Let \(T\) be a linear operator on the finite-dimensional space \(V\), and let \(R\) be the range of \(T\).
Prove that \(R\) has a complementary \(T\)-invariant subspace if and only if \(R\) is independent of the null space \(N\) of \(T\).
If \(R\) and \(N\) are independent, prove that \(N\) is the unique \(T\)-invariant subspace complementary to \(R\).
Solution
Let \[ R=\operatorname{range}T, \qquad N=\ker T. \] Recall that \(R\) is \(T\)-invariant because \(T(R)=T(T(V))\subseteq T(V)=R\), and \(N\) is \(T\)-invariant because \(T(N)=\{0\}\subseteq N\).
1. If \(R\) has a complementary \(T\)-invariant subspace, then \(R\cap N=\{0\}\).
Suppose \[ V=R\oplus W \] with \(W\) invariant under \(T\). Since \(T(W)\subseteq W\) and every value of \(T\) lies in \(R\), one has \[ T(W)\subseteq R\cap W=\{0\}. \] Hence \(W\subseteq N\).
Rank-nullity gives \[ \dim N=\dim V-\dim R=\dim W. \] Thus the inclusion \(W\subseteq N\) is equality. Consequently \[ V=R\oplus N, \] and in particular \(R\cap N=\{0\}\).
2. If \(R\) and \(N\) are independent, then \(N\) is a complementary invariant subspace.
Independence means \(R\cap N=\{0\}\). Rank-nullity gives \[ \dim R+\dim N=\dim V. \] Therefore the direct sum \(R\oplus N\) has dimension \(\dim V\), so \[ V=R\oplus N. \] As observed above, \(N\) is \(T\)-invariant. Hence \(R\) has a complementary \(T\)-invariant subspace. This proves the equivalence in part (a) [@HK71].
3. The invariant complement is unique.
Assume \(R\cap N=\{0\}\), and let \(W\) be any \(T\)-invariant complement of \(R\). The argument in step 1 applies verbatim and yields \(W\subseteq N\). Again \[ \dim W=\dim V-\dim R=\dim N, \] so \(W=N\). Thus \(N\) is the unique \(T\)-invariant subspace complementary to \(R\), as required in part (b).