Question

Let V,W be two vector spaces of dimension n and m, respectively. Let T:V⟶W be a linear transformation. (i) Show that T is one-one if and only if the columns of matrix [T] of T are linearly independent. (ii) By assuming m=n, show that T is one-one if and only if T is onto.

Answer 1

A linear transformation T:V⟶W is one-one if and only if the columns of the matrix [T] of T are linearly independent. If m=n, then T is one-one if and only if T is onto.

Step-by-step explanation:

A linear transformation T:V'⟶W is one-one if and only if the columns of the matrix [T] of T are linearly independent. To prove this, we can use the fact that T is one-one if and only if the null space of T contains only the zero vector (0). The null space of T is the set of all vectors in V that are mapped to the zero vector in W. If the columns of [T] are linearly independent, then the only solution to the equation [T][x]=0 is x=0, which implies that the null space of T is only the zero vector. Conversely, if T is not one-one, then there exists a non-zero vector x such that T(x)=0, which means that the columns of [T] are linearly dependent.

If m=n, then the dimension of the null space of T is n minus the rank of T (rank-nullity theorem). If T is one-one, then the rank of T is equal to the dimension of V, which implies that the null space of T is zero-dimensional. This means that the only solution to the equation T(x)=0 is x=0, which implies that T is onto. Conversely, if T is onto, then every vector in W has a preimage in V. Since the dimension of V is equal to the dimension of W, this implies that the dimension of the range of T is equal to the dimension of W, which implies that the rank of T is equal to the dimension of W. Therefore, the null space of T is zero-dimensional, which means that the only solution to the equation T(x)=0 is x=0, which implies that T is one-one.