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Unformatted text preview: MAT067 University of California, Davis Winter 2007 Solutions to Homework Set 4 1. Define the map T : R 2 R 2 by T ( x,y ) = ( x + y,x ). (a) Show that T is linear. (b) Show that T is surjective. (c) Find dim null T . (d) Find the matrix for T with respect to the canonical basis of R 2 . (e) Show that the map F : R 2 R 2 given by F ( x,y ) = ( x + y,x + 1) is not linear. Solution : (a) Let x,y,z,w, R . Checking scalar multiplication, T ( ( x,y )) = T ( x,y ) = ( x + y,x ) = ( x + y,x ) = T ( x,y ) . Now checking vector addition, T (( x,y ) + ( w,z )) = T ( x + w,y + z ) = ( x + w + y + z,x + w ) = ( x + y,x ) + ( w + z,z ) = T ( x,y ) + T ( w,z ) . Hence T is a linear map. (b) Let ( a,b ) R 2 . We need to check whether there is a ( x,y ) R 2 such that T ( x,y ) = ( x + y,x ) = ( a,b ). However, this relation implies that x = b and so y = a b . So T ( b,a b ) = ( a,b ). Since ( a,b ) is arbitrary, by definition T is surjective. (c) By Thm. 3.4, 2 = dim null T +dim range T . By the last part, dim range T = 2, since range T = R 2 . Hence, dim null T = 0. (d) M ( T ) = bracketleftbigg 1 1 1 0 bracketrightbigg (e) Note that F (0 , 0) = (0 + 0 , 0 + 1) = (0 , 1) negationslash = (0 , 0) . Hence F does not map the zero vector to the zero vector and cannot be linear. 2 2. Consider the complex vector spaces C 2 and C 3 with their canonical bases. Let S : C 3 C 2 be defined by the matrix M ( S ) = A = parenleftbigg i 1 1 2 i 1 1 parenrightbigg . Find a basis for null S . Solution : Using the Dimension Formula, dim(null( S )) + dim(range( S )) = dim( C 3 ) = 3. Moreover, range( S ) = span parenleftbiggparenleftbigg i 2 i parenrightbigg , parenleftbigg 1 1 parenrightbiggparenrightbigg . Since the vectors parenleftbigg i 2 i parenrightbigg , parenleftbigg 1 1 parenrightbigg are not scalar multiples of one another, they are linearly indepen- dent. It follows that dim(range( S )) = 2, and so dim(null( S )) = 1. Thus, given any non-zero vector v null( S ), the list ( v ) will be a basis for null( S ). By inspection, parenleftbigg i 1 1 2 i 1 1 parenrightbigg 1 1 = 0, so 1 1 is a basis for null( S ). 3. Give an example of a function f : R 2 R having the property that a R , v R ,f ( av ) = af ( v ) but such that f is not a linear map. Solution : Proposition. Define the function f : R 2 R by f ( x,y ) = ( x 3 + y 3 ) 1 / 3 for each ordered pair ( x,y ) R 2 . Then f is a homogeneous function, i.e., has the property that a R , v R ,f ( av ) = af ( v ) , but f is not a linear map....
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