Solution_3

# Solution_3 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310 Linear Algebra and Applications (Fall 2007) Suggested Solutions for Homework 3 Solution 1. (a) This is a vector space. We must check all ten properties. i. ( α ) and ( β ); ii. A 1 - A 4 ; iii. M 1 - M 4 . ( α ) If x and y are positive reals, so is x + y = xy ( A 1 ) x + y = xy = yx = y + x . ( A 2 ) x + ( y + z ) = x ( yz ) = ( xy ) z = ( x + y ) + z . ( A 3 ) There is an object 0, the positive real number 1, which is such that 1 + x = 1 x = x = x 1 = x + 1 for all positive real number x . ( A 4 ) For all real x , the positive real 1 x acts as a negative: x + 1 x = x 1 x = 1 = 0 = 1 = 1 x x = 1 x + x. ( β ) If k is a real and x is a positive real, then kx = x k is again a positive real. ( M 1 ) k ( x + y ) = ( xy ) k = x k y k = kx + ky . ( M 2 ) ( k + l ) x = x k + l = x k x l = x k + x l . ( M 3 ) k ( lx ) = ( lx ) k = ( x l ) k = x lk = x kl = ( kl ) x . ( M 4 ) 1 x = x 1 = x . (b) This is a vector space. We must check all ten properties. i. ( α ) and ( β ); ii. A 1 - A 4 ; iii. M 1 - M 4 . ( α ) If we add two matrices of this form, the result will again be a matrix of this form a 0 0 b + c 0 0 d = a + c 0 0 b + d . ( A 1 ) a 0 0 b + c 0 0 d = a + c 0 0 b + d = c + a 0 0 d + b = c 0 0 d + a 0 0 b .

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2 ( A 2 ) a 0 0 b + ˆ c 0 0 d + e 0 0 f ! = a + ( c + e ) 0 0 b + ( d + f ) = ( a + c ) + e 0 0 ( b + d ) + f = ˆ a 0 0 b + c 0 0 d ! + e 0 0 f . ( A 3 ) The 2 × 2 zero matrix is of the appropriate form and has the desired properties. That is, a 0 0 b + 0 0 0 0 = a + 0 0 0 b + 0 = a 0 0 b = 0 + a 0 0 0 + b = 0 0 0 0 + a 0 0 b . ( A 4 ) If u is a matrix of the given form, then - u = - a 0 0 - b is again of the desired form and u + ( - u ) = ( - u ) + u = 0 . ( β ) If u is any matrix of desired form, then ku is k a 0 0 b = ka 0 0 kb and ku has the desired form. ( M 1 ) k ˆ a 0 0 b + c 0 0 d ! = k a + c 0 0 b + d = k ( a + c ) k (0 + 0) k (0 + 0) k ( b + d ) = ka + kc k 0 + k 0 k 0 + k 0 kb + kd = ka k 0 k 0 kb + kc k 0 k 0 kd = k a 0 0 b + k c 0 0 d .
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