 Solution

Use an inner product linear in its first argument. The adjoint operator ${\alpha }^{\ast }$ is the unique linear operator such thatIn a finite-dimensional inner product space, existence follows by choosing an orthonormal basis: the matrix of the adjoint is the conjugate transpose of the matrix of $\alpha$. Uniqueness follows from nondegeneracy of the inner product.

If $\alpha (W)\subseteq W$, then for $w\in W$ and $v\in W^{\perp }$,Thus ${\alpha }^{\ast }v\in W^{\perp }$. Conversely, if ${\alpha }^{\ast }(W^{\perp })\subseteq W^{\perp }$, the same equality gives $⟨\alpha w,v⟩=0$ for every $v\in W^{\perp }$, so $\alpha w\in (W^{\perp }{)}^{\perp }=W$. This proves the adjoint criterion for an invariant orthogonal complement.

The equality $(W^{\perp }{)}^{\perp }=W$ uses the finite-dimensional setting appropriate here. In a general Hilbert space it holds for closed subspaces, and an adjoint is assured for bounded operators. Without closedness the converse only implies invariance of the closure. For example in ${\ell }^2$, take $W$ to be the finite-support sequences and $\alpha v=⟨v,e_1⟩(2^{-j}{)}_{j\ge 1}$. Then $W^{\perp }=\{0\}$ is automatically adjoint-invariant, but $\alpha e_1\notin W$. Thus the finite-dimensional or closed-subspace hypothesis cannot be omitted from an infinite-dimensional reading.