Math 3013 Solutions to Problem Set 5

Transcription

1 Math 33 Solutions to Problem Set 5. Determine which of the following mappings are linear transformations. (a) T : R 3 R 2 : T ([x, x 2, x 3 ]) = [x + x 2, x 3x 2 ] This mapping is linear since if v = [x, x 2, x 3 ] T (λv) = T (λ [x, x 2, x 3 ]) = T ([λx, λx 2, λx 3 ]) = [λx + λx 2, λx 3λx 2 ] = λ [x + x 2, x 3x 2 ] = λt ([x, x 2, x 3 ]) = λt (v) (T preserves scalar multiplication) and if v = [x, x 2, x 3 ] and v = [x, x 2, x 3] T (v + v ) = T ([x + x, x 2 + x 2, x 3 + x 3]) = [x + x + x 2 + x 2, x + x 3(x 2 + x 2)] = [x + x 2, x 3x 2 ] + [x + x 2, x 3x 2] = T (v) + T (v ) (T preserves vector addition) (b) T : R 3 R 4 : T ([x, x 2, x 3 ]) = [,,, ] This mapping is linear since if v = [x, x 2, x 3 ] T (λv) = T ([λx, λx 2, λx 3 ]) = [,,, ] = λ [,,, ] = λt ([x, x 2, x 3 ]) = λt (v) (T preserves scalar multiplication) and if v = [x, x 2, x 3 ] and v = [x, x 2, x 3] T (v + v ) = T ([x + x, x 2 + x 2, x 3 + x 3]) = [,,, ] = [,,, ] + [,,, ] = T (v) + T (v ) (T preserves vector addition) (c) T : R 3 R 4 : T ([x, x 2, x 3 ]) = [,,, ] This mapping is not linear since if v = [x, x 2, x 3 ] T (v) = [,,, ] T (2v) = [,,, ] 2 [,,, ] = 2T (v) So the mapping does not preserve scalar multiplication. (d) T : R 2 R 3 : T ([x, x 2 ]) = [x x 2, x 2 +, 3x 2x 2 ]

2 2 This mapping is not linear since, e.g., if v = [, ] T (v) = [, 2, ] T (2v) = T ([2, 2]) = [, 3, 2] [, 4, 2] = 2T (v) So the mapping does not preserve scalar multiplication. 2. For each of the following, assume T is a linear transformation, from the data given, compute the specified value. (a) Given T ([, ]) = [3, ], and T ([, ]) = [ 2, 5], find T ([4, 6]). Because linear transformations preserve scalar multiplication and vector addition, they also preserve linear combinations: T (c v + c 2 v 2 ) = c T (v ) + c 2 T (v 2 ) Now take e = [, ] and e 2 = [, ]. Then T ([4, 6]) = T (4e 6e 2 ) = 4T (e ) 6T (e 2 ) = 4 [3, ] 6 [ 2, 5] = [2 + 2, 4 3] = [24, 34] (b) Given T ([,, ]) = [3,, 2], T ([,, ]) = [2,, 4], and T ([,, ]) = [6,, ], find T ([2, 5, ]). As in Part (a), we set e = [,, ], e 2 = [,, ], and e 3 = [,, ] and then compute T ([2, 5, ]) = T (2e 5e 2 + e 3 ) = 2T (e ) 5T (e 2 ) + T (e 3 ) = 2 [3,, 2] 5 [2,, 4] + [6,, ] = [6 + 6, , ] = [2, 7, 5] 3. Find the standard matrix representations of the following linear transformations. (a) T : R 2 R 2 : T ([x, x 2 ]) = [x + x 2T, x 3x 2 ] The standard matrix representations are computed by computing the action of the linear transformation T on the standard basis vectors, and then using results as the columns of the corresponding matrix. For the case at hand we have e = [, ] T (e ) = [ +, 3()] = [, ] e 2 = [, ] T (e 2 ) = [ +, 3()] = [, 3] So the matrix corresponding to T is [ 3 ] (b) T : R 3 R 2 : T ([x, x 2, x 3 ]) = [x + x 2 + x 3, x + x 2, x ]

3 3 We proceed as in Part (a). e = [,, ] T (e ) = [ + +, +, ] = [,, ] e 2 = [,, ] T (e 2 ) = [ + +, +, ] = [,, ] e 3 = [,, ] T (e 3 ) = [ + +, +, ] = [,, ] So the matrix corresponding to T is (c) T : R 3 R 2 : T ([x, x 2, x 3 ]) = [x + x 2 + x 3, 2x + 2x 2 + 2x 3 ] Proceeding as in Part (a) A T = [ ] 4. For each of the linear transformations T : R m R n in Problem 3, determine and (a) We have Range (T ) := {y R n y = T (x) for some x R m } Kernel (T ) := {x R m T (x) = R n} ([ Range (T ) = ColSp (A T ) = ColSp 3 Kernel (T ) = NullSp (A T ) = solution set of A T x = A T row reduces to the Reduced Row Echelon Form Since each column of this row echelon form contains a pivot, each column of A T is a basis vector for the column space of A T. Thus, ( ) Range (T ) = ColSp (A T ) = span, 3 ]) From the Reduced Row Echelon Form, we can also read of the solution set of A T x =. We must have } x = x = x 2 = Therefore, {} Ker (T ) = NullSp (A t ) = (b) We proceed as in Part (a). The matrix A T row reduces to a R.R.E.F. A T =

4 4 Each column of the R.R.E.F. contains a pivot, so each of the columns of A T is a basis vector for the column space of A T and Range (T ) = ColSp (A T ) = span, The solution set of A T x = is x = x 2 = Ker (T ) = NullSp (A T ) = x 3 =, (c) We proceed as in Part (a). The matrix A T row reduces to a R.R.E.F. A T = Only the first column of the R.R.E.F. contains a pivot and so () Range (T ) = ColSp (A T ) = span 2 The solutions of A T x = can be read off the R.R.E.F. of A T : } x x + x 2 + x 3 = 2 x 3 x = x = 2 = x 2 x 3 Ker (T ) = NullSp (A T ) = span, + x 3 5. If T : R 2 R 3 is defined by T ([x, x 2 ]) = [2x + x 2, x, x x 2 ] and T : R 3 R 2 is defined by T ([x, x 2, x 3 ]) = [x x 2 + x 3, x + x 2 ], find the standard matrix representation for the linear transformation T T that carries R 2 into R 2. Find a formula for (T T ) ([x, x 2 ]). The matrix representations corresponding to T and T are 2 M T =, M T = The matrix representation corresponding to T T will be given by the product of the corresponding matrices M T T = M T M T = 2 2 = 3 Hence (T T ) (x, x 2 ) = [2x, 3x + x 2 ] : 6. Determine whether the following statements are true or false. (a) Every linear transformation is a function.

5 5 (b) Every function mapping R n to R m is a linear transformation. False. In order to be a linear transformation a function f : R n R m must preserve scalar multiplication and vector addtion. (c) Composition of linear transformations corresponds to multiplication of their standard matrix representations. (d) Function composition is associative. (e) An invertible linear transformation mapping R n to itself has a unique inverse. (This follows from the corresponding theorem about invertible matrices.) (f) The same matrix may be the standard matrix representation for several different linear transformations. False. (Unless one allows more general vector spaces - but idea won t be broached until Chapter 3.) (g) A linear transformation having an m n matrix as its standard matrix representation maps R n into R m. (h) If T and T are different linear transformations mapping R n into R m, then we may have T (e i ) = T (e i ) for all standard basis vectors e i of R n. False. Linear transformations are determined uniquely by their standard matrix representations. (i) If T and T are different linear transformations mapping R n into R m, then we may have T (e i ) = T (e i ) for some standard basis vectors e i of R n. (So long as they are not all the same.) (j) If B = {b, b 2,..., b n } is a basis for R n and T and T are linear transformations from R n into R m, then T (x) = T (x) for all x R n if and only if T (b i ) = T (b i ) for i =, 2,..., n.