Created by potrace 1.14, written by Peter Selinger 2001-2017 Eύρηκω

Rank-nullity theorem

 
  • Nov 08, 2019

Rank-nullity theorem Let V, W be vector spaces, when V is finite dimensional. Let T:V\( \to \)W be a linear transformation. Then Rank(T)+Nullity(T)=dimV

$$\DeclareMathOperator{\ima}{Im}$$

Let the basis of \( \ker(T) \) be \( \{u_1,u_2,\ldots u_{n-1},u_n\} \) and the basis of vector space \( V \) be \( \{v_1,v_2,\ldots v_{r-1},v_r\} \). Based on the definition of \( \ker(T) \), \( \{u_1,u_2,\ldots u_{n-1},u_n\} \) is a set of linearly independent vectors in vector space \( V \).

Based on Steinitz exchange lemma, after rearrangement \( \{u_1,u_2,\ldots u_{n-1},u_n, v_{n+1},\ldots v_{r-1},v_r\} \) is a spanning set for \( V \) with size \( =\dim V \). Therefore, \( \{u_1,u_2, \ldots u_{n-1},u_n,v_{n+1},\ldots v_{r-1},v_r\} \) is a basis for \( V \).

Therefore, \( \{T(u_1),\ldots T(u_n),T(v_{n+1}),\ldots T(v_r)\} \) is a spanning set for \( \ima(T) \). Since \( \{u_1,\ldots u_n\} \) is a basis of \( \ker(T) \), \( T(u_1)=T(u_2)=\cdots =T(u_n)=\vec{0} \). So clearly, \( \{T(v_{n+1}),\ldots T(v_r)\} \) is a spanning set for \( \ima(T) \).

\begin{equation} \sum_{i=n+1}^r \lambda_i \cdot T(v_i)=T(\sum_{i=n+1}^r \lambda_iv_i)=\vec{0} \label{li} \end{equation}

Thus,

\begin{equation} \sum_{i=n+1}^r \lambda_iv_i \in \ker(T)\Leftrightarrow \sum_{i=n+1}^r \lambda_iv_i=\sum_{t=1}^n \lambda_tu_t \Leftrightarrow \sum_{i=n+1}^r \lambda_iv_i-\sum_{t=1}^n\lambda_tu_t=\vec{0} \label{li'} \end{equation}

Since \( \{u_1,u_2, \ldots u_{n-1},u_n,v_{n+1},\ldots v_{r-1},v_r\} \) is a basis for \( V \), the only solution to (1) and (2) is that \( \forall i, \lambda_i=0. \) Therefore \( \{T(v_{n+1}),\ldots T(v_r)\} \) is a set of linearly independent vectors in \( \ima(T) \).

\( \Rightarrow \{T(v_{n+1}),\ldots T(v_r)\} \) is a basis for \( \ima(T) \).

\( \Rightarrow Rank(T)=n ,~Nullity(T)=r-n,~dimV=r \)

Therefore, Rank(\( T \))+Nullity(\( T \))=dim\( V \).

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