Let \(T\) be a linear operator on a finite-dimensional vector space \(V\). Let \(R\) be the range of \(T\) and let \(N\) be the null space of \(T\). Prove that \(R\) and \(N\) are independent if and only if \(V = R \oplus N\).
Solution
Because \(V\) is finite dimensional, rank-nullity gives \[ \dim R+\dim N=\dim V \] [@HK71]. For two subspaces, \[ \dim(R+N)=\dim R+\dim N-\dim(R\cap N). \] Therefore \[ \dim(R+N)=\dim V-\dim(R\cap N). \]
If \(R\) and \(N\) are independent, then \(R\cap N=\{0\}\), so \(\dim(R+N)=\dim V\). Since \(R+N\subseteq V\), equality of dimensions gives \[ R+N=V. \] Together with the trivial intersection, this says \[ V=R\oplus N. \]
Conversely, if \(V=R\oplus N\), the definition of direct sum gives \(R\cap N=\{0\}\). Thus \(R\) and \(N\) are independent. Hence the two conditions are equivalent.