Let V and W be vector spaces and assume that T be a linear mapping from V to W.

(a) Show that ker(T) is a subspace of V .
(b) Show that range(T) is a subspace of W.

To show that ker(T) is a subspace of V, we need to prove three properties: closure under addition, closure under scalar multiplication, and the existence of the zero vector.

(a) Closure under addition:
Let u, v be vectors in ker(T). This means T(u) = 0 and T(v) = 0.
Now, we need to show that their sum, u + v, is also in ker(T).
Using linearity of T, we have:
T(u + v) = T(u) + T(v) = 0 + 0 = 0.
Therefore, u + v is also in ker(T), satisfying closure under addition.

(b) Closure under scalar multiplication:
Let u be a vector in ker(T) and c be a scalar.
We need to show that the scalar multiple, c * u, is also in ker(T).
Using linearity of T, we have:
T(c * u) = c * T(u) = c * 0 = 0.
Therefore, c * u is also in ker(T), satisfying closure under scalar multiplication.

(c) Existence of the zero vector:
Since T is a linear mapping, we know that T(0) = 0, where 0 is the zero vector in V.
Therefore, the zero vector is in ker(T).

Hence, we have shown that ker(T) is a subspace of V.

To show that range(T) is a subspace of W, we also need to prove the same three properties: closure under addition, closure under scalar multiplication, and the existence of the zero vector.

(a) Closure under addition:
Let w1, w2 be vectors in range(T). This means there exist vectors v1, v2 such that T(v1) = w1 and T(v2) = w2.
We need to show that their sum, w1 + w2, is also in range(T).
Using linearity of T, we have:
T(v1 + v2) = T(v1) + T(v2) = w1 + w2.
Therefore, w1 + w2 is also in range(T), satisfying closure under addition.

(b) Closure under scalar multiplication:
Let w be a vector in range(T) and c be a scalar.
We need to show that the scalar multiple, c * w, is also in range(T).
Using linearity of T, we have:
T(c * v) = c * T(v) = c * w.
Therefore, c * w is also in range(T), satisfying closure under scalar multiplication.

(c) Existence of the zero vector:
Since T is a linear mapping, we know that T(0) = 0, where 0 is the zero vector in V.
Therefore, the zero vector is in range(T).

Hence, we have shown that range(T) is a subspace of W.