Proof of Rank-Nullity Theorem

THEOREM 15.5.1   Let $T: V {\longrightarrow}W$ be a linear transformation and $\{ u_1, u_2, \ldots, u_n \}$ be a basis of $V$ . Then

1. ${\cal R}(T) = L( T(u_1), T(u_2), \ldots, T(u_n) ).$
2. $T$ is one-one $\Longleftrightarrow \; \; {\cal N}(T) = \{{\mathbf 0}\}$ is the zero subspace of $V \Longleftrightarrow \; \;$ $\{ T(u_i): 1 \leq i \leq n \}$ is a basis of ${\cal R}(T).$
3. If $V$ is finite dimensional vector space then $\dim ({\cal R}(T)) \leq \dim (V).$ The equality holds if and only if ${\cal N}(T) = \{{\mathbf 0}\}.$

Proof. Part $1)$ can be easily proved. For $2),$ let $T$ be one-one. Suppose $u \in {\cal N}(T).$ This means that $T(u) = {\mathbf 0}= T({\mathbf 0}).$ But then $T$ is one-one implies that $u = {\mathbf 0}.$ If ${\cal N}(T) = \{{\mathbf 0}\}$ then $T(u) = T(v) \; \Longleftrightarrow T(u - v) = {\mathbf 0}$ implies that $u = v.$ Hence, $T$ is one-one.

The other parts can be similarly proved. Part $3)$ follows from the previous two parts. height6pt width 6pt depth 0pt

The proof of the next theorem is immediate from the fact that $T({\mathbf 0}) = {\mathbf 0}$ and the definition of linear independence/dependence.

THEOREM 15.5.2   Let $T: V {\longrightarrow}W$ be a linear transformation. If $\{T(u_1), T(u_2), \ldots, T(u_n) \}$ is linearly independent in ${\cal R}(T)$ then $\{u_1, u_2, \ldots, u_n \} \subset V$ is linearly independent.

THEOREM 15.5.3 (Rank Nullity Theorem)   Let $T: V {\longrightarrow}W$ be a linear transformation and $V$ be a finite dimensional vector space. Then

$$\dim ({\mbox{ Range}}(T)) + \dim ({\cal N}(T)) = \dim (V),$$

or $\rho(T) + \nu (T) = n.$

Proof. Let $\dim(V) = n$ and $\dim ({\cal N}(T)) = r.$ Suppose $\{u_1, u_2, \ldots, u_r \}$ is a basis of ${\cal N}(T).$ Since $\{u_1, u_2, \ldots, u_r \}$ is a linearly independent set in $V,$ we can extend it to form a basis of $V.$ Now there exists vectors $\{u_{r+1}, u_{r+2}, \ldots, u_n \}$ such that the set $\{u_1, \ldots, u_r, u_{r+1}, \ldots, u_n \}$ is a basis of $V.$ Therefore,

$${\mbox{Range }} (T) = L ( T(u_1), T(u_2), \ldots, T(u_n) )$$

$$= L ( {\mathbf 0}, \ldots, {\mathbf 0}, T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) )$$

$$= L ( T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) )$$

which is equivalent to showing that ${\mbox{ Range }}(T)$ is the span of $\{ T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) \}.$

We now prove that the set $\{ T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) \}$ is a linearly independent set. Suppose the set is linearly dependent. Then, there exists scalars, $\alpha_{r+1}, \alpha_{r+2}, \ldots, \alpha_n,$ not all zero such that

$$\alpha_{r+1} T(u_{r+1}) + \alpha_{r+2} T(u_{r+2}) + \cdots + \alpha_n T(u_n) = {\mathbf 0}.$$

Or $T ( \alpha_{r+1} u_{r+1} + \alpha_{r+2} u_{r+2} + \cdots + \alpha_n u_n )= {\mathbf 0}$ which in turn implies $\alpha_{r+1} u_{r+1} + \alpha_{r+2} u_{r+2} + \cdots + \alpha_n u_n \in {\cal N}(T) = L ( u_1, \ldots, u_r).$ So, there exists scalars $\alpha_i, \; 1 \leq i \leq r$ such that

$$\alpha_{r+1} u_{r+1} + \alpha_{r+2} u_{r+2} + \cdots + \alpha_n u_n = \alpha_{1} u_{1} + \alpha_{2} u_{2} + \cdots + \alpha_r u_r.$$

That is,

$$\alpha_1 u_1 + + \cdots + \alpha_r u_r - \alpha_{r+1} u_{r+1} - \cdots - \alpha_n u_n = {\mathbf 0}.$$

Thus $\alpha_i = 0$ for $1 \leq i \leq n$ as $\{ u_1, u_2, \ldots, u_n \}$ is a basis of $V.$ In other words, we have shown that the set $\{ T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) \}$ is a basis of ${\mbox{Range }} (T).$ Now, the required result follows. height6pt width 6pt depth 0pt

we now state another important implication of the Rank-nullity theorem.

COROLLARY 15.5.4   Let $T: V {\longrightarrow}V$ be a linear transformation on a finite dimensional vector space $V.$ Then

$$T {\mbox{ is one-one }} \Longleftrightarrow T {\mbox{ is onto}} \Longleftrightarrow T {\mbox{ has an inverse}}.$$

Proof. Let $\dim(V) = n$ and let $T$ be one-one. Then $\dim ({\cal N}(T)) = 0.$ Hence, by the rank-nullity Theorem 14.5.3 $\dim ({\mbox{ Range }} (T)) = n = \dim (V).$ Also, ${\mbox{ Range }}(T)$ is a subspace of $V.$ Hence, ${\mbox{Range}}(T) = V.$ That is, $T$ is onto.

Suppose $T$ is onto. Then ${\mbox{Range}}(T) = V.$ Hence, $\dim ({\mbox{ Range }} (T)) = n.$ But then by the rank-nullity Theorem 14.5.3, $\dim ({\cal N}(T)) = 0.$ That is, $T$ is one-one.

Now we can assume that $T$ is one-one and onto. Hence, for every vector ${\mathbf u}$ in the range, there is a unique vectors ${\mathbf v}$ in the domain such that $T({\mathbf v}) = {\mathbf u}.$ Therefore, for every ${\mathbf u}$ in the range, we define

$$T^{-1}({\mathbf u}) = {\mathbf v}.$$

That is, $T$ has an inverse.

Let us now assume that $T$ has an inverse. Then it is clear that $T$ is one-one and onto. height6pt width 6pt depth 0pt