PUMaC 2017 · 加试 · 第 1 题
PUMaC 2017 — Power Round — Problem 1
题目详情
Problem 1.2.3. (2 points) If φ is an isomorphism, show that φ is also an isomorphism. 1.3 Matrices (4 points) Suppose we have the linear map f : v = (1 , 0 , 0) 7 → (2 , 0) 1 : v = (0 , 1 , 0) 7 → (1 , 7) 2 : v = (0 , 0 , 1) 7 → (1 , 1) . 3 [ ] 2 1 1 We can use a shorthand this function: . The first column represents the image of 0 7 1 the first basis vector under the map f ; the same holds for the second and third columns. In general, this format of storing the data of a linear mapping may be used interchangeably with the function itself, since it conveys precisely the same information. This will be important and convenient when it becomes cumbersome to work with the function as a list of operations and instead more useful to use a grid. In general, if f : V → W and we fix a basis for each of V and W , then we can represent the map f as a m × n array of numbers in the following way: Definition 1.3.A. An m × n matrix is a grid of values in F having m rows and n columns, where the i th column represents f ( v ), the image of V ’s i th basis vector. i 2 For another example, R has the bases { (0 , 1) , (1 , 0) } and { (1 , 1) , (1 , − 1) } . The map [ ] [ ] f : (0 , 1) 7 → 4, f : (1 , 0) 7 → 2 has matrix 4 2 with the first basis and 6 − 2 with the second. T We will denote by A the transpose of matrix A , having at the j th column and i th row T the entry in A ’s i th column and j th row. A is an n × m matrix if A is m × n . For instance, [ ] 1 4 T 1 2 3 = 2 5 . 4 5 6 3 6 We will not treat matrices independently of the context of the functions that they represent. Matrix addition is rather intuitive; simply add corresponding elements. For example, [ ] [ ] [ ] 2 0 1 7 1 7 3 8 3 7 4 15
- = . 1 7 4 6 1 1 2 2 2 8 6 8 Multiplication is less obvious, however. When multiplying, we think of the left mul- [ ] [ ] 2 0 1 v 1 tiplicand as a column vector of row vectors; for example, = where v = 1 1 7 4 v 2 PUMaC 2017 Power Round Page 9 [ ] 2 0 1 , etc., and the right multiplicand is a row vector of column vectors; for example, 1 1 1 [ ] 1 0 = w w where w = 1 , etc. We simply define the product, then, to be the 1 2 1 2 7 2 matrix { a } where i ranges along the height of the left multiplicand and j ranges along the i,j [ ] [ ] 1 1 2 0 1 4 9 width of the right multiplicand, and a = v · w . So, · 1 0 = . In i,j i j 1 7 4 16 29 2 7 general, note the rule that a m × n matrix multiplied by a n × p matrix is a m × p matrix. 1 · · · 0 . . . . Definition 1.3.B. The n × n matrix I = having 1’s along its main diagonal n . . 0 0 0 1 and 0’s elsewhere is the identity matrix, because I M = M I for all n × n matrices M . n n n Definition 1.3.C. The set of linear mappings between F = F × F × · · · × F and itself is denoted gl . n The elements of gl can (and often will) be thought of as n × n matrices. ( F is assumed n to be C unless otherwise specified.) Sometimes for notational convenience we use block matrices . For instance, we could 1 0 0 0 [ ] [ ] 0 1 0 0 I 0 0 0 2 write as , where the 0’s actually represent ; or we may say 0 0 1 0 0 I 0 0 2 0 0 0 1 [ ] 0 [ ] I 0 3 0 0 0 0 , where one 0 represents and the other , as to make the size of the 0 1 0 matrix’s elements make sense.
解析
Problem 1.2.1. (2 points each) Fix N ∈ N . Let a ( x ) take the form a ( x ) = a x and n n =0 let F be the set of degree- N polynomials in F [ x ]. For each of the following functions N f : F → F [ x ], determine with proof whether or not it is a linear map. i N (i) f : a ( x ) 7 → a ( x ) + a (1) 1 N +1 (ii) f : a ( x ) 7 → a ( x ) + a (1) x 2 N ∑ n (iii) f : a ( x ) 7 → ( a − a ) x 3 n N − n n =0 Solution : (i) Yes; f (( αa + βb )( x )) = ( αa + βb )( x ) + ( αa + βb )(1) = αa ( x ) + βb ( x ) + αa (1) + βb (1) = 1 αf ( a ( x )) + βf ( b ( x )) (where α, β ∈ F ; a, b ∈ F [ x ]). 1 1 N +1 N +1 (ii) Yes; f (( αa + βb )( x )) = αa ( x ) + βb ( x ) + ( αa (1) + βb (1)) x = α ( a ( x ) + a (1) x ) + 2 N +1 β ( b ( x ) + b (1) x ) = αf ( a ( x )) + βf ( b ( x )) (where α, β ∈ F ; a, b ∈ F [ x ]). 2 2 N ∑ n (iii) Yes; f (( αa + βb )( x )) = ( αa + βb − αa − βb ) x = αf ( a ( x )) + βf ( b ( x )) 3 n n N − n N − n 3 3 n =0 (where α, β ∈ F ; a, b ∈ F [ x ]).