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 . Then

${\cal R}(T) = L( T(u_1), T(u_2), \ldots, T(u_n) ).$
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 ${\cal R}(T).$
If 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

can be easily proved. For

let

be one-one. Suppose $u \in {\cal N}(T).$ This means that $T(u) = {\mathbf 0}= T({\mathbf 0}).$ But then

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

Hence,

is one-one.

The other parts can be similarly proved. Part 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.

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

we can extend it to form a basis of

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

Therefore,

$\displaystyle {\mbox{Range }} (T)$	$\displaystyle =$	$\displaystyle L ( T(u_1), T(u_2), \ldots, T(u_n) )$
	$\displaystyle =$	$\displaystyle L ( {\mathbf 0}, \ldots, {\mathbf 0}, T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) )$
	$\displaystyle =$	$\displaystyle L ( T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) )$

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

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

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

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

That is,

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

Thus $\alpha_i = 0$ for $1 \leq i \leq n$ as $\{ u_1, u_2, \ldots, u_n \}$ is a basis of

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

Proof. Let $\dim(V) = n$ and let

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

Hence, ${\mbox{Range}}(T) = V.$ That is,

is onto.

Suppose 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, is one-one.

Now we can assume that 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

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

That is,

has an inverse.

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