previous example, we obtain [ 2 T (u) = Au =. 7] 3 (b) Find a vector x whose image under T is. Is x unique? 4 3 Solution: Let b =. We wish to find a vector x such that T (x) = Ax = b. Note 4 that Ax = x x 2 = x = x = 3, x 2 = 4, x 3 = free Therefore, every vector x in R 3 that maps to b has the form x = x 3 }{{} p One such vector is p. Since the system has a free variable, x is not unique. Note that we could have simply used the fact that A projects every vector in R 3 onto the xy-plane. Thus, any vector x in R 3 with an x-coordinate of 3 and a y-coordinate of 4 works. (c) Determine if b = 3 is in the range of T. 4 Solution: We showed in part (b) that there exists an x in R 3 such that T (x) = b. Thus, b is in the range of T.

6 Example: Let A = (a) Find all x in R 4 that are mapped into the zero vector by the transformation x Ax. Solution: Note that the augmented matrix [ A ] has the following reduced echelon form 4 R = 3 Therefore, the system of equations is x 4x 3 = x 2 3x 3 = x 4 = where x 3 is a free variable. It follows that every x in R 4 that is mapped to has the form x 4x 3 4 x = x 2 x 3 = 3x 3 x 3 = x 3 3. x 4 (b) Determine if the vector b = is in the range of x Ax. Solution: To show that b is in the range of the transformation, it suffices to show that the columns of A span R 3. This is equivalent to showing that every row of A has a pivot. The coefficient matrix R in part (a) has a pivot in every row (and therefore this must also be the case for A). Hence, the columns of A span R 3, and so b must be in the range of the transformation.

7 Linear Transformations Definition: A transformation (or mapping) T is linear if: (i) T (u + v) = T (u) + T (v) for all u, v in the domain of T ; (ii) T (cu) = ct (u) for all scalars c and all u in the domain of T. Fact: Every matrix transformation is a linear transformation, and vice versa. Therefore, if you can find a matrix to implement the transformation, it is a linear transformation. Fact: If T is a linear transformation, then T () = and T (cu + dv) = ct (u) + dt (v) for all vectors u, v in the domain of T and all scalars c, d. In general, T (c v + + c p v p ) = c T (v ) + + c p T (v p ) which is known as a superposition principle. Example: Show that T (x, x 2, x 3 ) = (3x, 3, x 2 + x 3 ) is not a linear transformation. Solution: Note that the given transformation can be written as T x x 2 = 3x 3 x 3 x 2 + x 3 If c is any scalar, then T (cx) = T cx cx 2 = 3cx 3 ct (x) cx 3 cx 2 + cx 3 Thus, T is not a linear transformation.

8 Example: Show that T (x, x 2, x 3 ) = (x +x 2, 3x 2 2x 3 ) is a linear transformation by finding a matrix A that implements the mapping. Solution: To show that T is a linear transformation, it suffices to show that we can write the mapping as T (x) = Ax. x + x T (x) = 2 3x 2 2x 3 = x + x 2 + x = x x }{{} x 3 A }{{} x Example: Let x = x, v = x 2 x 3 [ [ [ 2, v ] 2 =, and v 3] 3 = and let T : R 2] 3 R 2 be a linear transformation that maps x into x v + x 2 v 2 + x 3 v 3. Find a matrix A so that T (x) = Ax for any x. Solution: The transformation that we are setting up is simply T (x) = [ v v 2 ] v 3 x, i.e. 2 T (x) = x x x 3 Definition: The j th column of the identity matrix is denoted by e j. For example, if I = then e =, e 2 =, e 3 =.