Let \(V\) be a finite-dimensional inner product space and \(T\) a linear operator on \(V\). Show that the range of \(T^*\) is the orthogonal complement of the null space of \(T\).
Solution
Let \[ N=\ker T, \qquad R^*=\operatorname{range}T^*. \] We prove \[ \boxed{R^*=N^\perp.} \]
1. Every vector in the range of \(T^*\) is orthogonal to the null space of \(T\).
Take \(y\in R^*\). Then \(y=T^*x\) for some \(x\in V\). If \(n\in N\), then \(Tn=0\), and the defining identity for the adjoint gives \[ (n\mid y)=(n\mid T^*x)=(Tn\mid x)=0. \] Thus \(y\) is orthogonal to every vector in \(N\), so \[ R^*\subseteq N^\perp. \]
2. The two subspaces have the same dimension.
In an orthonormal basis, the matrix of \(T^*\) is the conjugate transpose of the matrix of \(T\), so \[ \operatorname{rank}T^*=\operatorname{rank}T \] [@HK71]. Hence rank-nullity gives \[ \dim R^* =\operatorname{rank}T =\dim V-\dim\ker T. \] For a subspace of a finite-dimensional inner product space, \[ \dim N^\perp=\dim V-\dim N \] [@HK71]. Therefore \[ \dim R^*=\dim N^\perp. \] Combined with the inclusion from step 1, this yields \[ R^*=N^\perp. \]