10. Unitary spaces

Sesquilinear and hermitian forms

The definition of the inner product which was introduced so far considers only spaces over the real numbers. Let us generalize it to the case of complex numbers. Let every space considered here be a space over complex numbers. We have to notice the following rule. The distance of a real number a from zero equals a^2. Meanwhile the distance of a complex number a from zero is a\cdot \overline{a}. This is the idea of the following definition and all subsequent definitions.

We shall say that a form \xi\colon V\times V\to\mathbb{C} is sesquilinear, if




for any u,v,w\in V, a,b\in\mathbb{C}.

The matrix of a sesquilinear form \xi\colon V\times V\to\mathbb{C} with respect to basis (v_1,\ldots,v_n) of V is the matrix [a_{i,j}]_{1\leq i,j\leq n}=A\in M_{n\times n}(\mathbb{C}) such that

    \[\xi(v,w)=\sum_{i,j=1}^n a_{ij}x_i\overline{y}_j,\]

where v,w\in V are any vectors and (x_1,\ldots, x_n) and (y_1,\ldots, y_n) are their coordinates in basis (v_1,\ldots,v_n) .

A sesquilinear form \xi\colon V\times V\to\mathbb{C} is hermitian if


for any v,w\in V.

Hermitian inner products and unitary spaces

A hermitian form \xi\colon V\times V\to\mathbb{C} is a hermitian inner product if


for any non zero v\in V (observe that it is always a real number).

A space V over \mathbb{C} along with a fixed hermitian inner product is called an unitary space.

The standard hermitian inner product in C^n is defined by the following formula

    \[\langle (a_1,\ldots, a_n),(b_1,\ldots, b_n)\rangle = a_1\overline{b_1}+\ldots+ a_n\overline{b_n}.\]

Similarly as in the case of euclidean spaces one can define the norm, perpendicularity, orthogonal and orthonormal bases.

Hermitian matrices

A matrix A\in M_{n\times n}(\mathbb{C}) is hermitian if A=\overline{A}^T. One can notice that a matrix of a sesquilinear form is hermitian if and only if the form is hermitian.

Mappings of unitary spaces and unitary matrices

The counterpart of an isometry is an isomorphism of unitary spaces, i.e. a linear isomorphism which preserves the hermitian product (unitary isomorphism).

It is easy to prove that a linear isomorphism of unitary spaces is unitary if and only if its matrix A with respect to some orthonormal basis is such that A^T\cdot \overline{A}=I. Such matrices are called unitary matrices.

Adjoint space to an unitary space

Notice that if (V,\xi) is an unitary space, we get an easy identification between V and V^*. Consider a mapping \xi'\colon V\to V^* defined as (\xi''(v))(w)=\xi(w,v). One can prove that \xi' is a bijection and \xi'(av+bw)=\overline{a}v+\overline{b}w, so \xi' is an isomorphism V\to V^\star, where V^\star is the space V^* with modified operation of multiplication (a\cdot f)(v)=\overline{a}\cdot f(v). From this point on we will be identifying v\in V with \xi'(v)\in V^*.

It is easy to prove that assuming this identification for any endomorphisms \varphi,\phi\colon V\to V, \phi=\varphi^* if and only if \xi(\varphi(v),w)=\xi(v,\phi(w)) for any v,w\in V. Therefore, \varphi=\varphi^{**}. The mapping \varphi is self-adjoint if and only if \varphi=\varphi^*. Moreover, a mapping \varphi is self-adjoint if and only if its matrix in an orthonormal basis is hermitian.

Diagonalization of endomorphisms of an unitary space

Similarly as in the case of a space over reals, we may notice that every self-adjoint endomorphism of an unitary space has an orthonormal basis consisting of eigenvectors, and moreover the eigenvalues are real.

Another example of an mapping which is diagonalizable in an orthonormal way is an unitary automorphism. Indeed, let us prove it by induction. For one-dimensional space it is obvious. Let as assume that it is so for n-1-dimensional spaces. Let \dim V=n and let \varphi be an unitary automorphism. There exists an eigenvector of \varphi, v for eigenvalue a (a\neq 0, since it is an automorphism), thus the space (\lin(v))^{\bot} is n-1-dimensional and invariant under \varphi, because if w\bot v, then

    \[\langle v,\varphi(w)\rangle = \frac{1}{a}\langle \varphi(v),\varphi(v)\rangle=\frac{\langle v,w\rangle }{a}=0.\]

We then take the basis from the induction hypothesis and add v/\|v\|.

There is a common property of self-adjoint endomorphisms and unitary automorphism. For both of them we get \varphi\circ \varphi^*=\varphi^*\circ \varphi. The mappings satisfying this condition are called normal endomorphims. One can prove that an endomorphism has an orthonormal basis consisting of eigenvectors if and only if it is normal.