State the ranknullity theorem and use it to prove that any s

State the rank-nullity theorem and use it to prove that any surjective linear operator on R² is injective.

Solution

Rank-Nullity Theorem:

Let V and W be finite dimensional vector spaces over the field F. Let T: V W be a linear mapping. Then Ker T is a subspace of V. The dimension of Ker T is called the nullity of T. Im T is a subspace of W. The dimension of Im T is called the rank of T.

Both Ker T and Im T are finite dimensional as V and W are finite dimensional vector spaces. Then,

Nullity (T) + Rank (T) = dim V

Let T: R² R² be a linear mapping. Given that T is surjective. Therefore, Im T = R²

This implies, Rank (T) = dim R² = 2

Using Rank-Nullity Theorem, we get, Nullity (T) + 2 = dim R² = 2

Which implies, Nullity (T) = 2 – 2 = 0

Therefore, Ker T = {0} where 0 is the zero vector

This implies, T is injective. (Proved)