Definition. A vector space V is finite dimensional if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for V is called the dimension of V and is denoted by \text{dim}(V).

Definition. Let V and W be vector spaces and let T:V\to W be linear. The null space (or kernel) N(T) of T is the set of all vectors x\in V such that T(x)=0; that is, N(T)=\{x\in V:T(x)=0\}. The range (or image) R(T) of T is the subset of W consisting of all images, under T, of vectors in V; that is, R(T)=\{T(x):x\in V\}.

Definition. Let V and W be vector spaces and let T:V\to W be linear. The nullity of T is the dimension of N(T) denoted by \text{nullity}(T). The rank of T is the dimension of R(T) denoted by \text{rank}(T).

Rank Nullity Theorem. Let V and W be vector spaces and let T:V\to W be linear. If V is finite dimensional, then \text{dim}(V)=\text{rank}(T)+\text{nullity}(T).

Proof. Suppose that \{u_1,\ldots,u_m\} forms a basis for N(T). We can extend this to form a basis for V, \{u_1,\ldots,u_m,w_1,\ldots,w_n\}. Since \text{nullity}(T)=m and \text{dim}(V)=m+n, it suffices to show that \text{rank}(T)=n. I claim that S=\{Tw_1,\ldots,Tw_n\} forms a basis for R(T).

First we show that S spans R(T). Let v\in V. Then there exist unique scalars a_1,\ldots,a_m and b_1,\ldots,b_n such that

v=a_1u_1+\ldots+a_mu_m+b_1w_1+\ldots+b_nw_n.

Thus,

Tv=a_1Tu_1+\ldots+a_mTu_m+b_1Tw_1+\ldots+b_nTw_n.

Since each u_i is in N(T), we have that Tu_i=0 and so Tv=b_1Tw_1+\ldots+b_nTw_n.  This shows that S spans R(T).

Next we show that S is linearly independent. Suppose that

c_1Tw_1+\ldots+c_nTw_n=0

for scalars c_i. Since T is linear, we have that

T(c_1w_1+\ldots+c_nw_n)=0.

Thus, c_1w_1+\ldots+c_nw_n is in N(T). Since u_i span N(T), there exist scalars d_i such that

c_1w_1+\ldots+c_nw_n=d_1u_1+\ldots+d_mu_m.

Which implies that

c_1w_1+\ldots+c_nw_n-(d_1u_1+\ldots+d_mu_m)=0.

But, \{u_1,\ldots,u_m,w_1,\ldots,w_n\} is a basis for V which means that all the c_i and d_i must be zero. Thus, S is linearly independent and is indeed a basis for R(T). This proves that \text{rank}(T)=n as desired. \Box

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