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.