Answer:
To show that T<0,2> (0,1), we need to evaluate the transformation T at the point (0,1) and check if the result is in the range of T.
T<0,2> (0,1) means that the transformation T takes the point (0,1) in the input space to a point in the output space that has coordinates (0,2).
Let's evaluate T at (0,1):
T<0,2> (0,1) = D3(0,1) + (0,1)
= (0+0, 3+1)
= (0, 4)
The output point (0,4) is not equal to (0,2), which means that T<0,2> (0,1) is false. Therefore, T<0,2> does not map the point (0,1) to (0,2).
To determine if T<0,2> (x,y) = D3 (x,y) is a true statement for any point (x,y), we need to check if the transformation T always equals the function D3.
T<0,2> (x,y) = (x, 3+y)
D3(x,y) = (x,y,3)
Since T and D3 have different ranges (T has a range of R^2 and D3 has a range of R^3), the statement T<0,2> (x,y) = D3(x,y) is not true for any point (x,y).
Therefore, we cannot equate T<0,2> and D3, and the statement T<0,2> (x,y) = D3 (x,y) is false for any point (x,y).