[Linear Algebra] Linear Transformations, Kernels and Ranges

[iJake]

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

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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

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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.

 

[mathwonk]

look, when you see squared powers, think non linear; linear means first order. so c and e are immediately out. the others look ok.