mat67-Homework4_Solutions

# mat67-Homework4_Solutions - MAT067 University of California...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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,λ ∈ R . Checking scalar multiplication, T ( λ ( x,y )) = T ( λx,λy ) = ( λx + λy,λx ) = λ ( x + y,x ) = λT ( 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

mat67-Homework4_Solutions - MAT067 University of California...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online