Let T : R3 → R3 be a linear transformation such that T (1, 0, 0) = (1, 0, 0), T (1, 1, 0) =(1, 1, 1) and T (1, 1, 1) = (1, 1, 0). Find (a) T (x, y, z), (b) ker(T ), (c) R(T ). Alsoshow that T 3 = T.

(a) To find T(x, y, z), we first express (x, y, z) as a linear combination of the given vectors. We can write (x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1). Now, we know that T is a linear transformation, so T(x, y, z) = xT(1, 0, 0) + yT(0, 1, 0) + zT(0, 0, 1). But we don't know T(0, 1, 0) and T(0, 0, 1). We can express these vectors in terms of the given vectors and use the linearity of T. We have (0, 1, 0) = (1, 1, 0) - (1, 0, 0) and (0, 0, 1) = (1, 1, 1) - (1, 1, 0). So, T(0, 1, 0) = T(1, 1, 0) - T(1, 0, 0) = (1, 1, 1) - (1, 0, 0) = (0, 1, 1) and T(0, 0, 1) = T(1, 1, 1) - T(1, 1, 0) = (1, 1, 0) - (1, 1, 1) = (0, 0, -1). Therefore, T(x, y, z) = x(1, 0, 0) + y(0, 1, 1) - z(0, 0, 1) = (x, y, y - z).

(b) The kernel of T, ker(T), is the set of all vectors in R3 that T maps to the zero vector. From the expression for T(x, y, z) in part (a), we see that T(x, y, z) = (0, 0, 0) if and only if x = 0 and y = z. So, ker(T) = {(0, z, z) : z in R}.

(c) The range of T, R(T), is the set of all vectors in R3 that T can map to. From the expression for T(x, y, z) in part (a), we see that T can map to any vector of the form (x, y, y - z) for x, y, z in R. So, R(T) = {(x, y, z) : x, y, z in R and z = y}.

To show that T^3 = T, we need to show that applying T three times to any vector gives the same result as applying T once. Let's compute T^3(x, y, z). We have T^2(x, y, z) = T(T(x, y, z)) = T(x, y, y - z) = (x, y - z, y - 2z) and T^3(x, y, z) = T(T^2(x, y, z)) = T(x, y - z, y - 2z) = (x, y - 2z, y - 3z). But T(x, y, z) = (x, y, y - z), so T^3(x, y, z) = T(x, y, z) for all (x, y, z) in R3. Therefore, T^3 = T.