Let TRR be the linear transformation given by Tz, p.2)=(x+y.2+2) (a) Give a base and dimension of Ker(T); (b) Give a base and dimension of Im(T)

### Part (a): Kernel of T
Given the linear transformation T(x, y) = (x+y, y+2) , we want to find the kernel of T , denoted as \ker(T) .

To find the kernel, we need to solve for x and y such that T(x, y) = 0 :
T(x, y) = (x+y, y+2) = (0, 0)

This gives us the system of equations:
x + y = 0
y + 2 = 0

Solving these equations, we find:
y = -2
x = 2

Thus, the kernel \ker(T) is the set of all scalar multiples of the vector (2, -2) , and a basis for \ker(T) is \{ (2, -2) \} . The dimension of \ker(T) is 1.

### Part (b): Image of T
The image of T , denoted as \text{Im}(T) , is the set of all possible outputs of T(x, y) for (x, y) \in \mathbb{R}^2 .

Expressing any vector in the image as a linear combination of outputs:
(x+y, y+2) = x(1, 0) + y(1, 1) + (0, 2)

The vectors (1, 0) and (1, 1) span the image. To show they are linearly independent, we solve the system:
a(1, 0) + b(1, 1) = (0, 0)
a + b = 0
b = 0

This implies a = 0, b = 0 , confirming linear independence. Therefore, a basis for \text{Im}(T) is \{ (1, 0), (1, 1) \} , and the dimension of \text{Im}(T) is 2.

**Answer:**
- **Kernel of T **: Basis = \{ (2, -2) \} , Dimension = 1
- **Image of T **: Basis = \{ (1, 0), (1, 1) \} , Dimension = 2