[Linear Algebra] Linear Transformations, Kernels and Ranges

  • #1
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Homework Statement



Prove whether or not the following linear transformations are, in fact, linear. Find their kernel and range.

a) ## T : ℝ → ℝ^2, T(x) = (x,x)##
b) ##T : ℝ^3 → ℝ^2, T(x,y,z) = (y-x,z+y)##
c) ##T : ℝ^3 → ℝ^3, T(x,y,z) = (x^2, x, z-x) ##
d) ## T: C[a,b] → ℝ, T(f) = f(a)##
e) ## T: C[a,b] → C[a,b], T(f) = f^2##

Homework Equations


[/B]
Transformations are linear if ##T(a+b) = T(a) + T(b)## and if ##T(c \cdot a) = c \cdot T(a)##
##Ker(T) = \{T(x) = 0\}##
##Im(T) = \{T(x) \in W | x \in V\}##


The Attempt at a Solution


[/B]
a)
##T((x,y,z) + (a,b,c)) = ((y-x) + (b-a), (z+y) + (c+b)) = (y-x, z+y) + (b-a, c+b) = T(x,y,z) + T(a,b,c)##
##T(c \cdot (x,y,z)) = T((cx, cy, cz)) = (cy-cx, cz+cy) = (c \cdot (y-x), c \cdot (z+y)) = c \cdot (y-x, z+y) = c\cdot T(x,y,z)##

T is linear.

##Ker(T) = 0##
##Im(T) = \{(a,a) | a \in ℝ\} = <(1,1)>##

b)
##T((x,y,z) + (a,b,c)) = ((y-x) + (b-a), (z+y)+(c+b)) = (y-x, z+y) + (b-a, c+b) = T(x,y,z) + T(a,b,c)##
##T(c \cdot (x,y,z)) = T((cx,cy,cz)) = (cy-cx, cz+cy) = (c \cdot (y-x), c \cdot (z+y)) = c \cdot (y-x, z+y) = c \cdot T(x,y,z)##

T is linear.

I will try to save some space.

##Ker(T) = \{(x,y,z) \in \mathbb R^3 | T(x,y,z) = (0,0,0)\}##
##Ker(T) = \{(y,y,-y) | y \in ℝ\} = <(1,1,-1)>##
##Im(T) = y(1,1) + x(1,0) + z(0,1) = <(1,1), (1,0), (0,1)>##

c)
##T((x,y,z) + (a,b,c))## holds, but ##T(c \cdot (x,y,z))## does not hold. I am not sure if the correct notation would be that it works out to ##c^2 \cdot T(x) + c \cdot T(y,z)## but in any case it works out to a non-linear transformation.

##Ker(T) = \{(x,y,z) | (x^2, x, z-x) = (0,0,0)\}##
##(x,y,z) = (0,y,0) | y \in ℝ##
##Ker(T) = \{(0,y,0) | y \in ℝ\} = <(0,1,0)>##

##Im(T)## is not linear.

d)
##T(f+g) = (f(a) + g(a)) = f(a) + g(a) = T(f) + T(g)##
##T(c \cdot f) = (c \cdot f)(a) = c \cdot f(a) = c \cdot T(f)##

T is linear.

##Ker(T) = \{f \in C[a,b] | T(f) = 0, f \in C[a,b] | f(a) = 0\}##
##Im(T) = \{r \in ℝ | r = f(a), f \in C[a,b]\}##

e)
##T(f+g) = (f^2 + g^2) = f^2 + g^2 = T(f) + T(g)##
##T(c \cdot f) = (c \cdot f)^2 = c^2 \cdot f^2 = c^2 \cdot T(f)##

T is not linear.
##Ker(T) = \{f \in C[a,b] | f^2 = 0\}##
##Im(T)## is not linear.
 

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
2020 Award
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look, when you see squared powers, think non linear; linear means first order. so c and e are immediately out. the others look ok.
 

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