Given information:
We are given that V is the vector space R2 and T is the linear transformation on V defined by T(x, y) = (x y, y).

Definition:
The characteristic polynomial of a linear transformation T is defined as the polynomial obtained by substituting the eigenvalues of T into the characteristic equation det(T - λI) = 0, where λ is the eigenvalue and I is the identity matrix.

Characteristic equation:
To find the characteristic polynomial of T, we need to find the eigenvalues of T.
Let (a, b) be an eigenvector of T corresponding to the eigenvalue λ.
Then, we have T(a, b) = λ(a, b).

Evaluating T(a, b):
Using the definition of T, we have T(a, b) = (a b, b).
Setting this equal to λ(a, b), we get (a b, b) = λ(a, b).
This gives us two equations:
a = λa (equation 1)
b = λb (equation 2)

Case 1: λ = 1
If λ = 1, equation 1 becomes a = a, which is true for any value of a.
Equation 2 becomes b = b, which is also true for any value of b.
Therefore, any vector of the form (a, b) with a and b nonzero can be an eigenvector corresponding to λ = 1.

Case 2: λ ≠ 1
If λ ≠ 1, equation 1 becomes a = λa, which implies λ = 1.
Since we assumed λ ≠ 1, this case is not possible.

Eigenvalues:
From the analysis above, we conclude that the only eigenvalue of T is λ = 1.

Characteristic polynomial:
To find the characteristic polynomial, we substitute λ = 1 into the characteristic equation det(T - λI) = 0.
The matrix T - λI is given by:
T - λI = [(x y, y)] - [(1 0, 0 1)] = [(x-1 y, y)].

Evaluating the determinant:
The determinant of T - λI is det(T - λI) = det([(x-1 y, y)]) = (x-1)y - y = xy - y.

Characteristic polynomial:
Setting det(T - λI) = 0, we have xy - y = 0.
Factoring out y, we get y(x - 1) = 0.
This equation holds true for all values of x and y when y = 0 or x = 1.

Therefore, the characteristic polynomial of T is given by (x - 1)y = 0, which can be written as y(x - 1) = 0.

Answer:
The characteristic polynomial of T is (x - 1)y = 0.