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Spectral Theorem for normal operators on finite dimensional vector spaces

May 15, 2012

In this post, I will follow up the discussion of my last post by proving the spectral theorem for normal operators and as corollaries the spectral theorem for hermitian, unitary and symmetric operators.

Theorem (Spectral Theorem for Normal Operators): For any normal operator \phi on a (finite dimensional) hermitian space V, there is an orthonormal basis of V consisting of eigenvectors for \phi. Equivalently, for any normal matrix A \in M_n(\mathbf{C}), there is a unitary matrix P such that P^*AP is diagonal.

Proof: Let v_1 be an eigenvector for \phi of length 1. By the last theorem of last post, v_1 is also an eigenvector for \phi^* so W = \mathbf{C}v_1 (the 1-dimensional space spanned by v_1) is \phi^*-invariant. Then by the proposition before that theorem, W^{\bot} is \phi-invariant. Since the restriction of \phi to any \phi-invariant subspace of V is also a normal operator on that subspace, we may use induction on dimension to prove the theorem:

Assume W^{\bot} has an orthonormal basis of eigenvectors (v_2, \ldots, v_n). Then since the form is non-degenerate (\langle u,u \rangle = 0 \Leftrightarrow u=0), we have V = W \oplus W^{\bot} thus the set \{v_1, v_2, \ldots, v_n\} is an orthonormal basis of V of eigenvectors for \phi.

To prove the theorem in matrix form, assume first that the form is the standard hermitian form (which we can do without loss of generality since every positive definite form is conjugate to this one) so that the canonical basis is orthonormal. Let A be the normal matrix of the normal operator \phi in the canonical basis. Then there exists an orthonormal basis B consisting of eigenvectors for \phi. The the change of basis matrix P from the canonical basis to B is unitary and the matrix of \phi in this new basis is P^*AP. Since the elements of B are eigenvectors for \phi, this matrix is diagonal.

QED

To see why a change of basis matrix P between two orthonormal basis B, B' is unitary, let u = (u_1b_1, \ldots, u_nb_n), v = (v_1b_1, \ldots v_nb_n) be two vectors in your space. Then

u \langle v \rangle = \overline{u^i} v^j \langle b_i, b_j \rangle = \overline{u^i} v^j \delta_{ij}

and

\langle Pu, Pv \rangle = \overline{u^i} v^j \langle Pb_i, Pb_j \rangle = \overline{u^i} v^j \langle b'_i, b'_j \rangle = \overline{u^i}v^j \delta_{ij}.

But we have seen in last post that a matrix M is unitary exactly when \langle u,v \rangle = \langle Mu, Mv \rangle for all u,v \in V.

Corollary (Spectral Theorem for Hermitian Operators): For any Hermitian operator \phi on a Hermitian space V there is an orthonormal basis of V consisting of eigenvectors for \phi, moreover, the eigenvalues of \phi are real. Equivalently, for any hermitian matrix A, there is a unitary matrix P such that P^*AP is a real diagonal matrix.

Proof: Since a hermitian operator is a normal operator, we only need to show that its eigenvalues are all real. Let v be an eigenvector for \phi with eigenvalue \lambda and A the matrix of \phi in some basis. Then

v^*Av = v^*\lambda v = \lambda v^*v.

Note that (\lambda v )^* = \overline{\lambda}v^* and that A^* = A. Then also

 v^*Av = v^*A^*v = (Av)^*v = (\lambda v)^*v = \overline{\lambda}v^* v.

So we get \lambda v^* v = \overline{\lambda} v^*v and since v^*v \in \textbf{R} - \{0\}, we must have \lambda = \overline{\lambda}, i.e. \lambda \in \textbf{R}.

QED

Corollary (Spectral Theorem for Unitary Matrix): For any unitary matrix A, there is a unitary matrix P such that P^*AP is diagonal. Equivalently, every conjugacy class in the unitary group U_n contains a diagonal matrix.

Proof: Since unitary matrices are normal, there is nothing to prove.

QED

Corollary (Spectral Theorem for Symmetric Operators): For any symmetric operator \phi on a Euclidean space V, there is an orthonormal basis of V consisting of eigenvectors for \phi, moreover, the eigenvalues of \phi are all real. Equivalently, for any real symmetric matrix A, there is an orthogonal matrix P such that P^tAP is a real diagonal matrix and the eigenvalues of A are real.

Proof: A Symmetric operator on an Euclidean space V can be seen as an hermitian operator on a hermitian space by extending V to a complex vector space and by extending the inner product to a hermitian product.

QED

Reference: M. Artin, Algebra, 2nd edition.

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